Chapter 2.6, Problem 29E

To determine

To find: inverse of given matrix.

Answer to Problem 29E

  $\left[\begin{array}{cc}\begin{array}{l}\frac{2}{7}\\ \frac{3}{7}\end{array}& \begin{array}{l}\frac{-1}{7}\\ \frac{2}{7}\end{array}\end{array}\right]$

Explanation of Solution

Given information:

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

Calculation:

The inverse of $2\times 2$ matrix is evaluated as follows:

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

Put the values in the above formula and evaluate the inverse of the given matrix as follows:

  $\begin{array}{c}{\left[\begin{array}{cc}\begin{array}{l}2\\ -3\end{array}& \begin{array}{l}1\\ 2\end{array}\end{array}\right]}^{-1}=\frac{1}{\left(2\right)\left(2\right)-\left(1\right)\left(-3\right)}\left[\begin{array}{cc}\begin{array}{l}2\\ 3\end{array}& \begin{array}{l}-1\\ 2\end{array}\end{array}\right]\\ =\frac{1}{4+3}\left[\begin{array}{cc}\begin{array}{l}2\\ 3\end{array}& \begin{array}{l}-1\\ 2\end{array}\end{array}\right]\\ =\frac{1}{7}\left[\begin{array}{cc}\begin{array}{l}2\\ 3\end{array}& \begin{array}{l}-1\\ 2\end{array}\end{array}\right]\\ =\left[\begin{array}{cc}\begin{array}{l}\frac{2}{7}\\ \frac{3}{7}\end{array}& \begin{array}{l}\frac{-1}{7}\\ \frac{2}{7}\end{array}\end{array}\right]\end{array}$