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Study Guides > College Algebra CoRequisite Course

Find the Inverse of a Matrix

Learning Outcomes

  • Verify that multiplying a matrix by its inverse results in 1.
  • Use matrix multiplication to find the inverse of a matrix.
  • Find an inverse by augmenting with an identity matrix.
We know that the multiplicative inverse of a real number aa is a−1{a}^{-1} and aa−1=a−1a=(1a)a=1a{a}^{-1}={a}^{-1}a=\left(\frac{1}{a}\right)a=1. For example, 2−1=12{2}^{-1}=\frac{1}{2} and (12)2=1\left(\frac{1}{2}\right)2=1. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix AA and its inverse A−1{A}^{-1} equals the identity matrix. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by In{I}_{n} where nn represents the dimension of the matrix. The equations below are the identity matrices for a 2×22\text{}\times \text{}2 matrix and 3×33\text{}\times \text{}3 matrix, respectively.

I2=[1001]{I}_{2}=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]

I3=[100010001]{I}_{3}=\left[\begin{array}{rrrrr}\hfill 1& \hfill & \hfill 0& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 1\end{array}\right]

The identity matrix acts as a 1 in matrix algebra. For example, AI=IA=AAI=IA=A. A matrix that has a multiplicative inverse has the properties

AA−1=IA−1A=I\begin{array}{l}A{A}^{-1}=I\\ {A}^{-1}A=I\end{array}

A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, AA−1=A−1A=IA{A}^{-1}={A}^{-1}A=I, is a requirement. Not all square matrices have an inverse, but if AA is invertible, then A−1{A}^{-1} is unique. We will look at two methods for finding the inverse of a 2×22\text{}\times \text{}2 matrix and a third method that can be used on both 2×22\text{}\times \text{}2 and 3×33\text{}\times \text{}3 matrices.

A General Note: The Identity Matrix and Multiplicative Inverse

The identity matrix, In{I}_{n}, is a square matrix containing ones down the main diagonal and zeros everywhere else.

I2=[1001]I3=[100010001]2×23×3\begin{array}{l}\begin{array}{l}\hfill \\ {I}_{2} & =\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]\begin{array}{cccc}& & & \end{array} & {I}_{3} & =\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]\hfill \\ & \text{}2\times 2 & & \text{3}\times 3\hfill \end{array}\hfill \end{array}

If AA is an n×nn\times n matrix and BB is an n×nn\times n matrix such that AB=BA=InAB=BA={I}_{n}, then B=A−1B={A}^{-1}, the multiplicative inverse of a matrix AA.

tip for success

Work through each of the examples below on paper, then apply the information in the General Note and How To boxes. Repeating this process a few times will help you to gain familiarity with the material. Remember, practice and patient effort enable understanding. Don't be discouraged if you don't "get it" right away!

Example: Showing That the Identity Matrix Acts as a 1

Given matrix A, show that AI=IA=AAI=IA=A. A=[34−25]A=\left[\begin{array}{cc}3& 4\\ -2& 5\end{array}\right]

Answer: Use matrix multiplication to show that the product of AA and the identity is equal to the product of the identity and A.

AI=[34−25][1001]=[3⋅1+4⋅03⋅0+4⋅1−2⋅1+5⋅0−2⋅0+5⋅1]=[34−25]AI=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]=\left[\begin{array}{rrrr}\hfill 3\cdot 1+4\cdot 0& \hfill & \hfill & \hfill 3\cdot 0+4\cdot 1\\ \hfill -2\cdot 1+5\cdot 0& \hfill & \hfill & \hfill -2\cdot 0+5\cdot 1\end{array}\right]=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right]

AI=[1001][34−25]=[1⋅3+0⋅(−2)1⋅4+0⋅50⋅3+1⋅(−2)0⋅4+1⋅5]=[34−25]AI=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right]=\left[\begin{array}{rrrr}\hfill 1\cdot 3+0\cdot \left(-2\right)& \hfill & \hfill & \hfill 1\cdot 4+0\cdot 5\\ \hfill 0\cdot 3+1\cdot \left(-2\right)& \hfill & \hfill & \hfill 0\cdot 4+1\cdot 5\end{array}\right]=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill 4\\ \hfill -2& \hfill & \hfill 5\end{array}\right]

How To: Given two matrices, show that one is the multiplicative inverse of the other

  1. Given matrix AA of order n×nn\times n and matrix BB of order n×nn\times n multiply ABAB.
  2. If AB=IAB=I, then find the product BABA. If BA=IBA=I, then B=A−1B={A}^{-1} and A=B−1A={B}^{-1}.

Example: Showing That Matrix A Is the Multiplicative Inverse of Matrix B

Show that the given matrices are multiplicative inverses of each other.

A=[15−2−9],B=[−9−521]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 5\\ \hfill -2& \hfill & \hfill -9\end{array}\right],B=\left[\begin{array}{rrr}\hfill -9& \hfill & \hfill -5\\ \hfill 2& \hfill & \hfill 1\end{array}\right]

Answer: Multiply ABAB and BABA. If both products equal the identity, then the two matrices are inverses of each other.

AB=[15−2−9]⋅[−9−521]=[1(−9)+5(2)1(−5)+5(1)−2(−9)−9(2)−2(−5)−9(1)]=[1001]\begin{array}{l}AB & =\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 5\\ \hfill -2& \hfill & \hfill -9\end{array}\right]\cdot \left[\begin{array}{rrr}\hfill -9& \hfill & \hfill -5\\ \hfill 2& \hfill & \hfill 1\end{array}\right]\hfill \\ & =\left[\begin{array}{rrr}\hfill 1\left(-9\right)+5\left(2\right)& \hfill & \hfill 1\left(-5\right)+5\left(1\right)\\ \hfill -2\left(-9\right)-9\left(2\right)& \hfill & \hfill -2\left(-5\right)-9\left(1\right)\end{array}\right]\hfill \\ & =\left[\begin{array}{ccc}1& & 0\\ 0& & 1\end{array}\right]\hfill \end{array}

BA=[−9−521]⋅[15−2−9]=[−9(1)−5(−2)−9(5)−5(−9)2(1)+1(−2)2(−5)+1(−9)]=[1001]\begin{array}{l}BA & =\left[\begin{array}{rrr}\hfill -9& \hfill & \hfill -5\\ \hfill 2& \hfill & \hfill 1\end{array}\right]\cdot \left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 5\\ \hfill -2& \hfill & \hfill -9\end{array}\right]\hfill \\ & =\left[\begin{array}{rrr}\hfill -9\left(1\right)-5\left(-2\right)& \hfill & \hfill -9\left(5\right)-5\left(-9\right)\\ \hfill 2\left(1\right)+1\left(-2\right)& \hfill & \hfill 2\left(-5\right)+1\left(-9\right)\end{array}\right]\hfill \\ & =\left[\begin{array}{ccc}1& & 0\\ 0& & 1\end{array}\right]\hfill \end{array}

AA and BB are inverses of each other.

Try It

Show that the following two matrices are inverses of each other.

A=[14−1−3],B=[−3−411]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right],B=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]

Answer:

AB=[14−1−3][−3−411]=[1(−3)+4(1)1(−4)+4(1)−1(−3)+−3(1)−1(−4)+−3(1)]=[1001]BA=[−3−411][14−1−3]=[−3(1)+−4(−1)−3(4)+−4(−3)1(1)+1(−1)1(4)+1(−3)]=[1001]\begin{array}{l}AB=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]=\left[\begin{array}{rrr}\hfill 1\left(-3\right)+4\left(1\right)& \hfill & \hfill 1\left(-4\right)+4\left(1\right)\\ \hfill -1\left(-3\right)+-3\left(1\right)& \hfill & \hfill -1\left(-4\right)+-3\left(1\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \\ BA=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]=\left[\begin{array}{rrr}\hfill -3\left(1\right)+-4\left(-1\right)& \hfill & \hfill -3\left(4\right)+-4\left(-3\right)\\ \hfill 1\left(1\right)+1\left(-1\right)& \hfill & \hfill 1\left(4\right)+1\left(-3\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \end{array}

tip for success

The following sections offer examples of various efficient techniques that show the power of matrix operations. Again, work through them slowly and carefully as needed to gain familiarity. These techniques, using matrix multiplication, augmenting a matrix with the identity then reducing, and using a formula to find the inverse matrix appear frequently in scientific application and are worth taking time to obtain.

Finding the Multiplicative Inverse Using Matrix Multiplication

We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication.

Example: Finding the Multiplicative Inverse Using Matrix Multiplication

Use matrix multiplication to find the inverse of the given matrix.

A=[1−22−3]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill -2\\ \hfill 2& \hfill & \hfill -3\end{array}\right]

Answer: For this method, we multiply AA by a matrix containing unknown constants and set it equal to the identity.

[1−22−3] [abcd]=[1001]\left[\begin{array}{rr}\hfill 1& \hfill -2\\ \hfill 2& \hfill -3\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill a& \hfill b\\ \hfill c& \hfill d\end{array}\right]=\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]

Find the product of the two matrices on the left side of the equal sign.

[1−22−3] [abcd]=[1a−2c1b−2d2a−3c2b−3d]\left[\begin{array}{rr}\hfill 1& \hfill -2\\ \hfill 2& \hfill -3\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill a& \hfill b\\ \hfill c& \hfill d\end{array}\right]=\left[\begin{array}{rr}\hfill 1a - 2c& \hfill 1b - 2d\\ \hfill 2a - 3c& \hfill 2b - 3d\end{array}\right]

Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.

1a−2c=1 R12a−3c=0 R2\begin{array}{c}1a - 2c=1\text{ }{R}_{1}\\ 2a - 3c=0\text{ }{R}_{2}\end{array}

Using row operations, multiply and add as follows: (−2)R1+R2→R2\left(-2\right){R}_{1}+{R}_{2}\to {R}_{2}. Add the equations, and solve for cc.

1a−2c=10+1c=−2c=−2\begin{array}{r}\hfill 1a - 2c=1\\ \hfill 0+1c=-2\\ \hfill c=-2\end{array}

Back-substitute to solve for aa.

a−2(−2)=1a+4=1a=−3\begin{array}{r}\hfill a - 2\left(-2\right)=1\\ \hfill a+4=1\\ \hfill a=-3\end{array}

Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.

1b−2d=0R12b−3d=1R2\begin{array}{rr}\hfill 1b - 2d=0& \hfill {R}_{1}\\ \hfill 2b - 3d=1& \hfill {R}_{2}\end{array}

Using row operations, multiply and add as follows: (−2)R1+R2=R2\left(-2\right){R}_{1}+{R}_{2}={R}_{2}. Add the two equations and solve for dd.

1b−2d=00+1d=1d=1\begin{array}{r}\hfill 1b - 2d=0\\ \hfill \frac{0+1d=1}{d=1}\\ \hfill \end{array}

Once more, back-substitute and solve for bb.

b−2(1)=0b−2=0b=2\begin{array}{r}\hfill b - 2\left(1\right)=0\\ \hfill b - 2=0\\ \hfill b=2\end{array}

A−1=[−32−21]{A}^{-1}=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill 2\\ \hfill -2& \hfill & \hfill 1\end{array}\right]

Finding the Multiplicative Inverse by Augmenting with the Identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix AA is transformed into II, the augmented matrix II transforms into A−1{A}^{-1}. For example, given

A=[2153]A=\left[\begin{array}{rrr}\hfill 2& \hfill & \hfill 1\\ \hfill 5& \hfill & \hfill 3\end{array}\right]

augment AA with the identity

[21105301]\left[\begin{array}{cc|cc}\hfill 2& \hfill 1& \hfill 1& \hfill 0\\ \hfill 5& \hfill 3& \hfill 0& \hfill 1\\ \end{array}\right]

Perform row operations with the goal of turning AA into the identity.
  1. Switch row 1 and row 2. [53012110]\left[\begin{array}{cc|cc}\hfill 5& \hfill 3& \hfill 0& \hfill 1\\ \hfill 2& \hfill 1& \hfill 1& \hfill 0\\ \end{array}\right]
  2. Multiply row 2 by −2-2 and add to row 1. [11−212110]\left[\begin{array}{cc|cc}\hfill 1& \hfill 1& \hfill -2& \hfill 1\\ \hfill 2& \hfill 1& \hfill 1& \hfill 0\\ \end{array}\right]
  3. Multiply row 1 by −2-2 and add to row 2. [11−210−15−2]\left[\begin{array}{cc|cc}\hfill 1& \hfill 1& \hfill -2& \hfill 1\\ \hfill 0& \hfill -1& \hfill 5& \hfill -2\\ \end{array}\right]
  4. Add row 2 to row 1. [103−10−15−2]\left[\begin{array}{cc|cc}\hfill 1& \hfill 0& \hfill 3& \hfill -1\\ \hfill 0& \hfill -1& \hfill 5& \hfill -2\\ \end{array}\right]
  5. Multiply row 2 by −1-1. [103−101−52]\left[\begin{array}{cc|cc}\hfill 1& \hfill 0& \hfill 3& \hfill -1\\ \hfill 0& \hfill 1& \hfill -5& \hfill 2\\ \end{array}\right]
The matrix we have found is A−1{A}^{-1}.

A−1=[3−1−52]{A}^{-1}=\left[\begin{array}{rrr}\hfill 3& \hfill & \hfill -1\\ \hfill -5& \hfill & \hfill 2\end{array}\right]

Finding the Multiplicative Inverse of 2×2 Matrices Using a Formula

When we need to find the multiplicative inverse of a 2×22\times 2 matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity. If AA is a 2×22\times 2 matrix, such as

A=[abcd]A=\left[\begin{array}{rrr}\hfill a& \hfill & \hfill b\\ \hfill c& \hfill & \hfill d\end{array}\right]

the multiplicative inverse of AA is given by the formula

A−1=1ad−bc[d−b−ca]{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{rrr}\hfill d& \hfill & \hfill -b\\ \hfill -c& \hfill & \hfill a\end{array}\right]

where ad−bc≠0ad-bc\ne 0. If ad−bc=0ad-bc=0, then AA has no inverse.

Example: Using the Formula to Find the Multiplicative Inverse of Matrix A

Use the formula to find the multiplicative inverse of

A=[1−22−3]A=\left[\begin{array}{cc}1& -2\\ 2& -3\end{array}\right]

Answer: Using the formula, we have

A−1=1(1)(−3)−(−2)(2)[−32−21]=1−3+4[−32−21]=[−32−21]\begin{array}{l}{A}^{-1} & =\frac{1}{\left(1\right)\left(-3\right)-\left(-2\right)\left(2\right)}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ & =\frac{1}{-3+4}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ & =\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \end{array}

Analysis of the Solution

We can check that our formula works by using one of the other methods to calculate the inverse. Let’s augment AA with the identity.

[1−2102−301]\left[\begin{array}{cc|cc}\hfill 1& \hfill -2& \hfill 1& \hfill 0\\ \hfill 2& \hfill -3& \hfill 0& \hfill 1\\ \end{array}\right]

Perform row operations with the goal of turning AA into the identity.
  1. Multiply row 1 by −2-2 and add to row 2. [1−21001−21]\left[\begin{array}{cc|cc}\hfill 1& \hfill -2& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill -2& \hfill 1\\ \end{array}\right]
  2. Multiply row 1 by 2 and add to row 1. [10−3201−21]\left[\begin{array}{cc|cc}\hfill 1& \hfill 0& \hfill -3& \hfill 2\\ \hfill 0& \hfill 1& \hfill -2& \hfill 1\\ \end{array}\right]
So, we have verified our original solution.

A−1=[−32−21]{A}^{-1}=\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]

Try It

Use the formula to find the inverse of matrix AA. Verify your answer by augmenting with the identity matrix.

A=[1−123]A=\left[\begin{array}{cc} \hfill 1& \hfill -1\\ \hfill 2& \hfill 3\end{array}\right]

Answer: A−1=[3515−2515]{A}^{-1}=\left[\begin{array}{cc} \hfill \frac{3}{5}& \hfill \frac{1}{5}\\ \hfill -\frac{2}{5}& \hfill \frac{1}{5}\end{array}\right]

[ohm_question]6363[/ohm_question]

Example: Finding the Inverse of the Matrix, If It Exists

Find the inverse, if it exists, of the given matrix.

A=[3612]A=\left[\begin{array}{cc}3& 6\\ 1& 2\end{array}\right]

Answer: We will use the method of augmenting with the identity.

[36101201]\left[\begin{array}{cc|cc}\hfill 3& \hfill 6& \hfill 1& \hfill 0\\ \hfill 1& \hfill 2& \hfill 0& \hfill 1\\ \end{array}\right]

  1. Switch row 1 and row 2. [12013610]\left[\begin{array}{cc|cc}\hfill 1& \hfill 2& \hfill 0& \hfill 1\\ \hfill 3& \hfill 6& \hfill 1& \hfill 0\\ \end{array}\right]
  2. Multiply row 1 by −3 and add it to row 2. [1210001−3]\left[\begin{array}{cc|cc}\hfill 1& \hfill 2& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill -3\\ \end{array}\right]
  3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

Finding the Multiplicative Inverse of 3×3 Matrices

Unfortunately, we do not have a formula similar to the one for a 2×22\text{}\times \text{}2 matrix to find the inverse of a 3×33\text{}\times \text{}3 matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse. Given a 3×33\text{}\times \text{}3 matrix

A=[231331241]A=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right]

augment AA with the identity matrix

A∣I=[231100331010241001]A|I=\left[\begin{array}{ccc|ccc}\hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 3& \hfill 3& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 2& \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 1\end{array}\right]

To begin, we write the augmented matrix with the identity on the right and AA on the left. Performing elementary row operations so that the identity matrix appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example.

How To: Given a 3×33\times 3 matrix, find the inverse

  1. Write the original matrix augmented with the identity matrix on the right.
  2. Use elementary row operations so that the identity appears on the left.
  3. What is obtained on the right is the inverse of the original matrix.
  4. Use matrix multiplication to show that AA−1=IA{A}^{-1}=I and A−1A=I{A}^{-1}A=I.

Example: Finding the Inverse of a 3 × 3 Matrix

Given the 3×33\times 3 matrix AA, find the inverse.

A=[231331241]A=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right]

Answer: Augment AA with the identity matrix, and then begin row operations until the identity matrix replaces AA. The matrix on the right will be the inverse of AA.

[231100331010241001]→Interchange R2 and R1[331010231100241001]\left[\begin{array}{ccc|ccc}\hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 3& \hfill 3& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 2& \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 1\end{array}\right]\stackrel{\text{Interchange }{R}_{2}\text{ and }{R}_{1}}{\to }\left[\begin{array}{ccc|ccc}\hfill 3& \hfill 3& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 2& \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 1\end{array}\right]

−R2+R1=R1→[100−110231100241001]-{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 2& \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 1\end{array}\right]

−R2+R3=R3→[100−110231100010−101]-{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1& \hfill 0& \hfill 1\end{array}\right]

R3↔R2→[100−110010−101231100]{R}_{3}\leftrightarrow {R}_{2}\to \left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1& \hfill 0& \hfill 1\\ \hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\end{array}\right]

−2R1+R3=R3→[100−110010−1010313−20]-2{R}_{1}+{R}_{3}={R}_{3}\to\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1& \hfill 0& \hfill 1\\ \hfill 0& \hfill 3& \hfill 1& \hfill 3& \hfill -2& \hfill 0\end{array}\right]

−3R2+R3=R3→[100−110010−1010016−2−3]-3{R}_{2}+{R}_{3}={R}_{3}\to\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1& \hfill 0& \hfill 1\\ \hfill 0& \hfill 0& \hfill 1& \hfill 6& \hfill -2& \hfill -3\end{array}\right]

Thus,

A−1=B=[−110−1016−2−3]{A}^{-1}=B=\left[\begin{array}{ccc}-1& 1& 0\\ -1& 0& 1\\ 6& -2& -3\end{array}\right]

Analysis of the Solution

To prove that B=A−1B={A}^{-1}, let’s multiply the two matrices together to see if the product equals the identity, if AA−1=IA{A}^{-1}=I and A−1A=I{A}^{-1}A=I.

AA−1=[231331241] [−110−1016−2−3]=[2(−1)+3(−1)+1(6)2(1)+3(0)+1(−2)2(0)+3(1)+1(−3)3(−1)+3(−1)+1(6)3(1)+3(0)+1(−2)3(0)+3(1)+1(−3)2(−1)+4(−1)+1(6)2(1)+4(0)+1(−2)2(0)+4(1)+1(−3)]=[100010001]\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\hfill \\ A{A}^{-1} & =\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 6& \hfill -2& \hfill -3\end{array}\right]\hfill \\ & =\left[\begin{array}{ccc}2\left(-1\right)+3\left(-1\right)+1\left(6\right)& 2\left(1\right)+3\left(0\right)+1\left(-2\right)& 2\left(0\right)+3\left(1\right)+1\left(-3\right)\\ 3\left(-1\right)+3\left(-1\right)+1\left(6\right)& 3\left(1\right)+3\left(0\right)+1\left(-2\right)& 3\left(0\right)+3\left(1\right)+1\left(-3\right)\\ 2\left(-1\right)+4\left(-1\right)+1\left(6\right)& 2\left(1\right)+4\left(0\right)+1\left(-2\right)& 2\left(0\right)+4\left(1\right)+1\left(-3\right)\end{array}\right]\hfill \\ & =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\hfill \end{array} A−1A=[−110−1016−2−3] [231331241]=[−1(2)+1(3)+0(2)−1(3)+1(3)+0(4)−1(1)+1(1)+0(1)−1(2)+0(3)+1(2)−1(3)+0(3)+1(4)−1(1)+0(1)+1(1)6(2)+−2(3)+−3(2)6(3)+−2(3)+−3(4)6(1)+−2(1)+−3(1)]=[100010001]\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\hfill \\ {A}^{-1}A & =\left[\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 6& \hfill -2& \hfill -3\end{array}\right]\text{ }\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right]\hfill \\ & =\left[\begin{array}{rrr}\hfill -1\left(2\right)+1\left(3\right)+0\left(2\right)& \hfill -1\left(3\right)+1\left(3\right)+0\left(4\right)& \hfill -1\left(1\right)+1\left(1\right)+0\left(1\right)\\ \hfill -1\left(2\right)+0\left(3\right)+1\left(2\right)& \hfill -1\left(3\right)+0\left(3\right)+1\left(4\right)& \hfill -1\left(1\right)+0\left(1\right)+1\left(1\right)\\ \hfill 6\left(2\right)+-2\left(3\right)+-3\left(2\right)& \hfill 6\left(3\right)+-2\left(3\right)+-3\left(4\right)& \hfill 6\left(1\right)+-2\left(1\right)+-3\left(1\right)\end{array}\right]\hfill \\ & =\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]\hfill \end{array}

Try It

Find the inverse of the 3×33\times 3 matrix.

A=[2−1711−111−703−2]A=\left[\begin{array}{ccc}\hfill 2&\hfill -17& \hfill 11\\ \hfill -1& \hfill 11& \hfill -7\\ \hfill 0& \hfill 3& \hfill -2\end{array}\right]

Answer: A−1=[11224−336−5]{A}^{-1}=\left[\begin{array}{ccc}\hfill 1& \hfill 1& \hfill 2\\ \hfill 2& \hfill 4& \hfill -3\\ \hfill 3& \hfill 6& \hfill -5\end{array}\right]

[ohm_question]127903[/ohm_question]

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