Chapter 8, Problem 87RE

To determine

Find the inverse of the given matrix.

Answer to Problem 87RE

The inverse of the matrix $A$ is $\left[\begin{array}{cc}2& \frac{20}{3}\\ \frac{1}{10}& \frac{1}{6}\end{array}\right]$ .

Explanation of Solution

Given information:

The given matrix is

  $\left[\begin{array}{cc}-\frac{1}{2}& 20\\ \frac{3}{10}& -6\end{array}\right]$ .

Calculation:

If $A$ is $2\times 2$ matrix

  $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Then $A$ is invertible if and only if $ad-bc\ne 0$ . Moreover, $ad-bc\ne 0$ , then the

  $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

Let $A=\left[\begin{array}{cc}-\frac{1}{2}& 20\\ \frac{3}{10}& -6\end{array}\right]$

  $\left|\begin{array}{cc}-\frac{1}{2}& 20\\ \frac{3}{10}& -6\end{array}\right|=-\frac{1}{2}\cdot \left(-6\right)-\frac{3}{10}\cdot 20=3-6=-3$

  $\begin{array}{l}{A}^{-1}=-\frac{1}{3}\left[\begin{array}{cc}-6& -20\\ -\frac{3}{10}& -\frac{1}{2}\end{array}\right]\\ =\left[\begin{array}{cc}-6\cdot \left(-\frac{1}{3}\right)& -20\cdot \left(-\frac{1}{3}\right)\\ -\frac{3}{10}\cdot \left(-\frac{1}{3}\right)& -\frac{1}{2}\cdot \left(-\frac{1}{3}\right)\end{array}\right]\\ =\left[\begin{array}{cc}2& \frac{20}{3}\\ \frac{1}{10}& \frac{1}{6}\end{array}\right]\end{array}$

Hence the inverse of thematrix $A$ is $\left[\begin{array}{cc}2& \frac{20}{3}\\ \frac{1}{10}& \frac{1}{6}\end{array}\right]$ .