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Mathematics LibreTexts

7.6: Matrices and Matrix Operations

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Learning Objectives

  • Find the sum and difference of two matrices.
  • Find scalar multiples of a matrix.
  • Find the product of two matrices.

Two club soccer teams, the Wildcats and the Mud Cats, are hoping to obtain new equipment for an upcoming season. Table \(\PageIndex{1}\) shows the needs of both teams.

A goal costs \($300\); a ball costs \($10\); and a jersey costs \($30\). How can we find the total cost for the equipment needed for each team? In this section, we discover a method in which the data in the soccer equipment table can be displayed and used for calculating other information. Then, we will be able to calculate the cost of the equipment.

CNX_Precalc_Figure_09_05_001n.jpg

Finding the Sum and Difference of Two Matrices

To solve a problem like the one described for the soccer teams, we can use a matrix, which is a rectangular array of numbers. A row in a matrix is a set of numbers that are aligned horizontally. A column in a matrix is a set of numbers that are aligned vertically. Each number is an entry, sometimes called an element, of the matrix. Matrices (plural) are enclosed in [ ] or ( ), and are usually named with capital letters. For example, three matrices named \(A\), \(B\), and \(C\) are shown below.

\[ \begin{align*} A&=\begin{bmatrix} 1& 2 \\ 3 & 4 \\ \end{bmatrix} \\[4pt] B &=\begin{bmatrix} 1 & 2 & 7 \\ 0 & -5 & 6 \\ 7 & 8 & 2 \end{bmatrix} \\[4pt] C &=\begin{bmatrix} -1 & 3 \\ 0 & 2 \\ 3 & 1 \end{bmatrix} \end{align*}\]

A matrix is often referred to by its size or dimensions: \(m×n\) indicating \(m\) rows and \(n\) columns. Matrix entries are defined first by row and then by column. For example, to locate the entry in matrix \(A\) identified as \(a_{ij}\),we look for the entry in row \(i\),column \(j\). In matrix \(A\), shown below, the entry in row \(2\), column \(3\) is \(a_{23}\).

\[A=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33} \end{bmatrix} \nonumber\]

  • A square matrix is a matrix with dimensions \(n × n\), meaning that it has the same number of rows as columns. The \(3×3\) matrix above is an example of a square matrix.
  • A row matrix is a matrix consisting of one row with dimensions \(1 × n\). \[\begin{bmatrix} a_{11} & a_{12} & a_{13} \end{bmatrix} \nonumber\]
  • A column matrix is a matrix consisting of one column with dimensions \(m × 1\). \[\begin{bmatrix} a_{11} \\ a_{21} \\a_{31} \end{bmatrix} \nonumber\]

A matrix may be used to represent a system of equations. In these cases, the numbers represent the coefficients of the variables in the system. Matrices often make solving systems of equations easier because they are not encumbered with variables. We will investigate this idea further in the next section, but first we will look at basic matrix operations .

Definition: MATRICES

A matrix is a rectangular array of numbers that is usually named by a capital letter: \(A\), \(B\), \(C\),and so on. Each entry in a matrix is referred to as \(a_{ij}\),such that \(i\) represents the row and \(j\) represents the column. Matrices are often referred to by their dimensions: \(m × n\) indicating \(m\) rows and \(n\) columns.

Example \(\PageIndex{1}\): Finding the Dimensions of the Given Matrix and Locating Entries

Given matrix \(A\):

  • What are the dimensions of matrix \(A\)?
  • What are the entries at \(a_{31}\) and \(a_{22}\)?

\[A=\begin{bmatrix} 2 & 1 & 0\\2 & 4 & 7\\3 & 1 & −2 \end{bmatrix} \nonumber\]

  • The dimensions are \(3 \times 3\) because there are three rows and three columns.
  • Entry \(a_{31}\) is the number at row 3, column 1, which is \(3\). The entry \(a_{22}\) is the number at row 2, column 2, which is \(4\). Remember, the row comes first, then the column.

Adding and Subtracting Matrices

We use matrices to list data or to represent systems. Because the entries are numbers, we can perform operations on matrices. We add or subtract matrices by adding or subtracting corresponding entries. To do this, the entries must correspond. Therefore, addition and subtraction of matrices is only possible when the matrices have the same dimensions . We can add or subtract a \(3 \times 3\) matrix and another \(3 \times 3\) matrix, but we cannot add or subtract a \(2 \times 3\) matrix and a \(3 \times 3\) matrix because some entries in one matrix will not have a corresponding entry in the other matrix.

ADDING AND SUBTRACTING MATRICES

Given matrices \(A\) and \(B\) of like dimensions, addition and subtraction of \(A\) and \(B\) will produce matrix \(C\) or matrix \(D\) of the same dimension.

such that \(a_{ij}+b_{ij}=c_{ij}\)

\[A−B=D\]

such that \(a_{ij}−b_{ij}=d_{ij}\)

Matrix addition is commutative .

\[A+B=B+A\]

It is also associative .

\[(A+B)+C=A+(B+C)\]

Example \(\PageIndex{2A}\): Finding the Sum of Matrices

Find the sum of \(A\) and \(B\), given

\[A=\begin{bmatrix}a & b\\c & d \end{bmatrix} \nonumber\]

\[B=\begin{bmatrix}e & f\\g & h\end{bmatrix} \nonumber\]

Add corresponding entries.

\[\begin{align} A+B &=\begin{bmatrix}a & b\\c & d\end{bmatrix}+\begin{bmatrix}e & f\\g & h\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}a+e & b+f\\c+g & d+h \end{bmatrix} \nonumber \end{align} \nonumber\]

Example \(\PageIndex{2B}\): Adding Matrix \(A\) and Matrix \(B\)

Find the sum of \(A\) and \(B\).

\[A=\begin{bmatrix}4 &1\\3 & 2 \end{bmatrix} \nonumber\]

\[B=\begin{bmatrix}5 & 9\\0 & 7\end{bmatrix} \nonumber\]

Add corresponding entries. Add the entry in row 1, column 1, \(a_{11}\), of matrix \(A\) to the entry in row 1, column 1, \(b_{11}\), of \(B\). Continue the pattern until all entries have been added.

\[\begin{align} A+B &=\begin{bmatrix}4&1\\3 &2\end{bmatrix}+\begin{bmatrix}5&9\\0&7\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}4+5&1+9\\3+0&2+7\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}9&10\\3&9\end{bmatrix} \nonumber \end{align} \nonumber\]

Example \(\PageIndex{2C}\): Finding the Difference of Two Matrices

Find the difference of \(A\) and \(B\).

\(A=\begin{bmatrix}−2&3\\0&1\end{bmatrix}\) and \(B=\begin{bmatrix}8&1\\5&4\end{bmatrix}\)

We subtract the corresponding entries of each matrix.

\[\begin{align} A−B &=\begin{bmatrix}−2&3\\0&1\end{bmatrix}−\begin{bmatrix}8&1\\5&4\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}−2−8&3−1\\0−5&1−4\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}−10&2\\−5&−3\end{bmatrix} \nonumber \end{align} \nonumber\]

Example \(\PageIndex{2D}\): Finding the Sum and Difference of Two 3 x 3 Matrices

Given \(A\) and \(B\):

  • Find the sum.
  • Find the difference.

\[A=\begin{bmatrix}2&−10&−2\\14&12&10\\4&−2&2\end{bmatrix} \nonumber\]

\[B=\begin{bmatrix}6&10&−2\\0&−12&−4\\−5&2&−2\end{bmatrix} \nonumber\]

  • Add the corresponding entries.

\[\begin{align} A+B & =\begin{bmatrix} 2& −10& −2\\14 & 12 & 10\\4 & −2 & 2\end{bmatrix}+\begin{bmatrix}6 & 10 & −2\\0 & −12 & −4\\−5 & 2 & −2\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}2+6 & −10+10 & −2−2\\14+0 & 12−12 & 10−4\\4−5 & −2+2 & 2−2\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix} 8 & 0 & −4\\14 & 0 & 6\\−1 & 0 & 0\end{bmatrix} \nonumber \end{align} \nonumber\]

  • Subtract the corresponding entries.

\[\begin{align} A−B &=\begin{bmatrix}2&−10&−2\\14&12&10\\4&−2&2\end{bmatrix}−\begin{bmatrix}6&10&−2\\0&−12&−4\\−5&2&−2\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}2−6 & −10−10 & −2+2\\14−0 & 12+12 & 10+4\\4+5 & −2−2 & 2+2\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}−4 & −20 & 0\\14 & 24 & 14\\9 & −4 & 4\end{bmatrix} \nonumber \end{align} \nonumber\]

Exercise \(\PageIndex{1}\)

Add matrix \(A\) and matrix \(B\).

\[A=\begin{bmatrix}2&6\\1&0\\1&−3\end{bmatrix} \nonumber\]

\[B=\begin{bmatrix}3&−2\\1&5\\−4&3\end{bmatrix} \nonumber\]

\[\begin{align} A+B&=\begin{bmatrix}2&6\\ 1 &0\\1&−3\end{bmatrix}+\begin{bmatrix} 3&-2 \\1&5 \\-4&3\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}2+3&6+(−2)\\1+1&0+5\\1+(-4)&−3+3\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}5&4\\2&5\\-3&0\end{bmatrix} \nonumber \end{align} \nonumber\]

Finding Scalar Multiples of a Matrix

Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. Recall that a scalar is a real number quantity that has magnitude, but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication involves multiplying each entry in a matrix by a scalar. A scalar multiple is any entry of a matrix that results from scalar multiplication.

Consider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment. They estimate that \(15%\) more equipment is needed in both labs. The school’s current inventory is displayed in Table \(\PageIndex{2}\).

Converting the data to a matrix, we have

\[C_{2013}=\begin{bmatrix}15 & 27\\16&34\\16&34\end{bmatrix} \nonumber\]

To calculate how much computer equipment will be needed, we multiply all entries in matrix \(C\) by \(0.15\).

\[(0.15)C_{2013}=\begin{bmatrix}(0.15)15&(0.15)27\\(0.15)16&(0.15)34\\(0.15)16 &(0.15)34\end{bmatrix}=\begin{bmatrix}2.25 &4.05\\2.4&5.1\\2.4&5.1\end{bmatrix} \nonumber\]

We must round up to the next integer, so the amount of new equipment needed is

\[\begin{bmatrix}3&5\\3&6\\3&6\end{bmatrix} \nonumber\]

Adding the two matrices as shown below, we see the new inventory amounts.

\[\begin{bmatrix}15&27\\16&34\\16&34\end{bmatrix}+\begin{bmatrix}3&5\\3&6\\3&6\end{bmatrix}=\begin{bmatrix}18&32\\19&40\\19&40\end{bmatrix} \nonumber\]

\[C_{2014}=\begin{bmatrix}18&32\\19&40\\19&40\end{bmatrix} \nonumber\]

Thus, Lab A will have \(18\) computers, \(19\) computer tables, and \(19\) chairs; Lab B will have \(32\) computers, \(40\) computer tables, and \(40\) chairs.

SCALAR MULTIPLICATION

Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given

\[A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} \nonumber\]

the scalar multiple \(cA\) is

\[cA=c\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} \nonumber\]

\[=\begin{bmatrix}ca_{11}&ca_{12}\\ca_{21}&ca_{22}\end{bmatrix} \nonumber\]

Scalar multiplication is distributive. For the matrices \(A\), \(B\),and \(C\) with scalars \(a\) and \(b\),

\[a(A+B)=aA+aB\]

\[(a+b)A=aA+bA\]

Example \(\PageIndex{3}\): Multiplying the Matrix by a Scalar

Multiply matrix \(A\) by the scalar \(3\).

\[A=\begin{bmatrix}8&1\\5&4\end{bmatrix} \nonumber\]

Multiply each entry in \(A\) by the scalar \(3\).

\[ \begin{align} 3A&=3\begin{bmatrix}8&1\\5&4\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}3⋅8&3⋅1\\3⋅5&3⋅4\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}24&3\\15&12\end{bmatrix} \nonumber \end{align} \nonumber\]

Exercise \(\PageIndex{2}\)

Given matrix \(B\), find \(−2B\) where

\[B=\begin{bmatrix}4&1\\3&2\end{bmatrix} \nonumber\]

\[−2B=\begin{bmatrix}−8&−2\\−6&−4\end{bmatrix} \nonumber\]

Example \(\PageIndex{4}\): Finding the Sum of Scalar Multiples

Find the sum \(3A+2B\).

\[A=\begin{bmatrix}1&−2&0\\0&−1&2\\4&3&−6\end{bmatrix} \nonumber\]

\[B=\begin{bmatrix}−1&2&1\\0&−3&2\\0&1&−4\end{bmatrix} \nonumber\]

First, find \(3A\), then \(2B\).

\[ \begin{align} 3A&=\begin{bmatrix}3⋅1&3(−2)&3⋅0\\3⋅0&3(−1)&3⋅2\\3⋅4&3⋅3&3(−6)\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}3&−6&0\\0&−3&6\\12&9&−18\end{bmatrix}\nonumber \end{align} \nonumber\]

\[ \begin{align} 2B&=\begin{bmatrix}2(−1)&2⋅2&2⋅1\\2⋅0&2(−3)&2⋅2\\2⋅0&2⋅1&2(−4)\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}−2&4&2\\0&−6&4\\0&2&−8\end{bmatrix}\nonumber \end{align} \nonumber\]

Now, add \(3A+2B\).

\[ \begin{align} 3A+2B&=\begin{bmatrix}3&−6&0\\0&−3&6\\12&9&−18\end{bmatrix}+\begin{bmatrix}−2&4&2\\0&−6&4\\0&2&−8\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}3−2&−6+4&0+2\\0+0&−3−6&6+4\\12+0&9+2&−18−8\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}1& −2&2\\0&−9&10\\12&11&−26\end{bmatrix} \nonumber \end{align} \nonumber\]

Finding the Product of Two Matrices

In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If \(A\) is an \(m × r\) matrix and \(B\) is an \(r × n\) matrix, then the product matrix \(AB\) is an \(m × n\) matrix. For example, the product \(AB\) is possible because the number of columns in \(A\) is the same as the number of rows in \(B\). If the inner dimensions do not match, the product is not defined.

alt

We multiply entries of \(A\) with entries of \(B\) according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers.

To obtain the entries in row \(i\) of \(AB\), we multiply the entries in row \(i\) of \(A\) by column \(j\) in \(B\) and add. For example, given matrices \(A\) and \(B\), where the dimensions of \(A\) are \(2 \times 3\) and the dimensions of \(B\) are \(3 \times 3\),the product of \(AB\) will be a \(2 \times 3\) matrix.

\[A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{bmatrix} \nonumber \]

\[B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{bmatrix} \nonumber\]

Multiply and add as follows to obtain the first entry of the product matrix \(AB\).

\[\begin{bmatrix}a_{11}&a_{12}&a_{13}\end{bmatrix} ⋅\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}=a_{11}⋅b_{11}+a_{12}⋅b_{21}+a_{13}⋅b_{31} \nonumber \]

\[\begin{bmatrix}a_{11}&a_{12}&a_{13}\end{bmatrix} ⋅\begin{bmatrix}b_{12}\\b_{22}\\b_{32}\end{bmatrix}=a_{11}⋅b_{12}+a_{12}⋅b_{22}+a_{13}⋅b_{32} \nonumber \]

\[\begin{bmatrix}a_{11}&a_{12}&a_{13}\end{bmatrix} ⋅\begin{bmatrix}b_{13}\\b_{23}\\b_{33}\end{bmatrix}=a_{11}⋅b_{13}+a_{12}⋅b_{23}+a_{13}⋅b_{33} \nonumber \]

We proceed the same way to obtain the second row of \(AB\). In other words, row 2 of \(A\) times column 1 of \(B\); row 2 of \(A\) times column 2 of \(B\); row 2 of \(A\) times column 3 of \(B\). When complete, the product matrix will be

\[AB=\begin{bmatrix}a_{11}⋅b_{11}+a_{12}⋅b_{21}+a_{13}⋅b_{31} &a_{11}⋅b_{12}+a_{12}⋅b_{22}+a_{13}⋅b_{32}&a_{11}⋅b_{13}+a_{12}⋅b_{23}+a_{13}⋅b_{33} \\a_{21}⋅b_{11}+a_{22}⋅b_{21}+a_{23}⋅b_{31}&a_{21}⋅b_{12}+a_{22}⋅b_{22}+a_{23}⋅b_{32}&a_{21}⋅b_{13}+a_{22}⋅b_{23}+a_{23}⋅b_{33}\end{bmatrix} \nonumber\]

PROPERTIES OF MATRIX MULTIPLICATION

For the matrice \(A, B\),and \(C\) the following properties hold.

  • Matrix multiplication is associative : \[(AB)C=A(BC).\]
  • Matrix multiplication is distributive : \[C(A+B)=CA+CB\] \[(A+B)C=AC+BC.\]

Note that matrix multiplication is not commutative.

Example \(\PageIndex{5A}\): Multiplying Two Matrices

Multiply matrix \(A\) and matrix \(B\).

\[A=\begin{bmatrix}1&2\\3&4\end{bmatrix} \nonumber\]

\[B=\begin{bmatrix}5&6\\7&8\end{bmatrix} \nonumber\]

First, we check the dimensions of the matrices. Matrix \(A\) has dimensions \(2 × 2\) and matrix \(B\) has dimensions \(2 × 2\). The inner dimensions are the same so we can perform the multiplication. The product will have the dimensions \(2 × 2\).

We perform the operations outlined previously.

alt

Example \(\PageIndex{5B}\): Multiplying Two Matrices

  • Find \(AB\).
  • Find \(BA\).

\[A=\begin{bmatrix}−1&2&3\\ 4&0&5\end{bmatrix} \nonumber\]

\[B=\begin{bmatrix}5&−1\\-4&0\\2&3\end{bmatrix} \nonumber\]

  • As the dimensions of \(A\) are \(2 \times 3\) and the dimensions of \(B\) are \(3 \times 2\),these matrices can be multiplied together because the number of columns in \(A\) matches the number of rows in \(B\). The resulting product will be a \(2 \times 2\) matrix, the number of rows in \(A\) by the number of columns in \(B\).

\[ \begin{align}AB&=\begin{bmatrix}−1&2&3\\4&0&5\end{bmatrix} \begin{bmatrix}5&−1\\−4&0\\2&3\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}−1(5)+2(−4)+3(2)&−1(−1)+2(0)+3(3)\\4(5)+0(−4)+5(2)&4(−1)+0(0)+5(3)\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}−7&10\\30&11\end{bmatrix} \nonumber \end{align} \nonumber\]

  • The dimensions of \(B\) are \(3 \times 2\) and the dimensions of \(A\) are \(2 \times 3\). The inner dimensions match so the product is defined and will be a \(3 \times 3\) matrix.

\[ \begin{align}BA&=\begin{bmatrix}5&−1\\−4&0\\2&3\end{bmatrix} \begin{bmatrix} −1&2&3\\4&0&5\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}5(−1)+−1(4)&5(2)+−1(0)&5(3)+−1(5)\\−4(−1)+0(4)&−4(2)+0(0)&−4(3)+0(5)\\2(−1)+3(4)& 2(2)+3(0)&2(3)+3(5)\end{bmatrix} \nonumber \\[4pt] &=\begin{bmatrix}−9&10&10\\4&−8&−12\\10&4&21\end{bmatrix} \nonumber \end{align} \nonumber\]

Notice that the products \(AB\) and \(BA\) are not equal.

\[AB=\begin{bmatrix}−7&10\\30&11\end{bmatrix}≠ \begin{bmatrix}−9&10&10\\4&−8&−12\\10&4&21\end{bmatrix}=BA \nonumber\]

This illustrates the fact that matrix multiplication is not commutative.

Q&A: Is it possible for AB to be defined but not BA?

Yes, consider a matrix \(A\) with dimension \(3 × 4\) and matrix \(B\) with dimension \(4 × 2\). For the product \(AB\) the inner dimensions are \(4\) and the product is defined, but for the product \(BA\) the inner dimensions are \(2\) and \(3\) so the product is undefined.

Example \(\PageIndex{6}\): Using Matrices in Real-World Problems

Let’s return to the problem presented at the opening of this section. We have Table \(\PageIndex{3}\), representing the equipment needs of two soccer teams.

We are also given the prices of the equipment, as shown in Table \(\PageIndex{4}\).

We will convert the data to matrices. Thus, the equipment need matrix is written as

\[E=\begin{bmatrix}6&10\\30&24\\14&20\end{bmatrix} \nonumber\]

The cost matrix is written as

\[C=\begin{bmatrix}300&10&30\end{bmatrix} \nonumber\]

We perform matrix multiplication to obtain costs for the equipment.

\[ \begin{align} CE&=\begin{bmatrix}300&10&30\end{bmatrix}⋅\begin{bmatrix}6&10\\30&24\\14&20\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}300(6)+10(30)+30(14)&300(10)+10(24)+30(20)\end{bmatrix} \nonumber \\[4pt] &= \begin{bmatrix}2,520&3,840\end{bmatrix} \nonumber \end{align} \nonumber\]

The total cost for equipment for the Wildcats is \($2,520\), and the total cost for equipment for the Mud Cats is \($3,840\).

How to: Given a matrix operation, evaluate using a calculator

  • Save each matrix as a matrix variable \([A], [B], [C],...\)
  • Enter the operation into the calculator, calling up each matrix variable as needed.
  • If the operation is defined, the calculator will present the solution matrix; if the operation is undefined, it will display an error message.

Example \(\PageIndex{7}\): Using a Calculator to Perform Matrix Operations

Find \(AB−C\) given

\(A=\begin{bmatrix}−15&25&32\\41&−7&−28\\10&34&−2\end{bmatrix}\), \(B=\begin{bmatrix}45&21&−37\\−24&52&19\\6&−48&−31\end{bmatrix}\), and \(C=\begin{bmatrix}−100&−89&−98\\25&−56&74\\−67&42&−75\end{bmatrix}\)

On the matrix page of the calculator, we enter matrix \(A\) above as the matrix variable \([ A ]\), matrix \(B\) above as the matrix variable \([ B ]\), and matrix \(C\) above as the matrix variable \([ C ]\).

On the home screen of the calculator, we type in the problem and call up each matrix variable as needed.

\[[A]×[B]−[C] \nonumber\]

The calculator gives us the following matrix.

\[\begin{bmatrix}−983&−462&136\\1,820&1,897&−856\\−311&2,032&413\end{bmatrix} \nonumber\]

Access these online resources for additional instruction and practice with matrices and matrix operations.

  • Dimensions of a Matrix
  • Matrix Addition and Subtraction
  • Matrix Operations
  • Matrix Multiplication

Key Concepts

  • A matrix is a rectangular array of numbers. Entries are arranged in rows and columns.
  • The dimensions of a matrix refer to the number of rows and the number of columns. A \(3×2\) matrix has three rows and two columns. See Example \(\PageIndex{1}\).
  • We add and subtract matrices of equal dimensions by adding and subtracting corresponding entries of each matrix. See Example \(\PageIndex{2}\), Example \(\PageIndex{3}\), Example \(\PageIndex{4}\), and Example \(\PageIndex{5}\).
  • Scalar multiplication involves multiplying each entry in a matrix by a constant. See Example \(\PageIndex{6}\).
  • Scalar multiplication is often required before addition or subtraction can occur. See Example \(\PageIndex{7}\).
  • Multiplying matrices is possible when inner dimensions are the same—the number of columns in the first matrix must match the number of rows in the second.
  • The product of two matrices, \(A\) and \(B\),is obtained by multiplying each entry in row 1 of \(A\) by each entry in column 1 of \(B\); then multiply each entry of row 1 of \(A\) by each entry in columns 2 of \(B\),and so on. See Example \(\PageIndex{8}\) and Example \(\PageIndex{9}\).
  • Many real-world problems can often be solved using matrices. See Example \(\PageIndex{10}\).
  • We can use a calculator to perform matrix operations after saving each matrix as a matrix variable. See Example \(\PageIndex{11}\).
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How to Solve Matrices

Last Updated: December 9, 2023 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 9 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 183,602 times.

A matrix is a very useful way of representing numbers in a block format, [1] X Research source which you can then use to solve a system of linear equations. If you only have two variables, you will probably use a different method. See Solve a System of Two Linear Equations and Solve Systems of Equations for examples of these other methods. But when you have three or more variables, a matrix is ideal. By using repeated combinations of multiplication and addition, you can systematically reach a solution.

Setting up the Matrix for Solving

Step 1 Verify that you have sufficient data.

  • If you have fewer equations than the number of variables, you will be able to learn some limiting information about the variables (such as x = 3y and y = 2z), but you cannot get a precise solution. For this article, we will be working toward getting a unique solution only.

Step 2 Write your equations in standard form.

  • If you have more variables, you will just continue the line as long as necessary. For example, if you are trying to solve a system with six variables, your standard form would look like Au+Bv+Cw+Dx+Ey+Fz =G. For this article, we will focus on systems with only three variables. Solving a larger system is exactly the same, but just takes more time and more steps.
  • Note that in standard form, the operations between the terms is always addition. If your equation has subtraction instead of addition, you will need to work with this later my making your coefficient negative. If it helps you remember, you can rewrite the equation and make the operation addition and the coefficient negative. For example, you can rewrite the equation 3x-2y+4z=1 as 3x+(-2y)+4z=1.

Step 3 Transfer the numbers from the system of equations into a matrix.

  • For example, suppose you have a system that consists of the three equations 3x+y-z=9, 2x-2y+z=-3, and x+y+z=7. The top row of your matrix will contain the numbers 3,1,-1,9, since these are the coefficients and solution of the first equation. Note that any variable that has no coefficient showing is assumed to have a coefficient of 1. The second row of the matrix will be 2,-2,1,-3, and the third row will be 1,1,1,7.
  • Be sure to align the x-coefficients in the first column, the y-coefficients in the second, the z-coefficients in the third, and the solution terms in the fourth. When you finish working with the matrix, these columns will be important in writing your solution.

Step 4 Draw a large square bracket around your full matrix.

  • You can indicate any specific position in a matrix by using a combination of R and C. For example, to pinpoint the term in the second row, third column, you could call it R2C3.

Learning the Operations for Solving a System with a Matrix

Step 1 Recognize the form of the solution matrix.

  • You will be working with some basic operations to create the “solution matrix.” The solution matrix will look like this: [6] X Research source
  • Notice that the matrix consists of 1’s in a diagonal line with 0’s in all other spaces, except the fourth column. The numbers in the fourth column will be your solution for the variables x, y and z.

Step 2 Use scalar multiplication.

  • It is common to use fractions in scalar multiplication, because you often want to create that diagonal row of 1s. Get used to working with fractions. It will also be easier, for most steps in solving the matrix, to be able to write your fractions in improper form, and then convert them back to mixed numbers for the final solution. Therefore, the number 1 2/3 is easier to work with if you write it as 5/3.
  • For example, the first row (R1) of our sample problem begins with the terms [3,1,-1,9]. The solution matrix should contain a 1 in the first position of the first row. In order to “change” our 3 into a 1, we can multiply the entire row by 1/3. Doing this will create the new R1 of [1,1/3,-1/3,3].
  • Be careful to keep any negative signs where they belong.

Step 3 Use row-addition or row-subtraction.

  • You can use some shorthand and indicate this operation as R2-R1=[0,-1,2,6].
  • Recognize that adding and subtracting are merely opposite forms of the same operation. You can either think of adding two numbers or subtracting the opposite. For example, if you begin with the simple equation 3-3=0, you could consider this instead as an addition problem of 3+(-3)=0. The result is the same. This seems basic, but it is sometimes easier to think of a problem in one form or the other. Just keep track of your negative signs.

Step 4 Combine row-addition and scalar multiplication in a single step.

  • Suppose you have a Row 1 of [1,1,2,6] and a Row 2 of [2,3,1,1]. You want to create a 0 term in the first column of R2. That is, you want to change the 2 into a 0. To do this, you need to subtract a 2. You can get a 2 by first multiplying Row 1 by the scalar multiplication 2, and then subtract the first row from the second row. In shorthand, you can think of this as R2-2*R1. First multiply R1 by 2 to get [2,2,4,12]. Then subtract this from R2 to get [(2-2),(3-2),(1-4),(1-12)]. Simplify this and your new R2 will be [0,1,-3,-11].

Step 5 Copy down rows that are unchanged as you work.

  • A common mistake occurs when conducting a combined multiplication and addition step in one move. Suppose, for example, you need to subtract double R1 from R2. When you multiply R1 by 2 to do this step, remember that you are not changing R1 in the matrix. You are only doing the multiplication to change R2. Copy R1 first in its original form, then make the change to R2.

Step 6 Work from top to bottom first.

  • 1. Create a 1 in the first row, first column (R1C1).
  • 2. Create a 0 in the second row, first column (R2C1).
  • 3. Create a 1 in the second row, second column (R2C2).
  • 4. Create a 0 in the third row, first column (R3C1).
  • 5. Create a 0 in the third row, second column (R3C2).
  • 6. Create 1 in the third row, third column (R3C3).

Step 7 Work back up from bottom to top.

  • Create a 0 in the second row, third column (R2C3).
  • Create a 0 in first row, third column (R1C3).
  • Create a 0 in the first row, second column (R1C2).

Step 8 Check that you have created the solution matrix.

Putting the Steps Together to Solve the System

Step 1 Begin with a sample system of linear equations.

  • Notice that multiplication and division are merely inverse functions of each other. We can say we are multiplying by 1/3 or dividing by 3, and the result is the same.

Step 3 Create a 0 in the second row, first column (R2C1).

  • Copy down the unaffected row 3 as R3=[1,1,1,7].
  • Be very careful with subtracting negative numbers, to make sure you keep the signs correct.
  • For now, leave the fractions in their improper forms. This will make later steps of the solution easier. You can simplify fractions in the final step of the problem.

Step 4 Create a 1 in the second row, second column (R2C2).

  • Notice that as the left half of the row starts looking like the solution with the 0 and 1, the right half may start looking ugly, with improper fractions. Just carry them along for now.
  • Remember to continue copying the unaffected rows, so R1=[1,1/3,-1/3,3] and R3=[1,1,1,7].

Step 5 Create a 0 in the third row, first column (R3C1).

  • Continue to copy along R1=[1,1/3,-1/3,3] and R2=[0,1,-5/8,27/8]. Remember that you only change one row at a time.

Step 6 Create a 0 in the third row, second column (R3C2).

  • Notice that the fractions, which appeared quite complicated in the previous step, have already begun to resolve themselves.
  • Continue to carry along R1=[1,1/3,-1/3,3] and R2=[0,1,-5/8,27/8].
  • Notice that at this point, you have the diagonal of 1’s for your solution matrix. You just need to transform three more items of the matrix into 0’s to find your solution.

Step 8 Create a 0 in the second row, third column.

  • Copy along R1=[1,1/3,-1/3,3] and R3=[0,0,1,1].

Step 9 Create a 0 in the first row, third column (R1C3).

  • Copy the unchanged R2=[0,1,0,4] and R3=[0,0,1,1].

Step 10 Create a 0 in the first row, second column (R1C2).

Verifying Your Solution

Step 1 Replace the solution values into each variable in each equation.

  • Recall that the original equations for this problem were 3x+y-z=9, 2x-2y+z=-3, and x+y+z=7. When you replace the variables with their solved values, you get 3*2+4-1=9, 2*2-2*4+1=-3, and 2+4+1=7.

Step 2 Simplify each equation.

  • Because each equation simplifies to a true mathematical statement, your solutions are correct. If any of them did not resolve correctly, you would have to go back through your work and look for any errors. Some common mistakes occur in dropping negative signs along the way or confusing the multiplication and addition of fractions.

Step 3 Write out your final solutions.

Community Q&A

Community Answer

  • If your system of equations is very complicated, with many variables, you may be able to use a graphing calculator instead of doing the work by hand. For information on this, see Use a Graphing Calculator to Solve a System of Equations. Thanks Helpful 0 Not Helpful 0

how solve matrix problems

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Find the Inverse of a 3x3 Matrix

  • ↑ https://www.mathsisfun.com/algebra/matrix-introduction.html
  • ↑ https://www.mathsisfun.com/algebra/systems-linear-equations-matrices.html
  • ↑ https://www.khanacademy.org/math/precalculus/precalc-matrices/representing-systems-with-matrices/a/representing-systems-with-matrices
  • ↑ https://www.cuemath.com/algebra/solve-matrices/
  • ↑ https://www.khanacademy.org/math/precalculus/precalc-matrices/multiplying-matrices-by-scalars/a/multiplying-matrices-by-scalars
  • ↑ https://people.richland.edu/james/lecture/m116/matrices/operations.html
  • ↑ https://www.varsitytutors.com/hotmath/hotmath_help/topics/solving-matrix-equations
  • ↑ https://math.libretexts.org/Bookshelves/Algebra/Intermediate_Algebra_(OpenStax)/04%3A_Systems_of_Linear_Equations/4.06%3A_Solve_Systems_of_Equations_Using_Matrices
  • ↑ https://www.cliffsnotes.com/study-guides/algebra/algebra-ii/linear-equations-in-three-variables/linear-equations-solutions-using-matrices-with-three-variables

About This Article

Grace Imson, MA

By properly setting up a matrix, you can use them to solve a system of linear equations. Start by writing out your equations and then transfer the numbers from them into a matrix by copying the coefficients and results into a single row. Stack the rows one on top of each other to form a block-looking format. Add a large square bracket around your full matrix and use the abbreviation “R” for the rows and “C” for the columns. This allows you to refer to a specific position in the matrix with a combination of R and C, such as R4C1. To solve the matrix, you can use different operations. For instance, you could use row-addition or row-subtraction, which allows you to add or subtract any two rows of the matrix. To learn about other ways to create a solution matrix, keep reading! Did this summary help you? Yes No

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4.5 Solve Systems of Equations Using Matrices

Learning objectives.

By the end of this section, you will be able to:

  • Write the augmented matrix for a system of equations
  • Use row operations on a matrix
  • Solve systems of equations using matrices

Be Prepared 4.13

Before you get started, take this readiness quiz.

Solve: 3 ( x + 2 ) + 4 = 4 ( 2 x − 1 ) + 9 . 3 ( x + 2 ) + 4 = 4 ( 2 x − 1 ) + 9 . If you missed this problem, review Example 2.2 .

Be Prepared 4.14

Solve: 0.25 p + 0.25 ( p + 4 ) = 5.20 . 0.25 p + 0.25 ( p + 4 ) = 5.20 . If you missed this problem, review Example 2.13 .

Be Prepared 4.15

Evaluate when x = −2 x = −2 and y = 3 : 2 x 2 − x y + 3 y 2 . y = 3 : 2 x 2 − x y + 3 y 2 . If you missed this problem, review Example 1.21 .

Write the Augmented Matrix for a System of Equations

Solving a system of equations can be a tedious operation where a simple mistake can wreak havoc on finding the solution. An alternative method which uses the basic procedures of elimination but with notation that is simpler is available. The method involves using a matrix . A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix with m rows and n columns has order m × n . m × n . The matrix on the left below has 2 rows and 3 columns and so it has order 2 × 3 . 2 × 3 . We say it is a 2 by 3 matrix.

Each number in the matrix is called an element or entry in the matrix.

We will use a matrix to represent a system of linear equations. We write each equation in standard form and the coefficients of the variables and the constant of each equation becomes a row in the matrix. Each column then would be the coefficients of one of the variables in the system or the constants. A vertical line replaces the equal signs. We call the resulting matrix the augmented matrix for the system of equations.

Notice the first column is made up of all the coefficients of x , the second column is the all the coefficients of y , and the third column is all the constants.

Example 4.37

Write each system of linear equations as an augmented matrix:

ⓐ { 5 x − 3 y = −1 y = 2 x − 2 { 5 x − 3 y = −1 y = 2 x − 2 ⓑ { 6 x − 5 y + 2 z = 3 2 x + y − 4 z = 5 3 x − 3 y + z = −1 { 6 x − 5 y + 2 z = 3 2 x + y − 4 z = 5 3 x − 3 y + z = −1

ⓐ The second equation is not in standard form. We rewrite the second equation in standard form.

We replace the second equation with its standard form. In the augmented matrix, the first equation gives us the first row and the second equation gives us the second row. The vertical line replaces the equal signs.

ⓑ All three equations are in standard form. In the augmented matrix the first equation gives us the first row, the second equation gives us the second row, and the third equation gives us the third row. The vertical line replaces the equal signs.

Try It 4.73

ⓐ { 3 x + 8 y = −3 2 x = −5 y − 3 { 3 x + 8 y = −3 2 x = −5 y − 3 ⓑ { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3 { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3

Try It 4.74

ⓐ { 11 x = −9 y − 5 7 x + 5 y = −1 { 11 x = −9 y − 5 7 x + 5 y = −1 ⓑ { 5 x − 3 y + 2 z = −5 2 x − y − z = 4 3 x − 2 y + 2 z = −7 { 5 x − 3 y + 2 z = −5 2 x − y − z = 4 3 x − 2 y + 2 z = −7

It is important as we solve systems of equations using matrices to be able to go back and forth between the system and the matrix. The next example asks us to take the information in the matrix and write the system of equations.

Example 4.38

Write the system of equations that corresponds to the augmented matrix:

[ 4 −3 3 1 2 −1 −2 −1 3 | −1 2 −4 ] . [ 4 −3 3 1 2 −1 −2 −1 3 | −1 2 −4 ] .

We remember that each row corresponds to an equation and that each entry is a coefficient of a variable or the constant. The vertical line replaces the equal sign. Since this matrix is a 4 × 3 4 × 3 , we know it will translate into a system of three equations with three variables.

Try It 4.75

Write the system of equations that corresponds to the augmented matrix: [ 1 −1 2 3 2 1 −2 1 4 −1 2 0 ] . [ 1 −1 2 3 2 1 −2 1 4 −1 2 0 ] .

Try It 4.76

Write the system of equations that corresponds to the augmented matrix: [ 1 1 1 4 2 3 −1 8 1 1 −1 3 ] . [ 1 1 1 4 2 3 −1 8 1 1 −1 3 ] .

Use Row Operations on a Matrix

Once a system of equations is in its augmented matrix form, we will perform operations on the rows that will lead us to the solution.

To solve by elimination, it doesn’t matter which order we place the equations in the system. Similarly, in the matrix we can interchange the rows.

When we solve by elimination, we often multiply one of the equations by a constant. Since each row represents an equation, and we can multiply each side of an equation by a constant, similarly we can multiply each entry in a row by any real number except 0.

In elimination, we often add a multiple of one row to another row. In the matrix we can replace a row with its sum with a multiple of another row.

These actions are called row operations and will help us use the matrix to solve a system of equations.

Row Operations

In a matrix, the following operations can be performed on any row and the resulting matrix will be equivalent to the original matrix.

  • Interchange any two rows.
  • Multiply a row by any real number except 0.
  • Add a nonzero multiple of one row to another row.

Performing these operations is easy to do but all the arithmetic can result in a mistake. If we use a system to record the row operation in each step, it is much easier to go back and check our work.

We use capital letters with subscripts to represent each row. We then show the operation to the left of the new matrix. To show interchanging a row:

To multiply row 2 by −3 −3 :

To multiply row 2 by −3 −3 and add it to row 1:

Example 4.39

Perform the indicated operations on the augmented matrix:

ⓐ Interchange rows 2 and 3.

ⓑ Multiply row 2 by 5.

ⓒ Multiply row 3 by −2 −2 and add to row 1.

ⓐ We interchange rows 2 and 3.

ⓑ We multiply row 2 by 5.

ⓒ We multiply row 3 by −2 −2 and add to row 1.

Try It 4.77

Perform the indicated operations sequentially on the augmented matrix:

ⓐ Interchange rows 1 and 3.

ⓑ Multiply row 3 by 3.

ⓒ Multiply row 3 by 2 and add to row 2.

[ 5 −2 −2 4 −1 −4 −2 3 0 | −2 4 −1 ] [ 5 −2 −2 4 −1 −4 −2 3 0 | −2 4 −1 ]

Try It 4.78

ⓐ Interchange rows 1 and 2,

ⓑ Multiply row 1 by 2,

ⓒ Multiply row 2 by 3 and add to row 1.

[ 2 −3 −2 4 1 −3 5 0 4 | −4 2 −1 ] [ 2 −3 −2 4 1 −3 5 0 4 | −4 2 −1 ]

Now that we have practiced the row operations, we will look at an augmented matrix and figure out what operation we will use to reach a goal. This is exactly what we did when we did elimination. We decided what number to multiply a row by in order that a variable would be eliminated when we added the rows together.

Given this system, what would you do to eliminate x ?

This next example essentially does the same thing, but to the matrix.

Example 4.40

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [ 1 −1 4 −8 | 2 0 ] . [ 1 −1 4 −8 | 2 0 ] .

To make the 4 a 0, we could multiply row 1 by −4 −4 and then add it to row 2.

Try It 4.79

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [ 1 −1 3 −6 | 2 2 ] . [ 1 −1 3 −6 | 2 2 ] .

Try It 4.80

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [ 1 −1 −2 −3 | 3 2 ] . [ 1 −1 −2 −3 | 3 2 ] .

Solve Systems of Equations Using Matrices

To solve a system of equations using matrices, we transform the augmented matrix into a matrix in row-echelon form using row operations. For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

Row-Echelon Form

For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

Once we get the augmented matrix into row-echelon form, we can write the equivalent system of equations and read the value of at least one variable. We then substitute this value in another equation to continue to solve for the other variables. This process is illustrated in the next example.

Example 4.41

How to solve a system of equations using a matrix.

Solve the system of equations using a matrix: { 3 x + 4 y = 5 x + 2 y = 1 . { 3 x + 4 y = 5 x + 2 y = 1 .

Try It 4.81

Solve the system of equations using a matrix: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.82

Solve the system of equations using a matrix: { 2 x + y = −4 x − y = −2 . { 2 x + y = −4 x − y = −2 .

The steps are summarized here.

Solve a system of equations using matrices.

  • Step 1. Write the augmented matrix for the system of equations.
  • Step 2. Using row operations get the entry in row 1, column 1 to be 1.
  • Step 3. Using row operations, get zeros in column 1 below the 1.
  • Step 4. Using row operations, get the entry in row 2, column 2 to be 1.
  • Step 5. Continue the process until the matrix is in row-echelon form.
  • Step 6. Write the corresponding system of equations.
  • Step 7. Use substitution to find the remaining variables.
  • Step 8. Write the solution as an ordered pair or triple.
  • Step 9. Check that the solution makes the original equations true.

Here is a visual to show the order for getting the 1’s and 0’s in the proper position for row-echelon form.

We use the same procedure when the system of equations has three equations.

Example 4.42

Solve the system of equations using a matrix: { 3 x + 8 y + 2 z = −5 2 x + 5 y − 3 z = 0 x + 2 y − 2 z = −1 . { 3 x + 8 y + 2 z = −5 2 x + 5 y − 3 z = 0 x + 2 y − 2 z = −1 .

Try It 4.83

Solve the system of equations using a matrix: { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3 . { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3 .

Try It 4.84

Solve the system of equations using a matrix: { −3 x + y + z = −4 − x + 2 y − 2 z = 1 2 x − y − z = −1 . { −3 x + y + z = −4 − x + 2 y − 2 z = 1 2 x − y − z = −1 .

So far our work with matrices has only been with systems that are consistent and independent, which means they have exactly one solution. Let’s now look at what happens when we use a matrix for a dependent or inconsistent system.

Example 4.43

Solve the system of equations using a matrix: { x + y + 3 z = 0 x + 3 y + 5 z = 0 2 x + 4 z = 1 . { x + y + 3 z = 0 x + 3 y + 5 z = 0 2 x + 4 z = 1 .

Try It 4.85

Solve the system of equations using a matrix: { x − 2 y + 2 z = 1 −2 x + y − z = 2 x − y + z = 5 . { x − 2 y + 2 z = 1 −2 x + y − z = 2 x − y + z = 5 .

Try It 4.86

Solve the system of equations using a matrix: { 3 x + 4 y − 3 z = −2 2 x + 3 y − z = −12 x + y − 2 z = 6 . { 3 x + 4 y − 3 z = −2 2 x + 3 y − z = −12 x + y − 2 z = 6 .

The last system was inconsistent and so had no solutions. The next example is dependent and has infinitely many solutions.

Example 4.44

Solve the system of equations using a matrix: { x − 2 y + 3 z = 1 x + y − 3 z = 7 3 x − 4 y + 5 z = 7 . { x − 2 y + 3 z = 1 x + y − 3 z = 7 3 x − 4 y + 5 z = 7 .

Try It 4.87

Solve the system of equations using a matrix: { x + y − z = 0 2 x + 4 y − 2 z = 6 3 x + 6 y − 3 z = 9 . { x + y − z = 0 2 x + 4 y − 2 z = 6 3 x + 6 y − 3 z = 9 .

Try It 4.88

Solve the system of equations using a matrix: { x − y − z = 1 − x + 2 y − 3 z = −4 3 x − 2 y − 7 z = 0 . { x − y − z = 1 − x + 2 y − 3 z = −4 3 x − 2 y − 7 z = 0 .

Access this online resource for additional instruction and practice with Gaussian Elimination.

  • Gaussian Elimination

Section 4.5 Exercises

Practice makes perfect.

In the following exercises, write each system of linear equations as an augmented matrix.

ⓐ { 3 x − y = −1 2 y = 2 x + 5 { 3 x − y = −1 2 y = 2 x + 5 ⓑ { 4 x + 3 y = −2 x − 2 y − 3 z = 7 2 x − y + 2 z = −6 { 4 x + 3 y = −2 x − 2 y − 3 z = 7 2 x − y + 2 z = −6

ⓐ { 2 x + 4 y = −5 3 x − 2 y = 2 { 2 x + 4 y = −5 3 x − 2 y = 2 ⓑ { 3 x − 2 y − z = −2 −2 x + y = 5 5 x + 4 y + z = −1 { 3 x − 2 y − z = −2 −2 x + y = 5 5 x + 4 y + z = −1

ⓐ { 3 x − y = −4 2 x = y + 2 { 3 x − y = −4 2 x = y + 2 ⓑ { x − 3 y − 4 z = −2 4 x + 2 y + 2 z = 5 2 x − 5 y + 7 z = −8 { x − 3 y − 4 z = −2 4 x + 2 y + 2 z = 5 2 x − 5 y + 7 z = −8

ⓐ { 2 x − 5 y = −3 4 x = 3 y − 1 { 2 x − 5 y = −3 4 x = 3 y − 1 ⓑ { 4 x + 3 y − 2 z = −3 −2 x + y − 3 z = 4 − x − 4 y + 5 z = −2 { 4 x + 3 y − 2 z = −3 −2 x + y − 3 z = 4 − x − 4 y + 5 z = −2

Write the system of equations that corresponds to the augmented matrix.

[ 2 −1 1 −3 | 4 2 ] [ 2 −1 1 −3 | 4 2 ]

[ 2 −4 3 −3 | −2 −1 ] [ 2 −4 3 −3 | −2 −1 ]

[ 1 0 −3 1 −2 0 0 −1 2 | −1 −2 3 ] [ 1 0 −3 1 −2 0 0 −1 2 | −1 −2 3 ]

[ 2 −2 0 0 2 −1 3 0 −1 | −1 2 −2 ] [ 2 −2 0 0 2 −1 3 0 −1 | −1 2 −2 ]

In the following exercises, perform the indicated operations on the augmented matrices.

[ 6 −4 3 −2 | 3 1 ] [ 6 −4 3 −2 | 3 1 ]

ⓐ Interchange rows 1 and 2

ⓑ Multiply row 2 by 3

ⓒ Multiply row 2 by −2 −2 and add row 1 to it.

[ 4 −6 3 2 | −3 1 ] [ 4 −6 3 2 | −3 1 ]

ⓑ Multiply row 1 by 4

ⓒ Multiply row 2 by 3 and add row 1 to it.

[ 4 −12 −8 4 −2 −3 −6 2 −1 | 16 −1 −1 ] [ 4 −12 −8 4 −2 −3 −6 2 −1 | 16 −1 −1 ]

ⓐ Interchange rows 2 and 3

ⓒ Multiply row 2 by −2 −2 and add to row 3.

[ 6 −5 2 2 1 −4 3 −3 1 | 3 5 −1 ] [ 6 −5 2 2 1 −4 3 −3 1 | 3 5 −1 ]

ⓑ Multiply row 2 by 5

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: [ 1 2 −3 −4 | 5 −1 ] . [ 1 2 −3 −4 | 5 −1 ] .

Perform the needed row operations that will get the first entry in both row 2 and row 3 to be zero in the augmented matrix: [ 1 −2 3 3 −1 −2 2 −3 −4 | −4 5 −1 ] . [ 1 −2 3 3 −1 −2 2 −3 −4 | −4 5 −1 ] .

In the following exercises, solve each system of equations using a matrix.

{ 2 x + y = 2 x − y = −2 { 2 x + y = 2 x − y = −2

{ 3 x + y = 2 x − y = 2 { 3 x + y = 2 x − y = 2

{ − x + 2 y = −2 x + y = −4 { − x + 2 y = −2 x + y = −4

{ −2 x + 3 y = 3 x + 3 y = 12 { −2 x + 3 y = 3 x + 3 y = 12

{ 2 x − 3 y + z = 19 −3 x + y − 2 z = −15 x + y + z = 0 { 2 x − 3 y + z = 19 −3 x + y − 2 z = −15 x + y + z = 0

{ 2 x − y + 3 z = −3 − x + 2 y − z = 10 x + y + z = 5 { 2 x − y + 3 z = −3 − x + 2 y − z = 10 x + y + z = 5

{ 2 x − 6 y + z = 3 3 x + 2 y − 3 z = 2 2 x + 3 y − 2 z = 3 { 2 x − 6 y + z = 3 3 x + 2 y − 3 z = 2 2 x + 3 y − 2 z = 3

{ 4 x − 3 y + z = 7 2 x − 5 y − 4 z = 3 3 x − 2 y − 2 z = −7 { 4 x − 3 y + z = 7 2 x − 5 y − 4 z = 3 3 x − 2 y − 2 z = −7

{ x + 2 z = 0 4 y + 3 z = −2 2 x − 5 y = 3 { x + 2 z = 0 4 y + 3 z = −2 2 x − 5 y = 3

{ 2 x + 5 y = 4 3 y − z = 3 4 x + 3 z = −3 { 2 x + 5 y = 4 3 y − z = 3 4 x + 3 z = −3

{ 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1 { 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1

{ 3 x − z = −3 5 y + 2 z = −6 4 x + 3 y = −8 { 3 x − z = −3 5 y + 2 z = −6 4 x + 3 y = −8

{ 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20 { 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20

{ x + 2 y + 6 z = 5 − x + y − 2 z = 3 x − 4 y − 2 z = 1 { x + 2 y + 6 z = 5 − x + y − 2 z = 3 x − 4 y − 2 z = 1

{ x + 2 y − 3 z = −1 x − 3 y + z = 1 2 x − y − 2 z = 2 { x + 2 y − 3 z = −1 x − 3 y + z = 1 2 x − y − 2 z = 2

{ 4 x − 3 y + 2 z = 0 −2 x + 3 y − 7 z = 1 2 x − 2 y + 3 z = 6 { 4 x − 3 y + 2 z = 0 −2 x + 3 y − 7 z = 1 2 x − 2 y + 3 z = 6

{ x − y + 2 z = −4 2 x + y + 3 z = 2 −3 x + 3 y − 6 z = 12 { x − y + 2 z = −4 2 x + y + 3 z = 2 −3 x + 3 y − 6 z = 12

{ − x − 3 y + 2 z = 14 − x + 2 y − 3 z = −4 3 x + y − 2 z = 6 { − x − 3 y + 2 z = 14 − x + 2 y − 3 z = −4 3 x + y − 2 z = 6

{ x + y − 3 z = −1 y − z = 0 − x + 2 y = 1 { x + y − 3 z = −1 y − z = 0 − x + 2 y = 1

{ x + 2 y + z = 4 x + y − 2 z = 3 −2 x − 3 y + z = −7 { x + 2 y + z = 4 x + y − 2 z = 3 −2 x − 3 y + z = −7

Writing Exercises

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6 ⓐ by graphing and ⓑ by substitution. ⓒ Which method do you prefer? Why?

Solve the system of equations { 3 x + y = 12 x = y − 8 { 3 x + y = 12 x = y − 8 by substitution and explain all your steps in words.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
  • Authors: Lynn Marecek, Andrea Honeycutt Mathis
  • Publisher/website: OpenStax
  • Book title: Intermediate Algebra 2e
  • Publication date: May 6, 2020
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
  • Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/4-5-solve-systems-of-equations-using-matrices

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Solving Systems of Linear Equations Using Matrices

Hi there! This page is only going to make sense when you know a little about Systems of Linear Equations and Matrices , so please go and learn about those if you don't know them already.

The Example

One of the last examples on Systems of Linear Equations was this one:

Example: Solve

  • x + y + z = 6
  • 2y + 5z = −4
  • 2x + 5y − z = 27

We went on to solve it using "elimination", but we can also solve it using Matrices!

Using Matrices makes life easier because we can use a computer program (such as the Matrix Calculator ) to do all the "number crunching".

But first we need to write the question in Matrix form.

In Matrix Form?

OK. A Matrix is an array of numbers:

Well, think about the equations:

They could be turned into a table of numbers like this:

We could even separate the numbers before and after the "=" into:

Now it looks like we have 2 Matrices.

In fact we have a third one, which is [x y z] :

Why does [x y z] go there? Because when we Multiply Matrices we use the "Dot Product" like this:

Which is the first of our original equations above (you might like to check that). Here it is for the second line.

 Try the third line for yourself.

The Matrix Solution

We can shorten this:

  • A is the 3x3 matrix of x, y and z coefficients
  • X is x, y and z , and
  • B is 6, −4 and 27

Then (as shown on the Inverse of a Matrix page) the solution is this:

What does that mean?

It means that we can find the X matrix (the values of x, y and z) by multiplying the inverse of the A matrix by the B matrix .

So let's go ahead and do that.

First, we need to find the inverse of the A matrix (assuming it exists!)

Using the Matrix Calculator we get this:

(I left the 1/determinant outside the matrix to make the numbers simpler)

Then multiply A -1 by B (we can use the Matrix Calculator again):

And we are done! The solution is:

x = 5 y = 3 z = −2

Just like on the Systems of Linear Equations page.

Quite neat and elegant, and the human does the thinking while the computer does the calculating.

Just For Fun ... Do It Again!

For fun (and to help you learn), let us do this all again, but put matrix "X" first.

I want to show you this way, because many people think the solution above is so neat it must be the only way.

So we will solve it like this:

And because of the way that matrices are multiplied we need to set up the matrices differently now. The rows and columns have to be switched over ("transposed"):

And XA = B looks like this:

Then (also shown on the Inverse of a Matrix page) the solution is this:

This is what we get for A -1 :

In fact it is just like the Inverse we got before, but Transposed (rows and columns swapped over).

Next we multiply B by A -1 :

And the solution is the same:

x = 5 , y = 3 and z = −2

It didn't look as neat as the previous solution, but it does show us that there is more than one way to set up and solve matrix equations. Just be careful about the rows and columns!

Matrix Equation

A system of equations can be solved using matrices by writing it in the form of a matrix equation. We can also determine whether a system has a unique solution or infinite number of solutions or no solution using the matrix equation.

Let us learn how to solve matrix equations in different methods along with examples.

What is Matrix Equation?

A matrix equation is of the form AX = B where A represents the coefficient matrix, X represents the column matrix of variables, and B represents the column matrix of the constants that are on the right side of the equations in a system. Let us consider a system of n nonhomogenous equations in n variables.

a₁₁ x₁ + a₁₂ x₂ + ... + a₁ₙ xₙ = b₁

a₂₁ x₁ + a₂₂ x₂ + ... + a₂ₙ xₙ = b₂

aₙ₁ x₁ + aₙ₂ x₂ + ... + aₙₙ xₙ = bₙ

Then the matrix equation that corresponds to the above system is:

AX = B, where

  • A = A matrix made of coefficients = \(\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{array}\right]\)
  • X = Column matrix of variables = \(\left[\begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array}\right]\)
  • B = Column matrix of constants = \(\left[\begin{array}{c} b_{1} \\ b_{2} \\ \vdots \\ b_{n} \end{array}\right]\)

How to Write Matrix Equation?

To write a system of equations as a matrix equation:

  • Make sure that the variables are in the same order in each of the equations.
  • Make sure that only the variable terms are on the left side and the constant terms are on the right side.
  • Then find the coefficient matrix A, variable matrix X, and the constant matrix B to write the given system as matrix equation AX = B.

Here is an example to understand these steps.

Example: Write the following system as a matrix equation.

3x + y + 2z - 10 = 0 z - 3y = -5 y + x - 2z = 7

Let us get all variable terms on one side (also in order) and constants on the right side.

3x + y + 2z = 10 0x - 3y + z = -5 x + y - 2z = 7

Now we can write this system as:

\(\left[\begin{array}{ccc} 3 & 1 & 2 \\ 0 & -3 & 1 \\ 1 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 10 \\ -5 \\ 7 \end{array}\right]\)

This is the matrix equation of the given system which alternatively can be written as AX = B where, A = \(\left[\begin{array}{ccc} 3 & 1 & 2 \\ 0 & -3 & 1 \\ 1 & 1 & -2 \end{array}\right]\), X = \(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\), and B = \(\left[\begin{array}{c} 10 \\ -5 \\ 7 \end{array}\right]\).

How to Solve Matrix Equation?

Let us solve the matrix equation AX = B for X. For this, we left multiply both sides of the equation by the inverse of A (that can be written as A -1 ).

A -1 (AX) = A -1 B

We know that A -1 A = I, where I is the identity matrix of the same order as A.

IX = A -1 B

We also know that IX = X.

This gives the solution of the matrix equation. This is also known as inverse matrix equation and hence the process of using the above formula to solve a system of equations is known as the " inverse matrix method". Thus, here are the steps to solve a system of equations using matrices:

  • Write the system as matrix equation AX = B.
  • Find the inverse, A -1 .
  • Multiply it by the constant matrix B to get the solution. i.e., X = A -1 B.

We can see the examples of solving a system using these steps in the " Matrix Equation Examples " section below.

Consistency of Solution of Matrix Equation

We know that we can find the inverse of a matrix only when it is nonsingular . i.e., A -1 exists only when det (A) ≠ 0. Thus, the solution (X = A -1 B) exists and it is unique only when det (A) ≠ 0.

In case, if det (A) = 0, how can we know how many solutions a system may have? In this case, we have to find (adj A) B, where adj A stands for " adjoint of A".

  • If (adj A) B ≠ O, then the system has no solution and hence the system is said to be inconsistent.
  • If (adj A) B = O, then the system may be "consistent with an infinite number of solutions" or "inconsistent (no solution)".

Here, 'O' is a null matrix . This is summarized in the flowchart below.

Number of solutions of a matrix equation

☛ Related Topics:

  • Homogeneous System of Linear Equations
  • Matrix Calculator
  • Matrix Formula
  • Matrix Multiplication Calculator

Matrix Equation Examples

Example 1: Write the following system of equations in terms of matrices: 2y - 3x = -2; 5x - 2y - 7 = 0.

Let us write the variables on the left and constants on the right side of the equation. Also, let us maintain the same order of variables in every equation.

-3x + 2y = -2 5x - 2y = 7

Now, the coefficient matrix is, A = \(\left[\begin{array}{ll} -3 & 2 \\ \\ 5 & -2 \end{array}\right]\).

The variable matrix is, X = \(\left[\begin{array}{ll} x \\ \\ y \end{array}\right]\).

The constant matrix is, B = \(\left[\begin{array}{ll} -2 \\ \\ 7 \end{array}\right]\).

Thus, the matrix equation is:

\(\left[\begin{array}{ll} -3 & 2 \\ \\ 5 & -2 \end{array}\right]\) \(\left[\begin{array}{ll} x \\ \\ y \end{array}\right]\) = \(\left[\begin{array}{ll} -2 \\ \\ 7 \end{array}\right]\).

Answer: The system of equations is written in matrix form.

Example 2: Using the matrix equation found in Example 1, find the solution of the given system.

From Example 1,

A = \(\left[\begin{array}{ll} -3 & 2 \\ \\ 5 & -2 \end{array}\right]\).

Using the formula of the inverse of 2x2 matrix ,

A -1 = 1 / (6 - 10) \(\left[\begin{array}{ll} -2 & -2 \\ \\ -5 & -3 \end{array}\right]\)

= -1/4 \(\left[\begin{array}{ll} -2 & -2 \\ \\ -5 & -3 \end{array}\right]\)

= \(\left[\begin{array}{ll} 1/2 & 1/2 \\ \\ 5/4 & 3/4 \end{array}\right]\)

The solution of the given system is,

=\(\left[\begin{array}{ll} 1/2 & 1/2 \\ \\ 5/4 & 3/4 \end{array}\right]\) \(\left[\begin{array}{ll} -2 \\ \\ 7 \end{array}\right]\)

= \(\left[\begin{array}{ll} (1/2) (-2) + (1/2) (7) \\ \\ (5/4)(-2) + (3/4) (7) \end{array}\right]\)

= \(\left[\begin{array}{ll} 5/2 \\ \\ 11/4 \end{array}\right]\)

= \(\left[\begin{array}{ll} x \\ \\ y \end{array}\right]\)

Answer: The solution is x = 5/4 and y = 11/4.

Example 3: Is the system that is solved in Example 2 consistent or inconsistent? How many solutions does it have? Justify your answer with another method.

In Example 2 , we have seen that the system has only one solution.

Since it has at least one solution, the system is consistent.

We can justify this using the determinant .

We have A = \(\left[\begin{array}{ll} -3 & 2 \\ \\ 5 & -2 \end{array}\right]\).

det (A) = (-3)(-2) - (5)(2) = 6 - 10 = -4 ≠ 0.

When det (A) ≠ 0, the system has a unique solution.

Answer: Consistent with a unique solution and the answer is justified using determinants.

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Practice Questions on Matrix Equation

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FAQs on Matrix Equation

What is the definition of matrix equation.

A matrix equation is of the form AX = B and it is writing the system of equations as a single equation in terms of matrices . Here,

  • A = A matrix formed by the coefficients
  • X = A column matrix formed by the variables
  • B = A column matrix formed by the constants

How to Solve Matrix Equation AX = B?

To solve a matrix equation AX = B:

  • Find A -1 (using the formula A -1 = (adj A) / (det A).
  • Find the solution using X = A -1 B.

How to Solve System of Equations Using Matrix Equation?

To solve a system of equations using matrices:

  • First, write all the variables on one side and the constants on the other side of the equations. Also, write the variables in the same order in every equation.
  • Write the system in the form AX = B by writing all coefficients of variables in matrix A, all variables as a column in matrix X, and all constants in a column in matrix B.

What are the Methods Used to Solve a Matrix Equation?

Here are the methods to solve a matrix equation AX = B.

  • Matrix inversion method (using the formula X = A -1 B)
  • Cramer's rule. Click here to learn about it.
  • Gauss Jordan method.

How to Write a System of Equations as a Matrix Equation?

We can write a system of equations as a matrix equation AX = B. Here are the steps for the same:

  • Maintain the order of the variables to be the same in all the equations.
  • Get all variables to the left side and send the constants to the right side of every equation.
  • Write the matrix A with all the coefficients of variables, where each row of A represents the coefficients of variables in one equation.
  • Write the matrix X with variables in order as a column.
  • Write the matrix B with constants in order as a column.

How Many Solutions a Matrix Equation Has?

A matrix equation AX = B has:

  • a unique solution (consistent) if det (A) ≠ 0.
  • no solution (inconsistent) if det (A) = 0 and (adj A) B ≠ O.
  • infinite solutions (consistent) or no solution (inconsistent) if det (A) = 0 and (adj A) B = O.

Here, O is the null matrix.

What is the Relation Between Matrix Equation and Rank of a Matrix?

For a matrix equation AX = B, [A B] represents the augmented matrix. Let r(A) and r(A, B) represent the ranks of matrices A and [A B].

  • If r(A) = r(A, B) = number of variables, then the system is consistent and has unique solution.
  • If r(A) = r(A, B) < number of variables, then the system is consistent and has infinitely many solutions.
  • If r(A) ≠ r(A, B), then the system is inconsistent and has no solution.

Matrices with Examples and Questions with Solutions

Examples and questions on matrices along with their solutions are presented .

Page Content

Definition of a matrix, matrix entry (or element), square matrix, identity matrix, diagonal matrix, triangular matrix, transpose of a matrix, symmetric matrix, questions on matrices: part a, questions on matrices: part b, solutions to the questions in part a, solutions to the questions in part b.

The following are examples of matrices (plural of matrix ).

Example 1 The following matrix has 3 rows and 6 columns.

The entry (or element) in a row i and column j of a matrix A (capital letter A) is denoted by the symbol \((A)_{ij} \) or \( a_{ij} \) (small letter a).

In the matrix A shown below, \(a_{11} = 5 \), \(a_{12} = 2 \), etc ... or \( (A)_{11} = 5 \), \( (A)_{12} = 2 \), etc ... \[ \textbf{A} = \begin{bmatrix} 5 & 2 & 7 & -3 \\ -9 & -2 & -7 & 11\\ \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ \end{bmatrix} \]

A square matrix has the number of rows equal to the number of columns.

For each matrix below, determine the order and state whether it is a square matrix. \[ a) \begin{bmatrix} -1 & 1 & 0 & 3 \\ 4 & -3 & -7 & -9\\ \end{bmatrix} \;\;\;\; b) \begin{bmatrix} -6 & 2 & 0 \\ 3 & -3 & 4 \\ -5 & -11 & 9 \end{bmatrix} \;\;\;\; \\ c) \begin{bmatrix} 1 & -2 & 5 & -2 \end{bmatrix} \;\;\;\; d) \begin{bmatrix} -2 & 0 \\ 0 & -3 \end{bmatrix} \;\;\;\; e) \begin{bmatrix} 3 \end{bmatrix} \] Solutions a) order: 2 × 4. Number of rows and columns are not equal therefore not a square matrix. b) order: 3 × 3. Number of rows and columns are equal therefore this matrix is a square matrix. c) order: 1 × 4. Number of rows and columns are not equal therefore not a square matrix. A matrix with one row is called a row matrix (or a row vector). d) order: 2 × 2. Number of rows and columns are equal therefore this is square matrix. e) order: 1 × 1. Number of rows and columns are equal therefore this matrix is a square matrix.

An identity matrix I n is an n×n square matrix with all its element in the diagonal equal to 1 and all other elements equal to zero. Example 4 The following are all identity matrices. \[I_1= \begin{bmatrix} 1 \\ \end{bmatrix} \quad , \quad I_2= \begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix} \quad , \quad I_3= \begin{bmatrix} 1 & 0 & 0\\ 0& 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

A diagonal matrix is a square matrix with all its elements (entries) equal to zero except the elements in the main diagonal from top left to bottom right. \[A = \begin{bmatrix} 6 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \]

An upper triangular matrix is a square matrix with all its elements below the main diagonal equal to zero. Matrix U shown below is an example of an upper triangular matrix. A lower triangular matrix is a square matrix with all its elements above the main diagonal equal to zero. Matrix L shown below is an example of a lower triangular matrix. \(U = \begin{bmatrix} 6 & 2 & -5 \\ 0 & -2 & 7 \\ 0 & 0 & 2 \end{bmatrix} \qquad L = \begin{bmatrix} 6 & 0 & 0 \\ -2 & -2 & 0 \\ 10 & 9 & 2 \end{bmatrix} \)

The transpose of an m×n matrix \( A \) is denoted \( A^T \) with order n×m and defined by \[ (A^T)_{ij} = (A)_{ji} \] Matrix \( A^T \) is obtained by transposing (exchanging) the rows and columns of matrix \( A \). Example 5 \[ \begin{bmatrix} 6 & 0 \\ -2 & -2\\ 10 & 9 \end{bmatrix} ^T = \begin{bmatrix} 6 & -2 & 10 \\ 0 & -2 &9\\ \end{bmatrix} \] Transpose a matrix an even number of times and you get the original matrix: \( ((A)^T)^T = A \). Transpose matrix an odd number of times and you get the transpose matrix: \( (((A)^T)^T)^T = A^T \). The transpose of any square diagonal matrix is the matrix itself. \[ \begin{bmatrix} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 6 \end{bmatrix} ^T = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 6 \end{bmatrix} \]

A square matrix is symmetric if its elements are such that \( A_{ij} = A_{ji} \) in other words \( A \) is symmetric if \(A = A^T \). Example 6 Symmetric matrices \[ \begin{bmatrix} 4 & -2 & 1 \\ -2 & 5 & 7 \\ 1 & 7 & 8 \end{bmatrix} \]

Given the matrices: \[ A = \begin{bmatrix} -1 & 23 & 10 \\ 0 & -2 & -11 \\ \end{bmatrix} ,\quad B = \begin{bmatrix} -6 & 2 & 10 \\ 3 & -3 & 4 \\ -5 & -11 & 9 \\ 1 & -1 & 9 \end{bmatrix} ,\quad C = \begin{bmatrix} -3 & 2 & 9 & -5 & 7 \end{bmatrix} \\ D = \begin{bmatrix} -2 & 6 \\ -5 & 2\\ \end{bmatrix} ,\quad E = \begin{bmatrix} 3 \end{bmatrix} ,\quad F = \begin{bmatrix} 3 \\ 5 \\ -11 \\ 7 \end{bmatrix} ,\quad G = \begin{bmatrix} -6 & -4 & 23 \\ -4 & -3 & 4 \\ 23 & 4 & 9 \\ \end{bmatrix} \] a) What is the dimension of each matrix? b) Which matrices are square? c) Which matrices are symmetric? d) Which matrix has the entry at row 3 and column 2 equal to -11? e) Which matrices has the entry at row 1 and column 3 equal to 10? f) Which are column matrices? g) Which are row matrices? h) Find \( A^T , C^T , E^T , G^T \).

1) Given the matrices: \[ A = \begin{bmatrix} 23 & 10 \\ 0 & -11 \\ \end{bmatrix} ,\quad B = \begin{bmatrix} -6 & 0 & 0 \\ -1 & -3 & 0 \\ -5 & 3 & -9 \\ \end{bmatrix} ,\quad C = \begin{bmatrix} -3 & 0\\ 0 & 2 \end{bmatrix} \\ ,\quad D = \begin{bmatrix} -7 & 3 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 9 \\ \end{bmatrix} ,\quad E = \begin{bmatrix} 12 & 0 & 0 \\ 0 & 23 & 0 \\ 0 & 0 & -19\\ \end{bmatrix} \] a) Which of the above matrices are diagonal? b) Which of the above matrices are lower triangular? c) Which of the above matrices are upper triangular?

a) A: 2 × 3, B: 4 × 3, C: 1 × 5, D: 2 × 2, E: 1 × 1, F: 4 × 1, G: 3 × 3, b) D, E and G c) E and G d) B e) A and B f) E and F g) E and C h) \[ A^T = \begin{bmatrix} -1 & 0 \\ 23 & -2 \\ 10 & -11 \end{bmatrix} ,\quad C^T = \begin{bmatrix} -3 \\ 2\\ 9\\-5\\7 \end{bmatrix} ,\quad E^T = \begin{bmatrix} 3 \end{bmatrix} ,\quad G^T = \begin{bmatrix} -6 & -4 & 23\\ -4 & -3 & 4\\ 23 & 4 & 9 \end{bmatrix} \]

More References and links

  • Add, Subtract and Scalar Multiply Matrices
  • Multiplication and Power of Matrices
  • Linear Algebra
  • Row Operations and Elementary Matrices
  • Matrix (mathematics)
  • Matrices Applied to Electric Circuits
  • The Inverse of a Square Matrix

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Solving Matrix Equations

Just like linear equations and quadratic equations, we may be asked to solve matrix equations. Although this might seem daunting at first, we will soon learn a few methods that can make this process relatively straightforward. Let's find out more:

What is a matrix equation?

A matrix equation is just like a normal equation. The only real difference is that we have swapped out our variables for numbers (we call these elements when we place them in a matrix). Consider the following system:

Note that this example is easy because all of our variables are lined up in the same order. Our "constraints" are also on the right side. If our system doesn't look like this, we might need to rearrange it a little before we move on. Each coefficient becomes an element on the resulting matrix:

Adding matrices

When we add matrices , all we need to do is add the numbers in the matching positions. For example:

Here we would add:

Note that the same basic principles apply to subtracting matrices.

Multiplying matrices by a scalar

Multiplying matrices is also pretty easy. All we need to do is multiply the matrix by a constant.

For example,

Note that this is called scalar multiplication.

Examples of solving matrices using addition and scalar multiplication.

Consider the following matrix :

Can we solve this matrix?

Now consider this matrix:

Can we solve this one?

Solving a system of linear equations using matrices

We can also use matrix equations to solve systems of linear equations. We can do this by working with the left and right sides of our equations:

Let's say we turned this system of linear equations:

Into this matrix:

Let's focus on the coefficient matrix (the left-hand matrix) and try to find the inverse:

Now we can multiply each side of the matrix equation by the inverse matrix. Remember that matrix multiplication is not commutative, so we need to place our inverse matrix on the left of each side of the equation:

The identity matrix on the left shows us that we calculated our inverse matrix correctly.

Now we can eliminate this value, leaving us with:

Therefore, our solution is 4 -5 .

Topics related to the Solving Matrix Equations

Distributive Property of Matrices

Adding and Subtracting Matrices

Square Matrix

Flashcards covering the Solving Matrix Equations

Linear Algebra Flashcards

Numerical Methods Flashcards

Practice tests covering the Solving Matrix Equations

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  • \begin{pmatrix}3 & 5 & 7 \\2 & 4 & 6\end{pmatrix}-\begin{pmatrix}1 & 1 & 1 \\1 & 1 & 1\end{pmatrix}
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  • To multiply two matrices together the inner dimensions of the matrices shoud match. For example, given two matrices A and B, where A is a m x p matrix and B is a p x n matrix, you can multiply them together to get a new m x n matrix C, where each element of C is the dot product of a row in A and a column in B.
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Jessica Grose

Loneliness is a problem that a.i. won’t solve.

A person’s hand reaching for a mobile device.

By Jessica Grose

Opinion Writer

When I was reporting my ed tech series , I stumbled on one of the most disturbing things I’ve read in years about how technology might interfere with human connection: an article on the website of the venture capital firm Andreessen Horowitz cheerfully headlined “ It’s Not a Computer, It’s a Companion! ”

It opens with this quote from someone who has apparently fully embraced the idea of having a chatbot for a significant other: “The great thing about A.I. is that it is constantly evolving. One day it will be better than a real [girlfriend]. One day, the real one will be the inferior choice.” The article goes on to breathlessly outline use cases for “A.I. companions,” suggesting that some future iteration of chatbots could stand in for mental health professionals, relationship coaches or chatty co-workers.

This week, OpenAI released an update to its ChatGPT chatbot, an indication that the inhuman future foretold by the Andreessen Horowitz story is fast approaching. According to The Washington Post, “The new model, called GPT-4o (“o” stands for “omni”), can interpret user instructions delivered via text, audio and image — and respond in all three modes as well.” GPT-4o is meant to encourage people to speak to it rather than type into it, The Post reports , as “The updated voice can mimic a wider range of human emotions, and allows the user to interrupt. It chatted with users with fewer delays, and identified an OpenAI executive’s emotion based on a video chat where he was grinning.”

There have been lots of comparisons between GPT-4o and the 2013 movie “Her,” in which a man falls in love with his A.I. assistant, voiced by Scarlett Johansson. While some observers, including the Times Opinion contributing writer Julia Angwin, who called ChatGPT’s recent update “ rather routine ,” weren’t particularly impressed, there’s been plenty of hype about the potential for humanlike chatbots to ameliorate emotional challenges, particularly loneliness and social isolation.

For example, in January, the co-founder of one A.I. company argued that the technology could improve quality of life for isolated older people, writing , “Companionship can be provided in the form of virtual assistants or chatbots, and these companions can engage in conversations, play games or provide information, helping to alleviate feelings of loneliness and boredom.”

Certainly, there are valuable and beneficial uses for A.I. chatbots — they can be life-changing for people who are visually impaired , for example. But the notion that bots will one day be an adequate substitute for human contact misunderstands what loneliness really is, and doesn’t account for the necessity of human touch.

There are disagreements among academics about the precise meaning of “loneliness,” but to come at it as a social problem, it’s worth trying to sharpen our definitions. Eric Klinenberg , a sociologist at New York University and the author of several books about social connectedness, including “Going Solo” and “Palaces for the People,” described the complexity of loneliness to me this way: “I think of loneliness as our bodies’ signal to us that we need better, more satisfying connections with other people.” And, he said, “the major issue I have with loneliness metrics is they often fail to distinguish between the ordinary healthy loneliness, which gets us off our couch and into the social world when we need it, and the chronic dangerous loneliness, which prevents us from getting off our couch, and spirals and leads us to spiral into depression and withdrawal.”

Why I worry about chatting with bots as a potential solution to loneliness is that it could be an approach that blunts the feeling just enough that it discourages or even prevents people from taking that step off the couch toward making connections with others. And some research indicates that a lack of human touch can exacerbate feelings of isolation. One 2023 paper by researchers at the University of Stirling expresses this more holistic view of loneliness quite eloquently, describing the emotion as “an embodied and contextualized sensory experience.”

Nick Gray, a co-author of that paper — which is about the effect of simulated versus real touch on feelings of loneliness — told me that he hasn’t seen any research yet on how realistic A.I. chatbots affect loneliness, noting that it is, obviously, a very new technology. But based on previous research in the field, including his own, he says “a realistic A.I. chatbot could give a temporary reprieve of the feelings of loneliness,” but “it’s a stretch to say it will reduce or get rid of loneliness.”

Klinenberg pointed out that we just had a natural experiment in forced isolation with the Covid-19 pandemic, and the results were quite clear: People, particularly people who lived alone, longed for human interaction. “If I told you that it’s a new pandemic that will hit this summer and we’ll all spend the next year alone or at home with our family, with everything in the public realm shut down, I don’t think the fact of A.I. would make us feel relieved,” he said, adding, “I think the prospect of a world without face-to-face interaction and human touch is terrifying.” He also noted that some of the companies that are pouring money into developing A.I. are among those that put ( unpopular ) return-to-office mandates into effect, so they certainly believe in the value of human interaction on some level.

I was struck by this passage in a story titled “Could A.I.-Powered Robot ‘Companions’ Combat Human Loneliness?” about work that researchers at Auckland, Duke and Cornell Universities are conducting with robots used as a means to try to help alleviate loneliness in older people:

“Right now, all the evidence points to having a real friend as the best solution,” said Murali Doraiswamy, M.B.B.S., F.R.C.P., professor of psychiatry and geriatrics at Duke University and member of the Duke Institute for Brain Sciences. “But until society prioritizes social connectedness and elder care, robots are a solution for the millions of isolated people who have no other solutions.”

What if even a tiny portion of the billions being spent developing A.I. chatbots could be spent on human and physical things we already know help loneliness? As Klinenberg put it, to help lonely and isolated people, we should be investing in things like collaborative housing, parks, libraries and other kinds of accessible social infrastructures that can help people of all ages build connectedness.

“The real social challenge and policy challenge and human challenge is for us to find ways to recognize these people and to attend to them, to care for them,” Klinenberg said. “But I also know that’s very hard work, and collectively we have failed to rise to that challenge.” We don’t want to spend the money or the time to support the most vulnerable among us. “In a way,” he added, “it’s our social failure that has created this opportunity for A.I. and technology to fill in the void.”

Jessica Grose is an Opinion writer for The Times, covering family, religion, education, culture and the way we live now.

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DEA Releases 2024 National Drug Threat Assessment

WASHINGTON – Today, DEA Administrator Anne Milgram announced the release of the 2024 National Drug Threat Assessment (NDTA), DEA’s comprehensive strategic assessment of illicit drug threats and trafficking trends endangering the United States.

For more than a decade, DEA’s NDTA has been a trusted resource for law enforcement agencies, policy makers, and prevention and treatment specialists and has been integral in informing policies and laws. It also serves as a critical tool to inform and educate the public.

DEA’s top priority is reducing the supply of deadly drugs in our country and defeating the two cartels responsible for the vast majority of drug trafficking in the United States. The drug poisoning crisis remains a public safety, public health, and national security issue, which requires a new approach.

“The shift from plant-based drugs, like heroin and cocaine, to synthetic, chemical-based drugs, like fentanyl and methamphetamine, has resulted in the most dangerous and deadly drug crisis the United States has ever faced,” said DEA Administrator Anne Milgram. “At the heart of the synthetic drug crisis are the Sinaloa and Jalisco cartels and their associates, who DEA is tracking world-wide. The suppliers, manufacturers, distributors, and money launderers all play a role in the web of deliberate and calculated treachery orchestrated by these cartels. DEA will continue to use all available resources to target these networks and save American lives.”

Drug-related deaths claimed 107,941 American lives in 2022, according to the Centers for Disease Control and Prevention (CDC). Fentanyl and other synthetic opioids are responsible for approximately 70% of lives lost, while methamphetamine and other synthetic stimulants are responsible for approximately 30% of deaths.

Fentanyl is the nation’s greatest and most urgent drug threat. Two milligrams (mg) of fentanyl is considered a potentially fatal dose. Pills tested in DEA laboratories average 2.4 mg of fentanyl, but have ranged from 0.2 mg to as high as 9 mg. The advent of fentanyl mixtures to include other synthetic opioids, such as nitazenes, or the veterinary sedative xylazine have increased the harms associated with fentanyl.   Seizures of fentanyl, in both powder and pill form, are at record levels. Over the past two years seizures of fentanyl powder nearly doubled. DEA seized 13,176 kilograms (29,048 pounds) in 2023. Meanwhile, the more than 79 million fentanyl pills seized by DEA in 2023 is almost triple what was seized in 2021. Last year, 30% of the fentanyl powder seized by DEA contained xylazine. That is up from 25% in 2022.  

Social media platforms and encrypted apps extend the cartels’ reach into every community in the United States and across nearly 50 countries worldwide. Drug traffickers and their associates use technology to advertise and sell their products, collect payment, recruit and train couriers, and deliver drugs to customers without having to meet face-to-face. This new age of digital drug dealing has pushed the peddling of drugs off the streets of America and into our pockets and purses.

The cartels have built mutually profitable partnerships with China-based precursor chemical companies to obtain the necessary ingredients to manufacturer synthetic drugs. They also work in partnership with Chinese money laundering organizations to launder drug proceeds and are increasingly using cryptocurrency.

Nearly all the methamphetamines sold in the United States today is manufactured in Mexico, and it is purer and more potent than in years past. The shift to Mexican-manufactured methamphetamine is evidenced by the dramatic decline in domestic clandestine lab seizures. In 2023, DEA’s El Paso Intelligence Center (EPIC) documented 60 domestic methamphetamine clandestine lab seizures, which is a stark comparison to 2004 when 23,700 clandestine methamphetamine labs were seized in the United States.

DEA’s NDTA gathers information from many data sources, such as drug investigations and seizures, drug purity, laboratory analysis, and information on transnational and domestic criminal groups.

It is available DEA.gov to view or download.

how solve matrix problems

The FAA bill President Joe Biden just signed might help Boston solve a big traffic problem. Here’s how

  • Published: May. 18, 2024, 3:22 p.m.

traffic jam or collapse in a city street road

(Adobe Stock Images) Kadmy - stock.adobe.com

See if this sounds familiar.

You’re driving down the narrow streets of Boston or some other big city when you suddenly find yourself forced to carefully navigate around someone who’s thrown on the hazards and double-parked in a travel lane, snarling traffic just when you need it the least.

Some relief — from the unlikeliest of places — might eventually be in sight.

Bipartisan legislation that President Joe Biden signed into law earlier this week reauthorizing the Federal Aviation Administration , includes language sponsored by a Massachusetts lawmaker intended to address parking nightmares outside the arrivals and departure gates at Boston Logan International Airport and other airports nationwide.

The language “empowers airports’ land operations to use cameras and other technologies to better manage their curbs,” the language’s sponsor, U.S. Rep. Jake Auchincloss, D-4th District, said in a post to X earlier this week . “That means that when you’re doing drop-off or pick-up, those double-park and triple-park situations that can make pedestrians and drivers feel unsafe, can be better managed.”

Auchincloss called the language “an important win” for his office, and travelers across the country.

The FAA Reauthorization Bill is headed to @POTUS ’ desk.✈️ I tucked in a win for walkability. Here’s what it might mean for pick-ups and drop-offs at airport curbs ⬇️ pic.twitter.com/mzs199pVAm — Rep. Jake Auchincloss 🟧 (@RepAuchincloss) May 16, 2024

In his post to X this week, Auchincloss said the language inserted into the FAA bill eventually could be “applied by downtowns in cities across the United States.”

  • Read More: Healey’s embattled transpo czar was right about one thing: Driving in Mass. is mostly misery | Analysis

Those jurisdictions have to “manage across a host of modalities ― public transit, walking, cycling, scooters, outdoor recreation, and of course, parking,” the Newton Democrat said.

“Giving cities the tools and technology they need to deconflict those different uses, and to maximize their curbs’ potential for public value will create more walkable and enjoyable downtowns,” he said.

City officials in Boston took a crack at the problem last year, by trying to reduce the ranks of food delivery drivers who routinely double- and triple-park on Boylston Street and other crowded thoroughfares in the city, NBC-10 in Boston reported .

“In some of the most congested areas, it has had a tremendous impact on the street,” the city’s streets chief told the station at the time. “What we see is rampant double or triple parking. We see vehicles, delivery drivers squeezing their car in ways that end up being terribly unsafe.”

The FAA bill that Biden signed into law renews the agency’s authority for the next five years, CNN reported. It authorizes more than $105 billion in funding for the FAA, as well as $738 million for the National Transportation Safety Board for fiscal years 2024 through 2028, CNN reported.

In a statement after the signing, Biden praised the bill, and touted a new U.S. Department of Transportation rule requiring automatic cash refunds instead of vouchers, CNN reported.

“The bipartisan Federal Aviation Administration reauthorization is a big win for travelers, the aviation workforce, and our economy. It will expand critical protections for air travelers, strengthen safety standards, and support pilots, flight attendants, and air traffic controllers,” Biden said, according to the statement released by the White House.

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‘you can solve anything’.

Priyanka Pillai wants to take on big problems — and has learned how good design can help

Christina Pazzanese

Harvard Staff Writer

Portrait of Priyanka Pillai inside the Design School.

Niles Singer/Harvard Staff Photographer

Part of the Commencement 2024 series

A collection of stories covering Harvard University’s 373rd Commencement.

Growing up in India, Priyanka Pillai witnessed the immense and varied struggles many impoverished people faced in their daily lives, such as getting prenatal care and protecting children from labor exploitation.

As an undergraduate in Bangalore studying industrial design, she wondered whether good design could help ease at least parts of these and other challenges. She came to Harvard Graduate School of Design two years ago and got her answer, discovering she could take on big problems “that you don’t even realize … could be tackled with design.”

Pillai wanted to do something to help address the refugee crisis in Uganda for her independent design engineering project. Those projects span two semesters and call for students seeking a master’s in design engineering (a joint GSD and John A. Paulson School of Engineering and Applied Sciences program) to identify complex, real-world problems and develop solution prototypes.

“For the first time, I truly felt like I was doing work that was very in touch with what GSD wants people to do, which is working with communities.”

Conducting fieldwork in Uganda, Pillai saw the difficulties that South Sudanese refugees were having reuniting with their families. The plight of those fleeing the ongoing civil war in the northeast African nation has become one of the largest refugee crises in the world, with more than half a million living in Uganda alone, mostly in camps.

More than 60 percent are children separated from parents who are looking for them, Pillai said, and need multiple layers of support. While non-governmental organizations (NGOs) are providing some assistance, much more help is needed.

“One thing that really stood out was agency. There’s currently a lack of agency when it comes to finding their family members on their own,” said Pillai, who graduates later this month. Many refugees use informal, ad hoc methods such as phone calls, WhatsApp, and photo sharing to try to find relatives.

“The second part, which is extremely critical, is that we need to move from a Western-centric way of finding a family member,” such as cataloguing names, ages, and date of separation done by NGOs, because it doesn’t capture vernacularor local geography, vital details that may speed up reunification, she said, noting that learning more about how to design for “the Indian context” and the Global South more generally was a key reason she came to Harvard.

“A lot of cultural nuances were missing in connection to the data to find missing family members,” she said. “And that’s the kind of solution that we’re moving toward.”

Given the ubiquity of cellphones there, Pillai and classmate Julius Stein designed and built an online platform for refugees to enter information about themselves using text, photos, and audio. The platform generates a series of questions that can lead to possible matches while minimizing the risk of exploitation by malign actors.

“For the first time, I truly felt like I was doing work that was very in touch with what GSD wants people to do, which is working with communities,” she said. “It was just a life-changing experience.”

Earlier this month, one startup Pillai is involved in, Alba, won an Ingenuity Award as part of the Harvard President’s Innovation Challenge. The team designed a special wipe so the visually impaired can better detect when their menstrual period has begun without relying on outside assistance.

In 2023, Pillai was part of a student project that won gold in the Spark International Design awards. The design team created Felt, a haptic armband that turns sound and visual clues into movement. The device assists people who are deaf blind to independently catch emotional nuances or subtexts in conversations, which often get lost in Braille or other translations.

During her time in the program, Pillai also jumped at the opportunity to take courses at the Harvard Kennedy School, Harvard Law School, and Harvard Graduate School of Education to learn more about things such as accessibility, ethical design, and negotiation.

“I knew that I was limiting myself because I didn’t know all these different things,” she said.

When not focused on her own studies, Pillai has been a teaching fellow for a design studio at GSD and at SEAS for a course led by her IDEP adviser, Krzysztof Gajos, Gordon McKay Professor of Computer Science.

“I love teaching,” she said. “It’s one of my favorite experiences.”

Reflecting on her time at GSD, Pillai has been deeply inspired by the faculty and her fellow students. This group from many different backgrounds with different interests and perspectives, working in many different disciplines, has been like a “dream” design studio where she’s been able to share and borrow ideas and practices from others and see how other fields look at things such as collaboration, sustainability and accessibility. It has been intellectually liberating to experience such fearlessness, she said, after years of feeling so “constrained” in her prior practice, which had been “rooted in ‘realistic goals.’”

“People tackling very huge issues that you don’t even realize 1) is a problem that could be tackled with design, and 2), they’re almost your age and they’re doing it somehow. That was very important to see,” she said.

“People really think that you can solve anything.”

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Longtime supporter of grads Kathy Hanley caps 13-year quest with a Commencement of her own

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Rehoming wild horses won't solve the brumby problem, but it transforms lives for horses and owners

Teenage girl stands with horse

At 14, Ruby Wild would rather demonstrate her connection with horses than explain it.

Whether she's prompting the more than 500-kilogram animal to lie down or waiting for the mare to be still enough to stand on her back, the bond between the teenager and her horses is undeniable.

Teenager on a horse jumping jump

"If you don't have a strong connection, it's hard to do anything," Ruby said.

But her herd of four aren't regular domestic horses, they were once wild brumbies.

A young woman and small girl with a horse

"We grew up together because she was only little. She was only two when we got her," she said.

Her mum Kylie Wild said, while Ruby was supervised, her young daughter had a natural intuition with horses.

"Ruby and Gidget built this connection together and it never really crossed my mind that she was a wild brumby," Ms Wild said.

A teenage girl hugs her horse while another girl squats down next to her

Ruby's love of brumbies is one she shares with her cousin, 17-year-old Brooke Wild.

"Brumbies, they're definitely underestimated, that's for sure," Brooke said.

"Brumbies are just so, so versatile, anything you put them to: jumps, cattle, whatever. They'll just pick it up like that and they just learn so quickly."

Where did brumbies come from?

Horses came to Australia on the First Fleet in 1788. Escaped animals become the country's first brumbies.

Over time, other domestic breeds have gone feral and now descendants of thoroughbreds, stock, quarter and heavy horses can be found in the wild.

Trapped horses

It is estimated there are more than 400,000 brumbies in Australia and when concentrated in high numbers, their hard hooves pose a risk to the environment and native species.

Horse trainer Anna Uhrig runs a brumby rehoming camp in south-east Queensland, using horses trapped in local forests and from central Queensland.

"We can't solve the brumby problem but we're doing what we can," Ms Uhrig said.

a woman and man behind a horse truck

"It's a very small part of the puzzle: the amount of horses we rehome. It's not thousands but I think the work adds quality of life to those horses."

Ecologist Dave Berman has been managing brumby populations for 40 years across the country.

For most of the past decade, he has managed the brumby herds of Tuan and Toolara state forests, which consist of pine plantations and native species north of Noosa.

man looking into a pen of horses

The growing population has brought wild horses close to roads, putting drivers at risk.

"We catch them and find homes for them to reduce the risk of collisions between horses and people," Dr Berman said.

"Usually it's the horses that get killed, but eventually, you know, there will be people killed.

An empty forestry road with a brumby hazard sign.

"Originally in about 2009 there were about four horses killed per year, we removed all the horses regularly crossing the road and then there were no collisions, so we showed that worked."

What's a brumby rehoming camp?

Ms Uhrig's 10-day intensive camp attracts participants from all over the country to choose, train and adopt a brumby.

"Yarding, drafting and then coming to these yards is totally new for the brumbies so it can be quite scary for them," Ms Uhrig said.

brumbies walking in a pen

In less than two weeks, some of the horses go from not letting a human within 100 metres to being ridden.

"I just love seeing the transformation and I love giving people the opportunity to learn those skills," Ms Uhrig said.

woman standing in front of a pen of brumbies

"I could break in a brumby continuously and I still wouldn't be able to rehome as many as we can through the camps."

Melissa Teunis's brumby Valour will join her herd of domestic horses used for equine therapy.

woman sitting in chair

"I've loved horses forever and I just wanted the opportunity to start [training] my own [horse]," she said. 

"It's the perfect place to come. Anna is amazing and just knows the process and talks and walks you through every step."

But Ms Uhrig said rehoming brumbies wouldn't work everywhere.

Anna Uhrig with horse

"In terms of the whole management debate, it's not a closed book. We don't have the answers necessarily," Ms Uhrig said.

"There's been some research done in some targeted areas and we think we have the answers and we can apply those but generally it's a nationwide issue and the management implications are different in each area."

How are the horses trapped?

Dr Berman said   the trapping process was slow and targeted brumbies who grazed near or regularly crossed roads.

Horses in a pen

"We use electric fences and we build a really large electric fence area that can be up to 4 kilometres long," he said.

"We'll build that around the horses and gradually make it smaller and then get them into a laneway and then into hessian panels, and then into panel yards."

He said patience and a respect for the animals were required.

"If you put too much pressure on, they go. They'll go through the fence anyway," he said.

Man standing in front of horse pens

Giving the horses a second chance can take weeks and sometimes months.

It's an emotional process for Dr Berman who creates a strong connection with the brumbies during the process.

"They're wonderful animals," he said.

A national problem

Brumby management differs across the country.

Dr Berman said rehoming was becoming more common, but wasn't the solution alone.

"In a lot of other parts of Australia where populations are getting too much, horses can be shot from the air or the ground," he said.

People rising horses down a hill

In New South Wales's largest national park, Kosciuszko, the government is culling horses in the hundreds . 

In 2022, the NSW government estimated about 18,000 horses lived in the park, with a population range between 14,501 and 23,535 horses .

Under the state government management plan, more than 15,000 brumbies could be killed or rehomed to reduce the population to 3,000 by mid-2027, to the despair of some of the locals.

Meanwhile, a Kosciuszko brumby rehoming program has been suspended  after hundreds of horse carcasses were found on a property near Wagga Wagga .

Further north in NSW, the officers from the Local Land Services have a different approach, working with landholders and rescue groups to trap and rehome wild horses near Grafton.

"It's not just about safety for people. It's the safety of the horses as a number of them have been hit on the road," senior biosecurity officer Tiffany Felton said.

"But it's not just that, they're a hard-hoofed animals so they do, and especially in these dry times, start affecting the waterways and in environmentally sensitive areas."

In the past three years more than 130 have been captured and rehomed under the state government's Biosecurity Act.

horses in a pen

"Previously, there have been probably far harsher approaches to wild horse management, and that's not what the community wanted," said Louise Orr of North Coast Local Land Services.

Untapped potential

After just seven days at brumby camp, some of the horses are almost unrecognisable for participants, including Melissa Teunis.

The back of a woman looking at a horse

"It's amazing to think that when we got here last Friday, he'd never been touched," Ms Teunis said.

"It shows you how resilient they are and just how willing — oh this will make me cry — how willing they are to try for us and, and give us all that they have."

woman standing next to horse

Lilly Anderson, 14, could put a saddle on her horse Rain by day four of the course and three days later, she was able to ride the animal.

"Like, for me, that's incredible. That's like just gone so quickly. But she's also so just amazing," Lilly said.

Girl standing in front of a horse pen

The camp is the first stage of the horses transitioning to domestic life with more time and effort required after they leave.

"Once you give them a chance, they'll do anything for you," Lilly said. 

"She's just so smart to catch on and she'll do anything for lucerne as well. Loves a bit of food."

Young woman with her brumby

Past participants of the camp said the biggest challenge with rehoming a brumby was fighting off the temptation to buy another.

"It's been hard but it gets better. The highs are really high," Ava Cloherty said.

Watch ABC TV's Landline at 12:30pm on Sunday or on  ABC iview .

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