Q29. - Algebra 2

Question

Find the inverse of each matrix, if it exists.29.

$\left[\begin{array}{cc}2& -5\\ 6& 1\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the matrix is $\left[\begin{array}{cc}\frac{1}{32}& \frac{5}{32}\\ \frac{-3}{16}& \frac{1}{16}\end{array}\right]$

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   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  A be the matrix $\left[\begin{array}{cc}2& -5\\ 6& 1\end{array}\right]$

That is, $A=\left[\begin{array}{cc}2& -5\\ 6& 1\end{array}\right]$

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

Then, $A^{-1}$  is:

$\begin{array}{c}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\\ =\frac{1}{(2\times 1)-(-5\times 6)}\left[\begin{array}{cc}1& -(-5)\\ -6& 2\end{array}\right]\\ =\frac{1}{2+30}\left[\begin{array}{cc}1& 5\\ -6& 2\end{array}\right]\\ =\frac{1}{32}\left[\begin{array}{cc}1& 5\\ -6& 2\end{array}\right]\end{array}$

Here,  $\begin{array}{c}ad-bc=2+30\\ =32\end{array}$

As  $ad-bc\ne 0$ here, inverse exist.

 Then,  $A^{-1}$ is:

$\begin{array}{c}{A}^{-1}=\frac{1}{32}\left[\begin{array}{cc}1& 5\\ -6& 2\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{32}& \frac{5}{32}\\ \frac{-6}{32}& \frac{2}{32}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{32}& \frac{5}{32}\\ \frac{-3}{16}& \frac{1}{16}\end{array}\right]\end{array}$

Step 3 - State the conclusion.

Therefore, the inverse of the matrix $\left[\begin{array}{cc}2& -5\\ 6& 1\end{array}\right]$is$\left[\begin{array}{cc}\frac{1}{32}& \frac{5}{32}\\ \frac{-3}{16}& \frac{1}{16}\end{array}\right]$