Question

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

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

Short answer

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

Step-by-step solution

- 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 thematrix A .

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

- 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:

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

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