Q25. - Algebra 2

Question

Find the inverse of each matrix, if it exists.25.

$\left[\begin{array}{cc}-3& 7\\ 2& -6\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the matrix is $\left[\begin{array}{cc}-\frac{3}{2}& -\frac{7}{4}\\ -\frac{1}{2}& -\frac{3}{4}\end{array}\right]$

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$\left[\begin{array}{cc}-3& 7\\ 2& -6\end{array}\right]$

That is,$A=\left[\begin{array}{cc}-3& 7\\ 2& -6\end{array}\right]$

Comparing with the standard form $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ , $a=-3,b=7,c=2,d=-6$ .

 

Then, $A^{-1}$  is:

$\begin{array}{c}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\\ =\frac{1}{(-3\times -6)-(7\times 2)}\left[\begin{array}{cc}-6& -7\\ -2& -3\end{array}\right]\\ =\frac{1}{18-14}\left[\begin{array}{cc}-6& -7\\ -2& -3\end{array}\right]\\ =\frac{1}{4}\left[\begin{array}{cc}-6& -7\\ -2& -3\end{array}\right]\end{array}$

Here, $\begin{array}{c}ad-bc=18-14\\ =4\end{array}$

As  $ad-bc\ne 0$ here, inverse exist.

 

Then, $A^{-1}$  is:

$\begin{array}{c}{A}^{-1}=\frac{1}{4}\left[\begin{array}{cc}-6& -7\\ -2& -3\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{6}{4}& -\frac{7}{4}\\ -\frac{2}{4}& -\frac{3}{4}\end{array}\right]\\ =\left[\begin{array}{cc}-\frac{3}{2}& -\frac{7}{4}\\ -\frac{1}{2}& -\frac{3}{4}\end{array}\right]\end{array}$

Step 3 - State the conclusion.

Therefore, the inverse of the matrix $\left[\begin{array}{cc}-3& 7\\ 2& -6\end{array}\right]$is$\left[\begin{array}{cc}-\frac{3}{2}& -\frac{7}{4}\\ -\frac{1}{2}& -\frac{3}{4}\end{array}\right]$