Chapter 7.6, Problem 31E

To determine

To calculate: the inverse of the matrix- $\left[\begin{array}{cc}\frac{7}{2}& -\frac{3}{4}\\ \frac{1}{5}& \frac{4}{5}\end{array}\right]$

Answer to Problem 31E

Inverse of the matrix is:

  $\left[\begin{array}{cc}\frac{16}{59}& \frac{15}{59}\\ -\frac{4}{59}& \frac{70}{59}\end{array}\right]$

Explanation of Solution

Given information:

The following matrix:

  $\left[\begin{array}{cc}\frac{7}{2}& -\frac{3}{4}\\ \frac{1}{5}& \frac{4}{5}\end{array}\right]$

Formula used:

Inverse of a $2\times 2$ matrix is given by,

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

when $ad-bc\ne 0$ .

Calculation:

For the given matrix, apply the formula for inverse of a $2\times 2$ matrix to obtain:

  $\begin{array}{l}ad-bc=\left(\frac{7}{2}\right)\left(\frac{4}{5}\right)-\left(-\frac{3}{4}\right)\left(\frac{1}{5}\right)\text{[Substitute]}\\ =\frac{14}{5}+\frac{3}{20}\text{[Multiply]}\\ =\frac{59}{20}\text{[Simplify]}\end{array}$

Because this quantity is not zero, the inverse is formed by interchanging the entries on the main diagonal, changing the signs of the other two entries, and multiplying by scalar $\frac{20}{59}$ as follows,

  $\begin{array}{l}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right][\text{Formula]}\\ =\frac{1}{\frac{59}{20}}\left[\begin{array}{cc}\frac{4}{5}& \frac{3}{4}\\ -\frac{1}{5}& \frac{7}{2}\end{array}\right]\text{[Substitute]}\\ \text{=}\frac{20}{59}\left[\begin{array}{cc}\frac{4}{5}& \frac{3}{4}\\ -\frac{1}{5}& \frac{7}{2}\end{array}\right][\text{Simplify]}\\ \text{=}\left[\begin{array}{cc}\frac{16}{59}& \frac{15}{59}\\ -\frac{4}{59}& \frac{70}{59}\end{array}\right]\left[\text{Multiply by scalar }\frac{20}{59}\right]\end{array}$

Hence, the inverse of the given matrix is-

  $\left[\begin{array}{cc}\frac{16}{59}& \frac{15}{59}\\ -\frac{4}{59}& \frac{70}{59}\end{array}\right]$