Chapter 11.4, Problem 36AYU

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

$\left[\begin{array}{l}\begin{array}{cc}b& 3\end{array}\\ \begin{array}{cc}b& 2\end{array}\end{array}\right]b\ne 0$

To determine

To find: The inverse of a nonsingular matrix.

Answer to Problem 36AYU

Solution:

$\left[\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]$

Explanation of Solution

Given:

A matrix $A=\left[\begin{array}{cc}b& 3\\ b& 2\end{array}\right]$

Calculation:

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

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

$\left[A|I_n\right]=[\begin{array}{cc}b& 3\\ b& 2\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}]$

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

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

$\underset{R_1=\frac{1}{b}*r_1}{\to }[\begin{array}{cc}1& \frac{3}{b}\\ b& 2\end{array}|\begin{array}{cc}\frac{1}{b}& 0\\ 0& 1\end{array}]$

$\underset{R_2=r_2-\left(b\right)r_1}{\to }[\begin{array}{cc}1& \frac{3}{b}\\ 0& -1\end{array}|\begin{array}{cc}\frac{1}{b}& 0\\ -1& 1\end{array}]$

$\underset{\begin{array}{c}{R}_1=r_1-\left(-\frac{3}{b}\right)r_2\\ R_2\to \left(-1\right)r_2\end{array}}{\to }[\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}]$

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

$A^{-1}=\left[\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]$

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

$A^{-1}A=\left[\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]\left[\begin{array}{cc}b& 3\\ b& 2\end{array}\right]$

$=\left[\begin{array}{cc}-2+3& -\frac{6}{b}+\frac{6}{b}\\ b-b& 3-2\end{array}\right]$

$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ which is $I_2$.

Hence, the inverse of $\left[\begin{array}{cc}b& 3\\ b& 2\end{array}\right]$ is $\left[\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]$.