Q23. - Algebra 2

Question

Find the inverse of each matrix, if it exists.23.

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

Step-by-Step Solution

Answer

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

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

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

  Comparing with the standard form $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ , $a=3,b=1,c=-4,d=1$ .Then, $A^{-1}$  is:

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

Here, $\begin{array}{c}ad-bc=3+4\\ =7\end{array}$

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

 Then,  $A^{-1}$ is:

Step 3 - State the conclusion.

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