Course Hero Logo
Upload to Study
Expert Help
College Algebra/Matrices/Matrix Operations

Matrices

Contents

Matrix Operations

Matrix Addition and Subtraction

Matrices of the same dimensions can be added or subtracted.

A matrix is a rectangular array of numbers, enclosed by brackets. The numbers in a matrix are arranged in horizontal rows and vertical columns. Each number in a matrix is an entry. The dimensions of a matrix are its numbers of rows and columns, written in the form m×nm\times n, where mm is the number of rows and nn is the number of columns.

Performing matrix calculations involves examining each entry in a row or column, locating entries, and writing and reading the notation for an entry. A general notation of ai,ja_{i,j} can represent any entry in a matrix, with ii as the row number and jj as the column number.

For example, matrix AA is a 3×43\times 4 matrix because it has 3 rows and 4 columns.
A=[1−5−252−2−340−7−00]←Row 1←Row 2←Row 3A=[↑↑↑↑ColumnsA=C1C2C3C4(Columns)\begin{aligned}A=\begin{bmatrix}1 & -5 & \phantom{-}2 & 5 \\ 2 & \phantom{-}2 & -3 & 4 \\ 0 & \phantom{-}7 & \phantom{-}0 & 0 \end{bmatrix} \begin{aligned} \hspace{10pt}\leftarrow\hspace{10pt}&\text{Row 1} \\\hspace{10pt}\leftarrow\hspace{10pt}&\text{Row 2}\\\hspace{10pt}\leftarrow\hspace{10pt}&\text{Row 3}\end{aligned}\\\begin{aligned} \phantom{A=\lbrack}\uparrow \hspace{5pt} \uparrow \hspace{5pt} \uparrow \hspace{5pt} \uparrow \phantom{\text{Columns}} \hspace{35pt}\\\phantom{A=}\text{C1}\hspace{10pt}\text{C2}\hspace{8pt} \text{C3}\hspace{8pt} \text{C4} \hspace{10pt} \text{(Columns)}\hspace{20pt}\end{aligned}\end{aligned}
The entry in row 3 and column 2 of matrix AA is 7. The notation for identifying the entry is a3,2a_{3,2} since the entry is in row 3 and column 2.

To add or subtract matrices, add or subtract corresponding entries. The matrices must have the same dimensions so that every entry in one matrix has a corresponding entry in the other matrix.

For example, matrix AA and matrix BB have the same dimensions. They are each a 3×43 \times 4 matrix.
A=[1−5−252−2−340−7−00]B=[2−1−0−14−3−7−41−4−2−3]A=\begin{bmatrix} {\color{#c42126}{1}} & -5 & \phantom{-}2 & 5 \\ 2 & \phantom{-}2 & -3 & 4 \\ 0 & \phantom{-}7 & \phantom{-}0 & 0 \end{bmatrix} \hspace{20pt} B=\begin{bmatrix} {\color{#c42126}{2}} & -1 & \phantom{-}0 & \phantom{-}1 \\ 4 & \phantom{-}3 & \phantom{-}7 & \phantom{-}4 \\ 1 & \phantom{-}4 & -2 & -3 \end{bmatrix}
Matrix AA and matrix BB can be added to create matrix A+BA+B. For the first entry in matrix A+BA+B, add the entry from row 1 and column 1 of matrix AA and matrix BB. Then, place the sum in row 1 and column 1 of matrix A+BA+B. Continue adding each corresponding entry until matrix A+BA+B is complete.
A+B=[1+2−5+(−1)2+05+12+42+3−3+74+40+17+40+(−2)0+(−3)]=[3−6−2−66−5−4−81−11−2−3]\begin{aligned}A+B &= \begin{bmatrix} {\color{#c42126}{1+2}} & -5+(-1) & 2+0 & 5+1 \\ 2+4 & 2+3 & -3+7 & 4+4 \\ 0+1 & 7+4 & 0+(-2) & 0+(-3) \end{bmatrix}\\\\&= \begin{bmatrix} {\color{#c42126}{3}} & -6 & \phantom{-}2 & \phantom{-}6 \\ 6 & \phantom{-}5 & \phantom{-}4 & \phantom{-}8 \\ 1 & \phantom{-}11 & -2 & -3 \end{bmatrix}\end{aligned}
Subtracting matrix AA and matrix BB follows the same process as adding the matrices. Subtract the entry from row 1 and column 1 of matrix AA and matrix BB. Then, place the difference in row 1 and column 1 of matrix A−BA-B. Continue subtracting until matrix A−BA-B is complete.
A−B=[1−2−5−(−1)2−05−12−42−3−3−74−40−17−40−(−2)0−(−3)]=[−1−4−24−2−1−100−1−3−23]\begin{aligned}A-B &= \begin{bmatrix}{\color{#c42126}{1 - 2}} & -5-(-1) & 2 - 0 & 5-1 \\ 2-4 & 2-3 & -3-7 & 4-4 \\ 0-1 & 7-4 & 0-(-2) & 0-(-3) \end{bmatrix}\\\\&= \begin{bmatrix} {\color{#c42126}{-1}} & -4 & \phantom{-}2 & 4 \\ -2 & -1 & -10 & 0 \\ -1 & \phantom{-}3 & \phantom{-}2 & 3 \end{bmatrix}\end{aligned}

Scalar Multiplication

A scalar product is the product of a real number and a matrix.

A scalar is a real number or a quantity that can be represented by a real number. To determine the product of a scalar and a matrix, multiply each entry in the matrix by the scalar. The result is called a scalar product.

For example, matrix AA can be multiplied by 4 to determine matrix 4A4A. Multiply the entry in row 1 and column 1 of matrix AA by the scalar, which is 4. Then, place the product in row 1 and column 1 of matrix 4A4A. Continue multiplying each entry by the scalar until matrix 4A4A is complete.
A=[1−5−252−2−340−7−00]A=\begin{bmatrix} {\color{#c42126}{1}} & -5 & \phantom{-}2 & 5 \\ 2 & \phantom{-}2 & -3 & 4 \\ 0 & \phantom{-}7 & \phantom{-}0 & 0 \end{bmatrix}
4A=[4(1)4(−5)4(2)4(5)4(2)4(2)4(−3)4(4)4(0)4(7)4(0)4(0)]=[4−20−82088−12160−28−00]\begin{aligned}{\color{#c42126}{4}}A&=\begin{bmatrix} {\color{#c42126}{4(1)}} & 4(-5) & 4(2) & 4(5) \\ 4(2) & 4(2) & 4(-3) & 4(4) \\ 4(0) & 4(7) & 4(0) & 4(0) \end{bmatrix}\\\\&=\begin{bmatrix} {\color{#c42126}{4}} & -20 & \phantom{-}8 & 20 \\ 8 & 8 & -12 & 16 \\ 0 & \phantom{-}28 & \phantom{-}0 & 0 \end{bmatrix}\end{aligned}
Matrices can be multiplied by any real number, including fractions. For example, matrix BB can be multiplied by a scalar of 13\frac{1}{3}.
B=[2−1−0−14−3−7−41−4−2−3]B=\begin{bmatrix} {\color{#c42126}{2}} & -1 & \phantom{-}0 & \phantom{-}1 \\ 4 & \phantom{-}3 & \phantom{-}7 & \phantom{-}4 \\ 1 & \phantom{-}4 & -2 & -3 \end{bmatrix}
13B=[13(2)13(−1)13(0)13(1)13(4)13(3)13(7)13(4)13(1)13(4)13(−2)13(−3)]=[23−13−0−1343−1−73−4313−43−23−1]\begin{aligned}{\color{#c42126}{\frac{1}{3}}}B&=\begin{bmatrix} {\color{#c42126}{\frac{1}{3}(2)}} & \frac{1}{3}(-1) & \frac{1}{3}(0) & \frac{1}{3}(1) \\[0.5em] \frac{1}{3}(4) & \frac{1}{3}(3) & \frac{1}{3}(7) & \frac{1}{3}(4) \\[0.5em] \frac{1}{3}(1) & \frac{1}{3}(4) & \frac{1}{3}(-2) & \frac{1}{3}(-3) \end{bmatrix}\\\\[0.5em]&=\begin{bmatrix} {\color{#c42126}{\frac{2}{3}}} & -\frac{1}{3} & \phantom{-}0 & \phantom{-}\frac{1}{3} \\[0.5em] \frac{4}{3} & \phantom{-}1 & \phantom{-}\frac{7}{3} & \phantom{-}\frac{4}{3} \\[0.5em] \frac{1}{3} & \phantom{-}\frac{4}{3} & -\frac{2}{3} & -1 \end{bmatrix}\end{aligned}
Scalar multiplication can also be combined with matrix addition and subtraction.
Step-By-Step Example
Multiplying a Matrix with a Scalar and with Subtraction
Determine 3A−2B3A - 2B.
A=[1−5−252−2−340−7−00]B=[2−1−0−14−3−7−41−4−2−3]A=\begin{bmatrix} 1 & -5 & \phantom{-}2 & 5 \\ 2 & \phantom{-}2 & -3 & 4 \\ 0 & \phantom{-}7 & \phantom{-}0 & 0 \end{bmatrix} \hspace{30pt} B=\begin{bmatrix} 2 & -1 & \phantom{-}0 & \phantom{-}1 \\ 4 & \phantom{-}3 & \phantom{-}7 & \phantom{-}4 \\ 1 & \phantom{-}4 & -2 & -3 \end{bmatrix}
Step 1
Identify 3A3A by multiplying each entry in matrix AA by the scalar, or 3.
A=[1−5−252−2−340−7−00]A=\begin{bmatrix} 1 & -5 & \phantom{-}2 & 5 \\ 2 & \phantom{-}2 & -3 & 4 \\ 0 & \phantom{-}7 & \phantom{-}0 & 0 \end{bmatrix}
3A=[3(1)3(−5)3(2)3(5)3(2)3(2)3(−3)3(4)3(0)3(7)3(0)3(0)]=[3−15−6156−6−9120−21−00]\begin{aligned}3A&=\begin{bmatrix} 3(1) & 3(-5) & 3(2) & 3(5) \\ 3(2) & 3(2) & 3(-3) & 3(4) \\ 3(0) & 3(7) & 3(0) &3(0) \end{bmatrix}\\\\&=\begin{bmatrix} 3 & -15 & \phantom{-}6 & 15 \\ 6 & \phantom{-}6 & -9 & 12 \\ 0 & \phantom{-}21 & \phantom{-}0 & 0 \end{bmatrix}\end{aligned}
Step 2
Identify 2B2B by multiplying each entry in matrix BB by the scalar, or 2.
B=[2−1−0−14−3−7−41−4−2−3]B=\begin{bmatrix} 2 & -1 & \phantom{-}0 & \phantom{-}1 \\ 4 & \phantom{-}3 & \phantom{-}7 & \phantom{-}4 \\ 1 & \phantom{-}4 & -2 & -3 \end{bmatrix}
2B=[2(2)2(−1)2(0)2(1)2(4)2(3)2(7)2(4)2(1)2(4)2(−2)2(−3)]=[4−2−0−28−6−14−82−8−4−6]\begin{aligned}2B&=\begin{bmatrix} 2(2) & 2(-1) & 2(0) & 2(1) \\ 2(4) & 2(3) & 2(7) & 2(4) \\ 2(1) & 2(4) & 2(-2) & 2(-3) \end{bmatrix}\\\\&=\begin{bmatrix} 4 & -2 & \phantom{-}0 & \phantom{-}2 \\ 8 & \phantom{-}6 & \phantom{-}14 & \phantom{-}8 \\ 2 & \phantom{-}8 & -4 & -6 \end{bmatrix}\end{aligned}
Step 3
Determine 3A−2B3A - 2B by subtracting each corresponding entry in matrix 3A3A and matrix 2B2B.
3A−2B=[3−15−6156−6−9120−21−00]−[4−2−0−28−6−14−82−8−4−6]=[3−4−15−(−2)6−015−26−86−6−9−1412−80−221−80−(−4)0−(−6)]\begin{aligned}3A-2B&=\begin{bmatrix} 3 & -15 & \phantom{-}6 & 15 \\ 6 & \phantom{-}6 & -9 & 12 \\ 0 & \phantom{-}21 & \phantom{-}0 & 0 \end{bmatrix}-\begin{bmatrix} 4 & -2 & \phantom{-}0 & \phantom{-}2 \\ 8 & \phantom{-}6 & \phantom{-}14 & \phantom{-}8 \\ 2 & \phantom{-}8 & -4 & -6 \end{bmatrix}\\\\&=\begin{bmatrix} 3-4 & -15-(-2) & 6-0 & 15-2 \\ 6-8 & 6-6 & -9-14 & 12-8 \\ 0-2 & 21-8 & 0-(-4) & 0-(-6) \end{bmatrix}\end{aligned}
Solution
Simplify each entry in the resulting matrix.
3A−2B=[3−4−15−(−2)6−015−26−86−6−9−1412−80−221−80−(−4)0−(−6)]=[−1−13−613−2−0−234−2−13−46]\begin{aligned}3A-2B&=\begin{bmatrix} 3-4 & -15-(-2) & 6-0 & 15-2 \\ 6-8 & 6-6 & -9-14 & 12-8 \\ 0-2 & 21-8 & 0-(-4) & 0-(-6) \end{bmatrix}\\\\&=\begin{bmatrix} -1 & -13 & \phantom{-}6 & 13 \\ -2 & \phantom{-}0 & -23 & 4 \\ -2 & \phantom{-}13 & \phantom{-}4 & 6 \end{bmatrix}\end{aligned}

Matrix Multiplication

An m×nm\times n matrix can be multiplied by an n×pn\times p matrix. The result is an m×pm\times p matrix.
Two matrices can be multiplied only if the number of columns of the first matrix is equal to the number of rows of the second matrix. The resulting matrix, called the matrix product, is the product of two or more matrices. It has the same number of rows as the first matrix and the same number of columns as the second matrix. If the number of columns of the first matrix is not equal to the number of rows of the second matrix, the product is undefined.

Matrix Product Dimensions

Matrix AA BB ABAB
Dimensions (Row×Column\text{Row} \times \text{Column})
3×2{\color{#c42126}3}\times{\color{#0047af}2}
2×4{\color{#0047af}2}\times{\color{#c42126}4}
3×4{\color{#c42126}3}\times{\color{#c42126}4}

When multiplying matrices, make sure that the number of columns in the first matrix and the number of rows in the second matrix are equal. For example, matrix AA is a 3×23 \times 2 matrix, while matrix BB is a 2×42 \times 4 matrix. They can be multiplied because matrix AA has 2 columns and matrix BB has 2 rows. The resulting product matrix, ABAB, will have the same number of rows as matrix AA (3 rows) and the same number of columns as matrix BB (4 columns).

Consider matrix AA.
A=[a1,1a1,2...a1,j...a1,na2,1a2,2...a2,j...a2,n⋮⋮⋮⋮ai,1ai,2...ai,j...ai,n⋮⋮⋮⋮am,1am,2...am,j...am,n]A=\begin{bmatrix} a_{1,1} & a_{1,2} & ... & a_{1,j} & ... & a_{1,n} \\[0.5em] a_{2,1} & a_{2,2} & ... & a_{2,j} & ... & a_{2,n} \\[0.5em] \vdots & \vdots & \: & \vdots & \: & \vdots \\[0.5em] a_{i,1} & a_{i,2} & ... & {\color{#c42126} a_{i,j}}& ... & a_{i,n} \\[0.5em] \vdots & \vdots & \: & \vdots & \: & \vdots \\[0.5em] a_{m,1} & a_{m,2} & ... & a_{m,j} & ... & a_{m,n} \end{bmatrix}
A row number of a matrix is represented by ii. A column number is represented by jj. The entry in row ii and column jj is written as ai,ja_{i,j}.

To multiply two matrices, multiply consecutive entries in row ii of the first matrix and column jj in the second matrix. The sum of those products will become the entry in row ii and column jj in the product matrix.

Unlike multiplication with real numbers, matrix multiplication is not commutative, so ABAB may not be equal to BABA, even if both matrices are defined.

Step-By-Step Example
Multiplying Matrices
Identify the matrix product ABAB.
A=[−5−0−2−3−1−4]B=[3−40−22−11−2]A=\begin{bmatrix} \phantom{-}5 & \phantom{-}0 \\ -2 & -3 \\ \phantom{-}1 & \phantom{-}4 \end{bmatrix}\hspace{20pt}B= \begin{bmatrix} 3 & \phantom{-}4 & 0 & \phantom{-}2 \\ 2 & -1 & 1 & -2 \end{bmatrix}
Step 1
To determine the entry in the first row and first column of the matrix product ABAB, multiply consecutive entries in the first row of matrix AA and the first column of matrix BB. Then add the products.
A=[−5−0−2−3−1−4]B=[3−40−22−11−2]A=\begin{bmatrix} {\color{#c42126}\phantom{-}5} & {\color{#c42126}\phantom{-}0 } \\ -2 & -3 \\ \phantom{-}1 & \phantom{-}4 \end{bmatrix} \hspace{20pt}B=\begin{bmatrix} {\color{#0047af}3} & \phantom{-}4 & 0 & \phantom{-}2 \\ {\color{#0047af}2} & -1 & 1 & -2 \end{bmatrix}
5⋅3+0⋅2=15{\color{#c42126}5}\cdot{\color{#0047af}3}+{\color{#c42126}0}\cdot{\color{#0047af}2}=15
AB=[15200−10−12−5−3−2−11−0−4−6]AB=\begin{bmatrix} 15 & \phantom{20} & \phantom{0} & \phantom{-10} \\ \phantom{-12} & \phantom{-5} & \phantom{-3} & \phantom{-2} \\ \phantom{-11} & \phantom{-0} & \phantom{-4} & \phantom{-6} \end{bmatrix}
Step 2
To determine the entry in the first row and second column of ABAB, multiply consecutive entries in the first row of AA and the second column of BB. Then add the products.
A=[−5−0−2−3−1−4]B=[3−40−22−11−2]A=\begin{bmatrix} {\color{#c42126}\phantom{-}5} & {\color{#c42126}\phantom{-}0 } \\ -2 & -3 \\ \phantom{-}1 & \phantom{-}4 \end{bmatrix} \hspace{20pt}B=\begin{bmatrix} 3 & {\color{#0047af}\phantom{-}4} & 0 & \phantom{-}2 \\ 2 & {\color{#0047af}-1} & 1 & -2 \end{bmatrix}
5⋅4+0⋅(−1)=20{\color{#c42126}5}\cdot{\color{#0047af}4}+{\color{#c42126}0}\cdot ({\color{#0047af}-1})=20
AB=[15200−10−12−5−3−2−11−0−4−6]AB=\begin{bmatrix} 15 & 20 & \phantom{0} & \phantom{-10} \\ \phantom{-12} & \phantom{-5} & \phantom{-3} & \phantom{-2} \\ \phantom{-11} & \phantom{-0} & \phantom{-4} & \phantom{-6} \end{bmatrix}
Step 3
To determine the entry in the first row and third column of ABAB, multiply consecutive entries in the first row of AA and the third column of BB. Then add the products.
A=[−5−0−2−3−1−4]B=[3−40−22−11−2]A=\begin{bmatrix} {\color{#c42126}\phantom{-}5} & {\color{#c42126}\phantom{-}0 } \\ -2 & -3 \\ \phantom{-}1 & \phantom{-}4 \end{bmatrix} \hspace{20pt} B=\begin{bmatrix} 3 & \phantom{-}4 & {\color{#0047af}0} & \phantom{-}2 \\ 2 & -1 & {\color{#0047af}1} & -2 \end{bmatrix}
5⋅0+0⋅1=0{\color{#c42126}5}\cdot{\color{#0047af}0}+{\color{#c42126}0}\cdot {\color{#0047af}1}=0
AB=[15200−10−12−5−3−2−11−0−4−6]AB=\begin{bmatrix} 15 & 20 & 0 & \phantom{-10} \\ \phantom{-12} & \phantom{-5} & \phantom{-3} & \phantom{-2} \\ \phantom{-11} & \phantom{-0} & \phantom{-4} & \phantom{-6} \end{bmatrix}
Step 4
To determine the entry in the first row and fourth column of ABAB, multiply consecutive entries in the first row of AA and the fourth column of BB. Then add the products.
A=[−5−0−2−3−1−4]B=[3−40−22−11−2]A=\begin{bmatrix} {\color{#c42126}\phantom{-}5} & {\color{#c42126}\phantom{-}0 } \\ -2 & -3 \\ \phantom{-}1 & \phantom{-}4 \end{bmatrix} \hspace{20pt}B=\begin{bmatrix} 3 & \phantom{-}4 & 0 & {\color{#0047af}\phantom{-}2} \\ 2 & -1 & 1 & {\color{#0047af}-2} \end{bmatrix}
5⋅2+0⋅(−2)=10{\color{#c42126}5}\cdot{\color{#0047af}2}+{\color{#c42126}0}\cdot ({\color{#0047af}-2})=10
AB=[15200−10−12−5−3−2−11−0−4−6]AB=\begin{bmatrix} 15 & 20 & 0 & \phantom{-}10 \\ \phantom{-12} & \phantom{-5} & \phantom{-3} & \phantom{-2} \\ \phantom{-11} & \phantom{-0} & \phantom{-4} & \phantom{-6} \end{bmatrix}
Step 5
Repeat the process to determine each entry in the second row. Multiply consecutive entries in the second row of AA and entries in each column of BB. Then add the products.
A=[−5−0−2−3−1−4]B=[3−40−22−11−2]A=\begin{bmatrix} \phantom{-}5 & \phantom{-}0 \\ {\color{#c42126}-2} & {\color{#c42126}-3} \\ \phantom{-}1 & \phantom{-}4 \end{bmatrix} \hspace{20pt}B=\begin{bmatrix} 3 & \phantom{-}4 & 0 & \phantom{-}2 \\ 2 & -1 & 1 & -2 \end{bmatrix}
Row 2, Column 1:−2⋅3+(−3)⋅2=−12Row 2, Column 2:−2⋅4+(−3)⋅(−1)=−5Row 2, Column 3:−2⋅0+(−3)⋅1=−3Row 2, Column 4:−2⋅2+(−3)⋅(−2)=2\begin{aligned}\text{Row 2, Column 1:}\;\hspace{10pt}&{\color{#c42126}-2}\cdot3+{\color{#c42126}(-3)}\cdot 2=-12\\\text{Row 2, Column 2:}\hspace{10pt}&{\color{#c42126}-2}\cdot4+{\color{#c42126}(-3)}\cdot (-1)=-5\\\text{Row 2, Column 3:}\hspace{10pt}&{\color{#c42126}-2}\cdot0+{\color{#c42126}(-3)}\cdot 1=-3\\\text{Row 2, Column 4:}\hspace{10pt}&{\color{#c42126}-2}\cdot2+{\color{#c42126}(-3)}\cdot (-2)=2\end{aligned}
AB=[−1520−0−10−12−5−3−2−11−0−4−6]AB=\begin{bmatrix} \phantom{-}15 & 20 & \phantom{-}0 & \phantom{-}10 \\ -12 & -5 & -3 & \phantom{-}2 \\ \phantom{-11} & \phantom{-0} & \phantom{-4} & \phantom{-6} \end{bmatrix}
Solution
To determine each entry in the third row, multiply consecutive entries in the third row of AA and entries in each column of BB. Then add the products.
A=[−5−0−2−3−1−4]B=[3−40−22−11−2]A=\begin{bmatrix} \phantom{-}5 & \phantom{-}0 \\ -2 & -3 \\ {\color{#c42126}\phantom{-}1} & {\color{#c42126}\phantom{-}4} \end{bmatrix}\hspace{20pt} B=\begin{bmatrix} 3 & \phantom{-}4 & 0 & \phantom{-}2 \\ 2 & -1 & 1 & -2 \end{bmatrix}
Row 3, Column 1:1⋅3+4⋅2=11Row 3, Column 2:1⋅4+4⋅(−1)=0Row 3, Column 3:1⋅0+4⋅1=4Row 3, Column 4:1⋅2+4⋅(−2)=−6\begin{aligned}\text{Row 3, Column 1:}\; \hspace{10pt}&{\color{#c42126}1}\cdot3+{\color{#c42126}4}\cdot 2=11\\\text{Row 3, Column 2:}\hspace{10pt}&{\color{#c42126}1}\cdot4+{\color{#c42126}4}\cdot (-1)=0\\\text{Row 3, Column 3:} \hspace{10pt}&{\color{#c42126}1}\cdot0+{\color{#c42126}4}\cdot 1=4\\\text{Row 3, Column 4:} \hspace{10pt}&{\color{#c42126}1}\cdot2+{\color{#c42126}4}\cdot (-2)=-6\end{aligned}
AB=[−1520−0−10−12−5−3−2−11−0−4−6]AB=\begin{bmatrix} \phantom{-}15 & 20 & \phantom{-}0 & \phantom{-}10 \\ -12 & -5 & -3 & \phantom{-}2 \\ \phantom{-}11 & \phantom{-}0 & \phantom{-}4 & -6 \end{bmatrix}

Inverses

If a matrix has an inverse, the product of the matrix and its inverse is the identity matrix.

The numbers 2 and 12\frac{1}{2} are multiplicative inverses because when multiplied, the result is the multiplicative identity 1. Square matrices may also have multiplicative inverses. A square matrix is a matrix that has the same number of rows and columns.

An identity matrix is a square matrix in which the entries (or numbers) along the main diagonal, from the top left corner to the bottom right corner of the matrix, are ones and the other entries are zeros. The 3×33\times3 identity matrix has an entry of 1 in positions a1,1a_{1,1}, a2,2a_{2,2}, and a3,3a_{3,3}. All other entries are zero.
I=[100010001]I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
The inverse matrix of a square matrix AA, if the inverse exists, is the matrix A−1A^{-1}, such that AA−1=A−1A=IAA^{-1} = A^{-1}A = I, where II is the identity matrix.

Sometimes, it is useful to determine the inverse of a matrix. This is because there is no such thing as matrix division. With numbers, it is possible to multiply by the reciprocal of the number and get the same result as dividing by that number. Similarly, multiplying by the inverse of a matrix is equivalent to dividing by the matrix.

Step-By-Step Example
Multiplying a Matrix by the Identity Matrix
Compare the matrix product AIAI with IAIA.
A=[−204−323−212]A= \begin{bmatrix} \phantom{-}2 & 0 & 4 \\ -3 & 2 & 3 \\ -2 & 1 & 2 \end{bmatrix}
Step 1
Write the appropriate identity matrix II. Matrix AA is a 3×33\times3 matrix, so use the 3×33\times3 identity matrix.
I=[100010001]I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
Step 2
Determine AIAI.
AI=[−204−323−212][100010001]AI= \begin{bmatrix} \phantom{-}2 & 0 & 4 \\ -3 & 2 & 3 \\ -2 & 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
Determine the entries for the first row. Multiply each entry in row 1 of matrix\text{{\color{#c42126}{row 1 of matrix}}} A{\color{#c42126}{A}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} I{\color{#0047af}{I}}. Then add the products.
−2⋅1+0⋅0+4⋅0=22⋅0+0⋅1+4⋅0=02⋅0+0⋅0+4⋅1=4\begin{aligned}{\color{#c42126}\phantom{-}2}\cdot{\color{#0047af}1}+{\color{#c42126}0}\cdot {\color{#0047af}0}+{\color{#c42126}4}\cdot {\color{#0047af}0}=2\\{\color{#c42126}2}\cdot{\color{#0047af}0}+{\color{#c42126}0}\cdot {\color{#0047af}1}+{\color{#c42126}4}\cdot {\color{#0047af}0}=0\\{\color{#c42126}2}\cdot{\color{#0047af}0}+{\color{#c42126}0}\cdot {\color{#0047af}0}+{\color{#c42126}4}\cdot {\color{#0047af}1}=4\end{aligned}
Determine the entries for the second row. Multiply each entry in row 2 of matrix\text{{\color{#c42126}{row 2 of matrix}}} A{\color{#c42126}{A}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} I{\color{#0047af}{I}}. Then add the products.
(−3)⋅1+2⋅0+3⋅0=−3(−3)⋅0+2⋅1+3⋅0=2(−3)⋅0+2⋅0+3⋅1=3\begin{gathered}{\color{#c42126}(-3)}\cdot{\color{#0047af}1}+{\color{#c42126}2}\cdot {\color{#0047af}0}+{\color{#c42126}3}\cdot {\color{#0047af}0}=-3\\{\color{#c42126}(-3)}\cdot{\color{#0047af}0}+{\color{#c42126}2}\cdot {\color{#0047af}1}+{\color{#c42126}3}\cdot {\color{#0047af}0}=2\\{\color{#c42126}(-3)}\cdot{\color{#0047af}0}+{\color{#c42126}2}\cdot {\color{#0047af}0}+{\color{#c42126}3}\cdot {\color{#0047af}1}=3\end{gathered}
Determine the entries for the third row. Multiply each entry in row 3 of matrix\text{{\color{#c42126}{row 3 of matrix}}} A{\color{#c42126}{A}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} I{\color{#0047af}{I}}. Then add the products.
(−2)⋅1+1⋅0+2⋅0=−2(−2)⋅0+1⋅1+2⋅0=1(−2)⋅0+1⋅0+2⋅1=2\begin{gathered}{\color{#c42126}(-2)}\cdot{\color{#0047af}1}+{\color{#c42126}1}\cdot {\color{#0047af}0}+{\color{#c42126}2}\cdot {\color{#0047af}0}=-2\\{\color{#c42126}(-2)}\cdot{\color{#0047af}0}+{\color{#c42126}1}\cdot {\color{#0047af}1}+{\color{#c42126}2}\cdot {\color{#0047af}0}=1\\{\color{#c42126}(-2)}\cdot{\color{#0047af}0}+{\color{#c42126}1}\cdot {\color{#0047af}0}+{\color{#c42126}2}\cdot {\color{#0047af}1}=2\end{gathered}
AI=[−204−323−212]AI= \begin{bmatrix} \phantom{-}2 & 0 & 4 \\ -3 & 2 & 3 \\ -2 & 1 & 2 \end{bmatrix}
Step 3
Determine IAIA.
IA=[100010001][−204−323−212]IA= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \phantom{-}2 & 0 & 4 \\ -3 & 2 & 3 \\ -2 & 1 & 2 \end{bmatrix}
Calculate the entries for the first row. Multiply each entry in row 1 of matrix\text{{\color{#c42126}{row 1 of matrix}}} I{\color{#c42126}{I}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} A{\color{#0047af}{A}}. Then add the products.
1⋅2+0⋅(−3)+0⋅(−2)=21⋅0+0⋅2+0⋅1=01⋅4+0⋅3+0⋅2=4\begin{gathered}{\color{#c42126}1}\cdot{\color{#0047af}2}+{\color{#c42126}0}\cdot {\color{#0047af}(-3)}+{\color{#c42126}0}\cdot {\color{#0047af}(-2)}=2\\{\color{#c42126}1}\cdot{\color{#0047af}0}+{\color{#c42126}0}\cdot {\color{#0047af}2}+{\color{#c42126}0}\cdot {\color{#0047af}1}=0\\{\color{#c42126}1}\cdot{\color{#0047af}4}+{\color{#c42126}0}\cdot {\color{#0047af}3}+{\color{#c42126}0}\cdot {\color{#0047af}2}=4\end{gathered}
Calculate the entries for the second row. Multiply each entry in row 2 of matrix\text{{\color{#c42126}{row 2 of matrix}}} I{\color{#c42126}{I}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} A{\color{#0047af}{A}}. Then add the products.
0⋅2+1⋅(−3)+0⋅(−2)=−30⋅0+1⋅2+0⋅1=20⋅4+1⋅3+0⋅2=3\begin{gathered}{\color{#c42126}0}\cdot{\color{#0047af}2}+{\color{#c42126}1}\cdot {\color{#0047af}(-3)}+{\color{#c42126}0}\cdot {\color{#0047af}(-2)}=-3\\{\color{#c42126}0}\cdot{\color{#0047af}0}+{\color{#c42126}1}\cdot {\color{#0047af}2}+{\color{#c42126}0}\cdot {\color{#0047af}1}=2\\{\color{#c42126}0}\cdot{\color{#0047af}4}+{\color{#c42126}1}\cdot {\color{#0047af}3}+{\color{#c42126}0}\cdot {\color{#0047af}2}=3\end{gathered}
Calculate the entries for the third row. Multiply each entry in row 3 of matrix\text{{\color{#c42126}{row 3 of matrix}}} I{\color{#c42126}{I}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} A{\color{#0047af}{A}}. Then add the products.
0⋅2+0⋅(−3)+1⋅(−2)=−20⋅0+0⋅2+1⋅1=10⋅4+0⋅3+1⋅2=2\begin{gathered}{\color{#c42126}0}\cdot{\color{#0047af}2}+{\color{#c42126}0}\cdot {\color{#0047af}(-3)}+{\color{#c42126}1}\cdot {\color{#0047af}(-2)}=-2\\{\color{#c42126}0}\cdot{\color{#0047af}0}+{\color{#c42126}0}\cdot {\color{#0047af}2}+{\color{#c42126}1}\cdot {\color{#0047af}1}=1\\{\color{#c42126}0}\cdot{\color{#0047af}4}+{\color{#c42126}0}\cdot {\color{#0047af}3}+{\color{#c42126}1}\cdot {\color{#0047af}2}=2\end{gathered}
IA=[−204−323−212]IA= \begin{bmatrix} \phantom{-}2 & 0 & 4 \\ -3 & 2 & 3 \\ -2 & 1 & 2 \end{bmatrix}
Solution
AI=IA=AAI=IA=A
Step-By-Step Example
Verifying the Inverse of a Matrix
Verify that A−1A^{-1} is the inverse of AA.
A=[−204−323−212]A−1=[16−23−430−2−316−13−23]A= \begin{bmatrix} \phantom{-}2 & 0 & 4 \\ -3 & 2 & 3 \\ -2 & 1 & 2 \end{bmatrix}\hspace{30pt}A^{-1}=\begin{bmatrix}\frac{1}{6}&\phantom{-}\frac{2}{3}&-\frac{4}{3}\\[0.5em]0&\phantom{-}2&-3\\[0.5em] \frac{1}{6}&-\frac{1}{3}&\phantom{-}\frac{2}{3} \end{bmatrix}
Step 1
Determine AA−1AA^{-1}.
AA−1=[−204−323−212][16−23−430−2−316−13−23]AA^{-1}= \begin{bmatrix} \phantom{-}2 & 0 & 4 \\ -3 & 2 & 3 \\ -2 & 1 & 2 \end{bmatrix} \begin{bmatrix} \frac{1}{6}&\phantom{-}\frac{2}{3}&-\frac{4}{3}\\[0.5em]0&\phantom{-}2&-3\\[0.5em] \frac{1}{6}&-\frac{1}{3}&\phantom{-}\frac{2}{3} \end{bmatrix}
Determine the entries for the first row. Multiply each entry in row 1 of matrix\text{{\color{#c42126}{row 1 of matrix}}} A{\color{#c42126}{A}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} A−1{\color{#0047af}{A^{-1}}}. Then add the products.
2⋅16+0⋅0+4⋅16=12⋅23+0⋅2+4⋅(−13)=02⋅(−43)+0⋅(−3)+4⋅23=0\begin{gathered}{\color{#c42126}2}\cdot{\color{#0047af}\frac{1}{6}}+{\color{#c42126}0}\cdot {\color{#0047af}0}+{\color{#c42126}4}\cdot {\color{#0047af}\frac{1}{6}}=1\\{\color{#c42126}2}\cdot{\color{#0047af}\frac{2}{3}}+{\color{#c42126}0}\cdot {\color{#0047af}2}+{\color{#c42126}4}\cdot {\color{#0047af}\left (-\frac{1}{3} \right )}=0\\{\color{#c42126}2}\cdot{\color{#0047af}\left (-\frac{4}{3} \right )}+{\color{#c42126}0}\cdot {\color{#0047af}(-3)}+{\color{#c42126}4}\cdot {\color{#0047af}\frac{2}{3}}=0\end{gathered}
Calculate the entries for the second row. Multiply each entry in row 2 of matrix\text{{\color{#c42126}{row 2 of matrix}}} A{\color{#c42126}{A}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} A−1{\color{#0047af}{A^{-1}}}. Then add the products.
(−3)⋅16+2⋅0+3⋅16=0(−3)⋅23+2⋅2+3⋅(−13)=1(−3)⋅(−43)+2⋅(−3)+3⋅23=0\begin{gathered}{\color{#c42126}(-3)}\cdot{\color{#0047af}\frac{1}{6}}+{\color{#c42126}2}\cdot {\color{#0047af}0}+{\color{#c42126}3}\cdot {\color{#0047af}\frac{1}{6}}=0\\{\color{#c42126}(-3)}\cdot{\color{#0047af}\frac{2}{3}}+{\color{#c42126}2}\cdot {\color{#0047af}2}+{\color{#c42126}3}\cdot {\color{#0047af}\left (-\frac{1}{3} \right )}=1\\{\color{#c42126}(-3)}\cdot{\color{#0047af}\left (-\frac{4}{3} \right )}+{\color{#c42126}2}\cdot {\color{#0047af}\left (-3 \right )}+{\color{#c42126}3}\cdot {\color{#0047af}\frac{2}{3}}=0\end{gathered}
Calculate the entries for the third row. Multiply each entry in row 3 of matrix\text{{\color{#c42126}{row 3 of matrix}}} A{\color{#c42126}{A}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} A−1{\color{#0047af}{A^{-1}}}. Then add the products.
(−2)⋅16+1⋅0+2⋅16=0(−2)⋅23+1⋅2+2⋅(−13)=0(−2)⋅(−43)+1⋅(−3)+2⋅23=1\begin{gathered}{\color{#c42126}(-2)}\cdot{\color{#0047af}\frac{1}{6}}+{\color{#c42126}1}\cdot {\color{#0047af}0}+{\color{#c42126}2}\cdot {\color{#0047af}\frac{1}{6}}=0\\{\color{#c42126}(-2)}\cdot{\color{#0047af}\frac{2}{3}}+{\color{#c42126}1}\cdot {\color{#0047af}2}+{\color{#c42126}2}\cdot {\color{#0047af}\left (-\frac{1}{3} \right )}=0\\{\color{#c42126}(-2)}\cdot{\color{#0047af}\left (-\frac{4}{3} \right )}+{\color{#c42126}1}\cdot {\color{#0047af}\left (-3 \right )}+{\color{#c42126}2}\cdot {\color{#0047af}\frac{2}{3}}=1\end{gathered}
AA−1=[100010001]\begin{aligned}AA^{-1}&=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\end{aligned}
Step 2
Identify A−1AA^{-1}A.
A−1A=[16−23−430−2−316−13−23][−204−323−212]A^{-1}A= \begin{bmatrix} \frac{1}{6}&\phantom{-}\frac{2}{3}&-\frac{4}{3}\\[0.5em]0&\phantom{-}2&-3\\[0.5em] \frac{1}{6}&-\frac{1}{3}&\phantom{-}\frac{2}{3} \end{bmatrix}\begin{bmatrix} \phantom{-}2 & 0 & 4 \\ -3 & 2 & 3 \\ -2 & 1 & 2 \end{bmatrix}
Determine the entries for the first row. Multiply each entry in row 1 of matrix\text{{\color{#c42126}{row 1 of matrix}}} A−1{\color{#c42126}{A^{-1}}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} A{\color{#0047af}{A}}. Then add the products.
16⋅2+23⋅(−3)+(−43)⋅(−2)=116⋅0+23⋅2+(−43)⋅1=016⋅4+23⋅3+(−43)⋅2=0\begin{gathered}{\color{#c42126}\frac{1}{6}}\cdot{\color{#0047af}2}+{\color{#c42126}\frac{2}{3}}\cdot {\color{#0047af}(-3)}+{\color{#c42126}\left (-\frac{4}{3} \right )}\cdot {\color{#0047af}(-2)}=1\\{\color{#c42126}\frac{1}{6}}\cdot{\color{#0047af}0}+{\color{#c42126}\frac{2}{3}}\cdot {\color{#0047af}2}+{\color{#c42126}\left (-\frac{4}{3} \right )}\cdot {\color{#0047af}1}=0\\{\color{#c42126}\frac{1}{6}}\cdot{\color{#0047af}4}+{\color{#c42126}\frac{2}{3}}\cdot {\color{#0047af}3}+{\color{#c42126}\left (-\frac{4}{3} \right )}\cdot {\color{#0047af}2}=0\end{gathered}
Determine the entries for the second row. Multiply each entry in row 2 of matrix\text{{\color{#c42126}{row 2 of matrix}}} A−1{\color{#c42126}{A^{-1}}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} A{\color{#0047af}{A}}. Then add the products.
0⋅2+2⋅(−3)+(−3)⋅(−2)=00⋅0+2⋅2+(−3)⋅1=10⋅4+2⋅3+(−3)⋅2=0\begin{gathered}{\color{#c42126}0}\cdot{\color{#0047af}2}+{\color{#c42126}2}\cdot {\color{#0047af}(-3)}+{\color{#c42126}(-3)}\cdot {\color{#0047af}(-2)}=0\\{\color{#c42126}0}\cdot{\color{#0047af}0}+{\color{#c42126}2}\cdot {\color{#0047af}2}+{\color{#c42126}(-3)}\cdot {\color{#0047af}1}=1\\{\color{#c42126}0}\cdot{\color{#0047af}4}+{\color{#c42126}2}\cdot {\color{#0047af}3}+{\color{#c42126}(-3)}\cdot {\color{#0047af}2}=0\end{gathered}
Calculate the entries for the third row. Multiply each entry in row 3 of matrix\text{{\color{#c42126}{row 3 of matrix}}} A−1{\color{#c42126}{A^{-1}}} by each entry in all columns of matrix\text{{\color{#0047af}{columns of matrix}}} A{\color{#0047af}{A}}. Then add the products.
16⋅2+(−13)⋅(−3)+23⋅(−2)=016⋅0+(−13)⋅2+23⋅1=016⋅4+(−13)⋅3+23⋅2=1\begin{gathered}{\color{#c42126}\frac{1}{6}}\cdot{\color{#0047af}2}+{\color{#c42126}\left (-\frac{1}{3} \right )}\cdot {\color{#0047af}(-3)}+{\color{#c42126}\frac{2}{3}}\cdot {\color{#0047af}(-2)}=0\\{\color{#c42126}\frac{1}{6}}\cdot{\color{#0047af}0}+{\color{#c42126}\left (-\frac{1}{3} \right )}\cdot {\color{#0047af}2}+{\color{#c42126}\frac{2}{3}}\cdot {\color{#0047af}1}=0\\{\color{#c42126}\frac{1}{6}}\cdot{\color{#0047af}4}+{\color{#c42126}\left (-\frac{1}{3} \right )}\cdot {\color{#0047af}3}+{\color{#c42126}\frac{2}{3}}\cdot {\color{#0047af}2}=1\end{gathered}