Q31. - Algebra 2

Question

Find the inverse of each matrix, if it exists.31.

$\left[\begin{array}{cc}\frac{3}{10}& \frac{5}{8}\\ \frac{1}{5}& \frac{3}{4}\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the given matrix is$\left[\begin{array}{cc}\frac{15}{2}& \mathbf{-}\frac{25}{4}\\ -2& 3\end{array}\right]$

Step 1 ­- Description of step.

A square matrixA does not have its inverse if .$\left|A\right|=0A$

A square matrixA have its inverse if .$\left|A\right|\ne 0$

Step 2 ­- Find the determinant of the given matrix.

The determinant of the given matrix is:

$\begin{array}{c}\left|\begin{array}{cc}\frac{3}{10}& \frac{5}{8}\\ \frac{1}{5}& \frac{3}{4}\end{array}\right|=\left(\frac{3}{10}\right)\left(\frac{3}{4}\right)-\left(\frac{1}{5}\right)\left(\frac{5}{8}\right)\\ =\frac{9}{40}-\frac{5}{40}\\ =\frac{9-5}{40}\\ =\frac{4}{40}\\ =\frac{1}{10}\end{array}$

As the determinant of the given matrix is not equal to zero, therefore the inverse of the given matrix exists. 

Step 3 ­- Description of step.

The inverse of$2\times 2$ matrix$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is given by$A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ and .$ad-bc\ne 0$

Step 4 ­- Description of step.

The inverse of the given matrix is given by:

$\begin{array}{c}\frac{1}{\left(\frac{3}{10}\right)\left(\frac{3}{4}\right)-\left(\frac{1}{5}\right)\left(\frac{5}{8}\right)}\left[\begin{array}{cc}\frac{3}{4}& -\left(\frac{5}{8}\right)\\ -\left(\frac{1}{5}\right)& \frac{3}{10}\end{array}\right]=\frac{1}{\frac{9}{40}-\frac{5}{40}}\left[\begin{array}{cc}\frac{3}{4}& -\frac{5}{8}\\ -\frac{1}{5}& \frac{3}{10}\end{array}\right]\\ =\frac{1}{\frac{4}{40}}\left[\begin{array}{cc}\frac{3}{4}& -\frac{5}{8}\\ -\frac{1}{5}& \frac{3}{10}\end{array}\right]\\ =\frac{1}{\frac{1}{10}}\left[\begin{array}{cc}\frac{3}{4}& -\frac{5}{8}\\ -\frac{1}{5}& \frac{3}{10}\end{array}\right]\\ =10\left[\begin{array}{cc}\frac{3}{4}& -\frac{5}{8}\\ -\frac{1}{5}& \frac{3}{10}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{30}{4}& -\frac{50}{8}\\ -\frac{10}{5}& \frac{30}{10}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{15}{2}& -\frac{25}{4}\\ -2& 3\end{array}\right]\end{array}$

Therefore, the inverse of the given matrix is $\left[\begin{array}{cc}\frac{15}{2}& -\frac{25}{4}\\ -2& 3\end{array}\right]$