Chapter 8.3, Problem 17E

To determine

Tofind:the inverse of matrix if it’s exists.

Answer to Problem 17E

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

Explanation of Solution

Given:

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

Concept used:

Step1:Form the augmented matrix $\left[\text{A|I}\right]$ .

Step2: Perform row operation on $\left[\text{A|I}\right]$ to get the matrix of the form $\left[\text{I|B}\right]$ .

Calculation:

Here,

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

To find the inverse of A let’s form the augmented matrix $\left[\text{A|I}\right]$ as

  $\left[\begin{array}{cc}3& 1\\ 4& 2\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Now perform row operation on $\left[\text{A|I}\right]$ to get the matrix of the form $\left[\text{I|B}\right]$ as

  $\left(\frac{1}{3}\right)R_1\to \left[\begin{array}{cc}1& \frac{1}{3}\\ 4& 2\end{array}|\begin{array}{cc}\frac{1}{3}& 0\\ 0& 1\end{array}\right]$

  $-4R_1+R_2\to \left[\begin{array}{cc}1& \frac{1}{3}\\ 0& \frac{2}{3}\end{array}|\begin{array}{cc}\frac{1}{3}& 0\\ -\frac{4}{3}& 1\end{array}\right]$

  $\left(\frac{3}{2}\right)R_2\to \left[\begin{array}{cc}1& \frac{1}{3}\\ 0& 1\end{array}|\begin{array}{cc}\frac{1}{3}& 0\\ -2& \frac{3}{2}\end{array}\right]$

  $\text{-}\left(\frac{\text{3}}{\text{2}}\right){\text{R}}_{\text{2}}{\text{+R}}_{\text{1}}\left[\begin{array}{cc}\text{1}& \text{0}\\ \text{0}& \text{1}\end{array}|\begin{array}{cc}\text{1}& \text{-}\frac{\text{1}}{\text{2}}\\ \text{-2}& \frac{\text{3}}{\text{2}}\end{array}\right]\text{=}\left[\text{I|B}\right]$

Hence, the matrix B is the last transformation is the desired inverse matrix.

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