Q27. - Algebra 2

Question

Find the inverse of each matrix, if it exists.27.

$\left[\begin{array}{cc}-2& 0\\ 5& 6\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the matrix is $\left[\begin{array}{cc}-\frac{1}{2}& 0\\ \frac{5}{12}& \frac{1}{6}\end{array}\right]$

Step 1 - Define inverse of a matrix.

For the matrix A ,

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

The inverse of matrix of the matrix  A is:

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

where $ad-bc\ne 0$ .  $ad-bc$  is the determinant of the matrix A .

 If $ad-bc=0$ , the inverse of a matrix doesn not exist.

Step 2 - Calculate the inverse.

Let   be the matrix$\left[\begin{array}{cc}-2& 0\\ 5& 6\end{array}\right]$

That is,$A=\left[\begin{array}{cc}-2& 0\\ 5& 6\end{array}\right]$

Comparing with the standard form $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ ,$a=-2,b=0,c=5,d=6$  .Then,  $A^{-1}$ is:

$\begin{array}{c}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\\ =\frac{1}{(-2\times 6)-(0\times 5)}\left[\begin{array}{cc}6& 0\\ -5& -2\end{array}\right]\\ =\frac{1}{-12-0}\left[\begin{array}{cc}6& 0\\ -5& -2\end{array}\right]\\ =\frac{1}{-12}\left[\begin{array}{cc}6& 0\\ -5& -2\end{array}\right]\end{array}$

Here,   $\begin{array}{c}ad-bc=-12-0\\ =-12\end{array}$

As  $ad-bc\ne 0$ here, inverse exist.

 Then,  $A^{-1}$ is:

$\begin{array}{c}{A}^{-1}=\frac{1}{-12}\left[\begin{array}{cc}6& 0\\ -5& -2\end{array}\right]\\ =\left[\begin{array}{cc}\frac{6}{-12}& \frac{0}{-12}\\ \frac{-5}{-12}& \frac{-2}{-12}\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{1}{2}& 0\\ \frac{5}{12}& \frac{1}{6}\end{array}\right]\end{array}$

Step 3 - State the conclusion.

Therefore, the inverse of the matrix $\left[\begin{array}{cc}-2& 0\\ 5& 6\end{array}\right]$is$\left[\begin{array}{cc}-\frac{1}{2}& 0\\ \frac{5}{12}& \frac{1}{6}\end{array}\right]$