Chapter 4, Problem 30SGR

To determine

To find: the inverse of the given matrix.

Answer to Problem 30SGR

  ${\text{A}}^{\text{-1}}\text{=}\left[\begin{array}{cc}\text{13}& 5\\ -5& -2\end{array}\right]$

Explanation of Solution

Given:

  $\left[\begin{array}{cc}2& 5\\ -5& -13\end{array}\right]$ .

Concept used:

Steps to solve the inverse of matrix by

Step1: calculating the matrix of minors.

Step2: turn that into the matrix of cofactors.

Step3: then ad jugate by transposing the cofactor.

Step4: last multiply by the $\frac{1}{\text{determinant}}$ .

Formula for inverse of matrix A is:

  ${\text{A}}^{\text{-1}}\text{=}\frac{\text{adjA}}{\left|\text{A}\right|}$ .

Calculation:

Here according to the formula:

  ${\text{A}}^{\text{-1}}\text{=}\frac{\text{adjA}}{\left|\text{A}\right|}$ .

Let $\text{A}$ = $\left[\begin{array}{cc}2& 5\\ -5& -13\end{array}\right]$ .

  $\text{adjA=}{\left[\begin{array}{cc}-13& 5\\ -5& 2\end{array}\right]}^{\text{T}}\text{=}\left[\begin{array}{cc}-13& \text{-5}\\ 5& 2\end{array}\right]$ .

  $\left|\text{A}\right|\text{=}\left|\text{-26+25}\right|\text{=-1}$ .

  ${\text{A}}^{\text{-1}}\text{=}\frac{\text{adjA}}{\left|\text{A}\right|}$

  ${\text{A}}^{\text{-1}}\text{=}\frac{\left[\begin{array}{cc}-13& \text{-5}\\ 5& 2\end{array}\right]}{-1}\text{=}\left[\begin{array}{cc}\text{13}& 5\\ -5& -2\end{array}\right]$ .

Hence ${\text{A}}^{\text{-1}}\text{=}\left[\begin{array}{cc}\text{13}& 5\\ -5& -2\end{array}\right]$ .