Question

Find the inverse of each matrix, if it exists.

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

Short answer

The inverse of the given matrix is:$\left[\begin{array}{cc}\frac{\mathbf{-}\mathbf{4}}{\mathbf{27}}& \frac{\mathbf{1}}{\mathbf{27}}\\ \frac{\mathbf{7}}{\mathbf{27}}& \frac{\mathbf{5}}{\mathbf{27}}\end{array}\right]$

Step-by-step solution

­- Description of step.

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

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

­- Find the determinant of the given matrix.

The determinant of the given matrix is:

$\begin{array}{c}\left|\begin{array}{cc}-5& 1\\ 7& 4\end{array}\right|=\left(-5\right)\left(4\right)-\left(7\right)\left(1\right)\\ =-20-7\\ =-27\end{array}$

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

­- Description of step. 

The inverse of $2\times 2$matrixlocalid="1648646974912" $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$is given by localid="1648646983272" $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$and $ad-bc\ne 0$.

­- Description of step.

The inverse of the given matrix is given by:

$\begin{array}{c}\frac{1}{\left(-5\right)\left(4\right)-\left(7\right)\left(1\right)}\left[\begin{array}{cc}4& -\left(1\right)\\ -\left(7\right)& -5\end{array}\right]=\frac{1}{-20-7}\left[\begin{array}{cc}4& -1\\ -7& -5\end{array}\right]\\ =\frac{1}{-27}\left[\begin{array}{cc}4& -1\\ -7& -5\end{array}\right]\\ =\left[\begin{array}{cc}\frac{4}{-27}& \frac{-1}{-27}\\ \frac{-7}{-27}& \frac{-5}{-27}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{-4}{27}& \frac{1}{27}\\ \frac{7}{27}& \frac{5}{27}\end{array}\right]\end{array}$

Therefore, the inverse of the given matrix is $\left[\begin{array}{cc}\frac{-4}{27}& \frac{1}{27}\\ \frac{7}{27}& \frac{5}{27}\end{array}\right]$.