Chapter 11.4, Problem 38AYU

In Problems 35-44, each matrix is nonsingular. Find the inverse of each matrix.

$\left[\begin{array}{l}\begin{array}{ccc}1& 0& 2\end{array}\\ \begin{array}{ccc}-1& 2& 3\end{array}\\ \begin{array}{ccc}1& -1& 0\end{array}\end{array}\right]$

To determine

To find: The inverse of a nonsingular matrix.

Answer to Problem 38AYU

Solution:

$\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$

Explanation of Solution

Given:

$A=\left[\begin{array}{ccc}1& 0& 2\\ -1& 2& 3\\ 1& -1& 0\end{array}\right]$

Calculation:

Since $A$ is a non singular matrix, inverse $A^{-1}$ exists.

Step 1: Construction of $\left[A|I_n\right]$.

$\left[A|I_3\right] = [\begin{array}{ccc}1& 0& 2\\ -1& 2& 3\\ 1& -1& 0\end{array} | \begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

Step 2: Transform the matrix $\left[A|I_n\right]$ into reduced row echelon form.

$= [\begin{array}{ccc}1& 0& 2\\ -1& 2& 3\\ 1& -1& 0\end{array} | \begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

$\underset{\begin{array}{c}{R}_2=r_2+r_3\\ R_3= r_3-r_1\end{array}}{\to } [\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& -1& -2\end{array} | \begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ -1& 0& 1\end{array}]$

$\underset{R_3=r_3+r_2}{\to } [\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& 0& 1\end{array} | \begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ -1& 1& 2\end{array}]$

$\underset{\begin{array}{c}{R}_1=r_1-2r_3\\ R_2=r_2-3r_3\end{array}}{\to } [\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array} | \begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}]$

Step 3: Separate the matrix on the right of the vertical bar which is the inverse of $A$.

$A^{-1} =\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$

Step 4: Check $A^{-1}A=I_n$.

$A{A}^{-1} =\left[\begin{array}{ccc}1& 0& 2\\ -1& 2& 3\\ 1& -1& 0\end{array}\right]\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$

$=\left[\begin{array}{ccc}3-2& -2+2& -4+4\\ -3+6-3& 2-4+3& 4-10+6\\ 3-3& -2+2& -4+5\end{array}\right]$

$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ which is $I_3$.

Hence, the inverse of $\left[\begin{array}{ccc}1& 0& 2\\ -1& 2& 3\\ 1& -1& 0\end{array}\right]$ is $\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$