Q26. - Algebra 2

Question

Find the inverse of each matrix, if it exists.26.

$\left[\begin{array}{cc}4& -3\\ 2& 7\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the matrix is $\left[\begin{array}{cc}\frac{7}{34}& \frac{3}{34}\\ -\frac{1}{17}& \frac{2}{17}\end{array}\right]$

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$\left[\begin{array}{cc}4& -3\\ 2& 7\end{array}\right]$

That is,$A=\left[\begin{array}{cc}4& -3\\ 2& 7\end{array}\right]$

Comparing with the standard form $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$a=4,b=-3,c=2,d=7$

Then,  $A^{-1}$ is:

$\begin{array}{c}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\\ =\frac{1}{(4\times 7)-(-3\times 2)}\left[\begin{array}{cc}7& -(-3)\\ -2& 4\end{array}\right]\\ =\frac{1}{28+6}\left[\begin{array}{cc}7& 3\\ -2& 4\end{array}\right]\\ =\frac{1}{34}\left[\begin{array}{cc}7& 3\\ -2& 4\end{array}\right]\end{array}$

Here, $\begin{array}{c}ad-bc=28+6\\ =34\end{array}$

As $ad-bc\ne 0$  here, inverse exist.

 

Then,  $A^{-1}$ is:

$\begin{array}{c}{A}^{-1}=\frac{1}{34}\left[\begin{array}{cc}7& 3\\ -2& 4\end{array}\right]\\ =\left[\begin{array}{cc}\frac{7}{34}& \frac{3}{34}\\ -\frac{2}{34}& \frac{4}{34}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{7}{34}& \frac{3}{34}\\ -\frac{1}{17}& \frac{2}{17}\end{array}\right]\end{array}$

Step 3 - State the conclusion.

Therefore, the inverse of the matrix $\left[\begin{array}{cc}4& -3\\ 2& 7\end{array}\right]$is$\left[\begin{array}{cc}\frac{7}{34}& \frac{3}{34}\\ -\frac{1}{17}& \frac{2}{17}\end{array}\right]$