Chapter 2.5, Problem 30E

To determine

To find: the inverse of given matrix, if it exists.

Answer to Problem 30E

  $\left[\begin{array}{cc}\begin{array}{l}\frac{1}{12}\\ \frac{1}{14}\end{array}& \begin{array}{l}-\frac{1}{12}\\ \frac{1}{14}\end{array}\end{array}\right]$

Explanation of Solution

Given information:

  $\left[\begin{array}{cc}\begin{array}{l}6\\ -6\end{array}& \begin{array}{l}7\\ 7\end{array}\end{array}\right]$

Calculation:

Check if the detreminant of the matrix is non-zero.

  $\begin{array}{c}\det =6\left(7\right)-\left(-6\right)\left(7\right)\\ =42+42\\ =84\end{array}$

Since the determinant is non-zero, so for matrix $A=\left[\begin{array}{cc}\begin{array}{l}a\\ c\end{array}& \begin{array}{l}b\\ d\end{array}\end{array}\right]$ the inverse of the matrix is as follows:

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

Thus, the inverse of given matrix will evaluated be as follows:

  $\begin{array}{c}{\left[\begin{array}{cc}\begin{array}{l}6\\ -6\end{array}& \begin{array}{l}7\\ 7\end{array}\end{array}\right]}^{-1}=\frac{1}{84}\left[\begin{array}{cc}\begin{array}{l}7\\ 6\end{array}& \begin{array}{l}-7\\ \text{ 6}\end{array}\end{array}\right]\\ =\left[\begin{array}{cc}\begin{array}{l}\frac{1}{12}\\ \frac{1}{14}\end{array}& \begin{array}{l}-\frac{1}{12}\\ \frac{1}{14}\end{array}\end{array}\right]\end{array}$