Q20. - Algebra 2

Question

Find the inverse of each matrix, if it exists.20.

$\left[\begin{array}{cc}5& 0\\ 0& 1\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the given matrix is: $\left[\begin{array}{cc}\frac{1}{5}& 0\\ 0& 1\end{array}\right]$

Step 1 ­- Description of step.

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

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:

$\begin{array}{c}\left|\begin{array}{cc}5& 0\\ 0& 1\end{array}\right|=\left(5\right)\left(1\right)-\left(0\right)\left(0\right)\\ =5-0\\ =5\end{array}$

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:

$\begin{array}{c}\frac{1}{\left(5\right)\left(1\right)-\left(0\right)\left(0\right)}\left[\begin{array}{cc}1& -\left(0\right)\\ -\left(0\right)& 5\end{array}\right]=\frac{1}{5-0}\left[\begin{array}{cc}1& 0\\ 0& 5\end{array}\right]\\ =\frac{1}{5}\left[\begin{array}{cc}1& 0\\ 0& 5\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{5}& \frac{0}{5}\\ \frac{0}{5}& \frac{5}{5}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{5}& 0\\ 0& 1\end{array}\right]\end{array}$

Therefore, the inverse of the given matrix is $\left[\begin{array}{cc}\frac{1}{5}& 0\\ 0& 1\end{array}\right]$