Chapter 11.4, Problem 36AYU

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

Calculation:

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

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

  

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

  

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

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

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