Chapter 11, Problem 11CT

To determine

The inverse of a nonsingular matrix $A=\left[\begin{array}{cc}3& 2\\ 5& 4\end{array}\right]$.

Answer to Problem 11CT

Solution:

The inverse of the matrix $A=\left[\begin{array}{cc}3& 2\\ 5& 4\end{array}\right]$ is $\left[\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right]$.

Explanation of Solution

Given information:

The matrix, $A=\left[\begin{array}{cc}3& 2\\ 5& 4\end{array}\right]$.

Explanation:

Consider a matrix, $A=\left[\begin{array}{cc}3& 2\\ 5& 4\end{array}\right]$.

To find inverse of matrix $A$:

Step 1: Form the matrix $\left[A|I_n\right]$.

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

Step 3: The reduced row echelon form of $\left[A|I_n\right]$ contain the identity matrix $I_n$ on the left of the vertical bar, the $n$ by $n$ matrix on the right vertical bar is an inverse of $A$.

Consider the matrix $\left[A|I_2\right]=\left[\begin{array}{cc}3& 2\\ 5& 4\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Use $R_1\to \frac{1}{3}{r}_1$,

$\left[A|I_2\right]=\left[\begin{array}{cc}1& \frac{2}{3}\\ 5& 4\end{array}|\begin{array}{cc}\frac{1}{3}& 0\\ 0& 1\end{array}\right]$

Use row operation $R_2\to -5r_1+r_2$,

$\left[A|I_2\right]=\left[\begin{array}{cc}1& \frac{2}{3}\\ 0& \frac{2}{3}\end{array}|\begin{array}{cc}\frac{1}{3}& 0\\ -\frac{5}{3}& 1\end{array}\right]$

Use $R_2\to \frac{3}{2}{r}_2$

$\left[A|I_2\right]=\left[\begin{array}{cc}1& \frac{2}{3}\\ 0& 1\end{array}|\begin{array}{cc}\frac{1}{3}& 0\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right]$

Use $R_1\to -\frac{2}{3}{r}_2+r_1$

$\left[A|I_2\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right]$

Thus, the inverse of the matrix $A$ is $\left[\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right]$.