Q30. - Algebra 2

Question

Find the inverse of each matrix, if it exists.30.

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

Step-by-Step Solution

Answer

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

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

That is,   $A=\left[\begin{array}{cc}\frac{1}{2}& -\frac{3}{4}\\ \frac{1}{6}& \frac{1}{4}\end{array}\right]$

Comparing with the standard form $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$a=\frac{1}{2},b=-\frac{3}{4},c=\frac{1}{6},d=\frac{1}{4}$

Then,  $A^{-1}$ is:

$\begin{array}{c}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\\ =\frac{1}{(\frac{1}{2}\times \frac{1}{4})-(-\frac{3}{4}\times \frac{1}{6})}\left[\begin{array}{cc}\frac{1}{4}& -(-\frac{3}{4})\\ -\frac{1}{6}& \frac{1}{2}\end{array}\right]\\ =\frac{1}{\frac{1}{8}+\frac{3}{24}}\left[\begin{array}{cc}\frac{1}{4}& \frac{3}{4}\\ -\frac{1}{6}& \frac{1}{2}\end{array}\right]\end{array}$

Here,   

$\begin{array}{c}ad-bc=\frac{1}{8}+\frac{3}{24}\\ =\frac{3+3}{24}\\ =\frac{6}{24}\\ =\frac{1}{4}\end{array}$

As $ad-bc\ne 0$  here, inverse exist.

 

Then, $A^{-1}$  is:

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

Step 3 - State the conclusion.

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