Q22. - Algebra 2

Question

Find the inverse of each matrix, if it exists. 22.

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

Step-by-Step Solution

Answer

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

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

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

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

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

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$Here,$\begin{array}{c}ad-bc=1-4\\ =-3\end{array}$

 As  $ad-bc\ne 0$ here, inverse exist.

 

Then, $A^{-1}$  is:

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

Step 3 - State the conclusion.

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