Chapter 8, Problem 75RE

To determine

To find the inverse of the matrix $\left[\begin{array}{l}-72\\ -82\end{array}\right]$ .

Answer to Problem 75RE

  $A^{-1}=\left[\begin{array}{l}1-1\\ 4-\frac{7}{2}\end{array}\right]$

Explanation of Solution

Given:

  $\left[\begin{array}{l}-72\\ -82\end{array}\right]$

Concept used:

Let $A=\left[\begin{array}{l}ab\\ cd\end{array}\right]$ be a $2\times 2$ matrix. Then A is invertible if and only if $ad-bc\ne 0$ and the inverse is given by,

  $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{l}d-b\\ -ca\end{array}\right]$

Calculation:

Let $A=\left[\begin{array}{l}-72\\ -82\end{array}\right]$ .

Comparing with $A=\left[\begin{array}{l}ab\\ cd\end{array}\right]$ , we get

  $\begin{array}{l}a=-7,b=2,c=-8\&d=2\\ \therefore ad-bc=-7\left(2\right)-2\left(-8\right)\\ =-14+16\\ =2\ne 0\end{array}$

Therefore matrix A is invertible and the inverse is given by,

  $\begin{array}{l}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{l}d-b\\ -ca\end{array}\right]\\ \Rightarrow A^{-1}=\frac{1}{2}\left[\begin{array}{l}2-\left(2\right)\\ -\left(-8\right)-7\end{array}\right]\\ \Rightarrow A^{-1}=\frac{1}{2}\left[\begin{array}{l}2-2\\ 8-7\end{array}\right]\\ \Rightarrow A^{-1}=\left[\begin{array}{l}\frac{1}{2}\left(2\right)\frac{1}{2}\left(-2\right)\\ \frac{1}{2}\left(8\right)\frac{1}{2}\left(-7\right)\end{array}\right]\\ \Rightarrow A^{-1}=\left[\begin{array}{l}1-1\\ 4-\frac{7}{2}\end{array}\right]\end{array}$