Chapter 11.4, Problem 32AYU

To determine

To find: The inverse of a nonsingular matrix.

Answer to Problem 32AYU

Solution:

  $\left[\begin{array}{ll}1\hfill & 1\hfill \\ 2\hfill & 3\hfill \end{array}\right]$

Explanation of Solution

Given:

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

Calculation:

since $\text{A}$ is a non singular matrix, invers e $A^{-1}$ exists.

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

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

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

  

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

  $A^{-1}=\left[\begin{array}{ll}1\hfill & 1\hfill \\ 2\hfill & 3\hfill \end{array}\right]$

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

  $A^{-1}A=\left[\begin{array}{ll}1\hfill & 1\hfill \\ 2\hfill & 3\hfill \end{array}\right]\left[\begin{array}{cc}3& -1\\ -2& 1\end{array}\right]$$\begin{array}{l}=\left[\begin{array}{ll}3-2\hfill & -1+1\hfill \\ 6-6\hfill & -2+3\hfill \end{array}\right]\hfill \\ =\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]\text{ which is }{I}_2\hfill \end{array}$

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