Q6. - Algebra 2

Question

Find the inverse of each matrix, if it exists.

6.

$\left[\begin{array}{cc}\mathbf{8}& \mathbf{-}\mathbf{5}\\ \mathbf{-}\mathbf{3}& \mathbf{2}\\ & \end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the given matrix is:

$\left[\begin{array}{cc}\mathbf{2}& \mathbf{5}\\ \mathbf{3}& \mathbf{8}\\ & \end{array}\right]$

Step 1 ­- 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 $\left|A\right|\ne 0$.

Step 2 ­- Find the determinant of the given matrix.

The determinant of the given matrix is:

$\begin{array}{c}\left|\begin{array}{cc}8& -5\\ -3& 2\end{array}\right|=\left(8\right)\left(2\right)-\left(-3\right)\left(-5\right)\\ =16-15\\ =1\end{array}$

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:

$\begin{array}{c}\frac{1}{\left(8\right)\left(2\right)-\left(-3\right)\left(-5\right)}\left[\begin{array}{cc}2& -\left(-5\right)\\ -\left(-3\right)& 8\end{array}\right]=\frac{1}{16-15}\left[\begin{array}{cc}2& 5\\ 3& 8\end{array}\right]\\ =\frac{1}{1}\left[\begin{array}{cc}2& 5\\ 3& 8\end{array}\right]\\ =\left[\begin{array}{cc}2& 5\\ 3& 8\end{array}\right]\end{array}$

Therefore, the inverse of the given matrix is $\left[\begin{array}{cc}2& 5\\ 3& 8\\ & \end{array}\right]$

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