Chapter 11, Problem 11CT

To determine

To find: The inverse of each nonsingular matrix.

Answer to Problem 11CT

The invers of nonsingular matrix is $\left(\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right)$

Explanation of Solution

Given information:

A = $\left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right)$

Calculation:

Calculate the matrix,

  $A=\left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right)$

Since $A$ is a 2 by 2 matrix, we have to first from the matrix $\left[A|I_2\right]$ where $I_2$ is second order identity matrix

  $\left[A|I_2\right]$ = $\left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right)$

  $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

In order to transform the matrix $\left[A|I_2\right]$ into reduced echelon from first perform the row operations

  $R_1=\frac{1}{3}{r}_1$ and then $R_2=r_2-5r_1$

  $\left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right)$

  $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$\to$$\left(\begin{array}{cc}1& \frac{2}{3}\\ 5& 4\end{array}\right)$

  $\left(\begin{array}{cc}\frac{1}{3}& 0\\ 0& 1\end{array}\right)$$\to$$\left(\begin{array}{cc}1& \frac{2}{3}\\ 0& \frac{2}{3}\end{array}\right)$

  $\left(\begin{array}{cc}\frac{1}{3}& 0\\ -\frac{5}{4}& 1\end{array}\right)$

Now, $R_1=r_1-r_2$ followed by $R_2=\frac{1}{3}{r}_2$

  $\left(\begin{array}{cc}1& \frac{2}{3}\\ 0& \frac{2}{3}\end{array}\right)$

  $\left(\begin{array}{cc}\frac{1}{3}& 0\\ -\frac{5}{4}& 1\end{array}\right)$$\to$$\left(\begin{array}{cc}1& 0\\ 0& \frac{2}{3}\end{array}\right)$

  $\left(\begin{array}{cc}2& -1\\ -\frac{5}{3}& 1\end{array}\right)$$\to$$\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$

  $$

  $\left(\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right)$

We can see that the reduced echelon from of $\left[A|I_2\right]$ contains the identity matrix $I_2$ on the left

Therefore, the inverse of the matrix is $\left(\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right)$