Chapter 8.3, Problem 20E

To determine

To find:the inverse of the matrix

Answer to Problem 20E

The inverse of the matrix is $\left[\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]$

Explanation of Solution

Given information: $A=\left[\begin{array}{ccc}1& 2& 2\\ 3& 7& 9\\ -1& -4& -7\end{array}\right]$

Concept Involved:

Definition of Matrix Multiplication:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $A=\left[b_{ij}\right]$ is an $n\times p$ matrix, then the product AB is an $m\times p$ matrix given by $A=\left[c_{ij}\right]$ , where $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\mathrm{....}+a_{in}{b}_{nj}$ .

The definition of matrix multiplication uses a row-by-column multiplication, where the entry in the $i^{th}$ row and $j^{th}$ column of the product AB is obtained by multiplying the entries in the $i^{th}$ row of A by the corresponding entries in the $j^{th}$ column of B and then adding the results.

So, for the product of two matrices to be defined, the number of columns of the first matrix must equal the number of rows of the second matrix. That is, the middletwo indices must be the same.

The outside two indices give the dimension of the product as show below

  $\begin{array}{l}\underset{\underset{\uparrow }{m}\times \underset{\uparrow }{n}}{A}\underset{\begin{array}{l}\\ \\ Equal\end{array}}{\times }\underset{\underset{\uparrow }{n}\times \underset{\uparrow }{p}}{B}=\underset{m\times p}{AB}\\ DimensionofAB\end{array}$

Definition of the Inverse of a Square Matrix:

Let A be $n\times n$ matrix and let $I_n$ be the $n\times n$ identity matrix. If there exists a matrix $A^{-1}$ such that $A{A}^{-1}=I_n=A^{-1}A$ then $A^{-1}$ is the inverse of A. The symbol $A^{-1}$ is read as “ $A$ inverse.”

To show that B is the inverse of A, we need to show that AB = I = BA.

Finding an Inverse Matrix:

Let A be a square matrix of dimension n × n.

1. Write the $n\times 2n$ matrix that consists of the given matrix A on the left andthe $n\times n$ identity matrix I on the right to obtain $\left[\begin{array}{ccc}A& ⋮& I\end{array}\right]$ .

2. If possible, row reduce A to I using elementary row operations on the entirematrix $\left[\begin{array}{ccc}A& ⋮& I\end{array}\right]$ . The result will be the matrix $\left[\begin{array}{ccc}I& ⋮& A^{-1}\end{array}\right]$ . If this is not possible, then A is not invertible.

3. Check your work by multiplying to see that $A{A}^{-1}=I=A^{-1}A$ .

Calculation:

Begin by adjoining the identity matrix to A to form the matrix

  $\left[\begin{array}{ccc}1& 2& 2\\ 3& 7& 9\\ -1& -4& -7\end{array}\begin{array}{c}⋮\\ ⋮\\ ⋮\end{array}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Use elementary row operations to obtain the form $\left[\begin{array}{ccc}I& ⋮& A^{-1}\end{array}\right]$

• Add -3 times the 1st row to the 2nd row
$\begin{array}{c}\\ -3R_1+R_2\to \\ \end{array}\left[\begin{array}{ccc}1& 2& 2\\ 0& 1& 3\\ -1& -4& -7\end{array}\begin{array}{c}⋮\\ ⋮\\ ⋮\end{array}\begin{array}{ccc}1& 0& 0\\ -3& 1& 0\\ 0& 0& 1\end{array}\right]$

• Add 1 times the 1st row to the 3rd row
$\begin{array}{c}\\ \\ 1R_1+R_3\to \end{array}\left[\begin{array}{ccc}1& 2& 2\\ 0& 1& 3\\ 0& -2& -5\end{array}\begin{array}{c}⋮\\ ⋮\\ ⋮\end{array}\begin{array}{ccc}1& 0& 0\\ -3& 1& 0\\ 1& 0& 1\end{array}\right]$

• Add 2 times the 2nd row to the 3rd row
$\begin{array}{c}\\ \\ 2R_2+R_3\to \end{array}\left[\begin{array}{ccc}1& 2& 2\\ 0& 1& 3\\ 0& 0& 1\end{array}\begin{array}{c}⋮\\ ⋮\\ ⋮\end{array}\begin{array}{ccc}1& 0& 0\\ -3& 1& 0\\ -5& 2& 1\end{array}\right]$

• Add -3 times the 3rd row to the 2nd row
$\begin{array}{c}\\ -3R_3+R_2\to \\ \end{array}\left[\begin{array}{ccc}1& 2& 2\\ 0& 1& 0\\ 0& 0& 1\end{array}\begin{array}{c}⋮\\ ⋮\\ ⋮\end{array}\begin{array}{ccc}1& 0& 0\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]$

• Add -2 times the 3rd row to the 1st row
$\begin{array}{c}-2R_3+R_1\to \\ \\ \end{array}\left[\begin{array}{ccc}1& 2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\begin{array}{c}⋮\\ ⋮\\ ⋮\end{array}\begin{array}{ccc}11& -4& -2\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]$

• Add -2 times the 2nd row to the 1st row
$\begin{array}{c}-2R_2+R_1\to \\ \\ \end{array}\left[\begin{array}{ccc}1& 2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\begin{array}{c}⋮\\ ⋮\\ ⋮\end{array}\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]$

Conclusion:

The inverse of the matrix $A=\left[\begin{array}{ccc}1& 2& 2\\ 3& 7& 9\\ -1& -4& -7\end{array}\right]$ is $A^{-1}=\left[\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]$