Question

The inverse of the matrix $A=\left[\begin{array}{cc}-3& 5\\ 2& 1\end{array}\right]$ is

• A. $\left[\begin{array}{cc}\frac{5}{13}& \frac{-1}{13}\\ \frac{2}{13}& \frac{3}{13}\end{array}\right]$
• B. $\left[\begin{array}{cc}\frac{2}{13}& \frac{5}{13}\\ \frac{-1}{13}& \frac{3}{13}\end{array}\right]$
• C. $\left[\begin{array}{cc}\frac{-1}{13}& \frac{5}{13}\\ \frac{2}{13}& \frac{3}{13}\end{array}\right]$
• D. $\left[\begin{array}{cc}\frac{1}{13}& \frac{-5}{13}\\ \frac{2}{13}& \frac{3}{13}\end{array}\right]$
  

Answer 1

The correct option is C $\left[\begin{array}{cc}\frac{-1}{13}& \frac{5}{13}\\ \frac{2}{13}& \frac{3}{13}\end{array}\right]$

$A=\left[\begin{array}{cc}-3& 5\\ 2& 1\end{array}\right]$

$\left|A\right|=\left|\begin{array}{cc}-3& 5\\ 2& 1\end{array}\right|=-3-10=-13\ne 0\Rightarrow A^{-1}\text{exists}$
$\text{Now Cof}\left(A\right)=\left[\begin{array}{cc}1& -2\\ -5& -3\end{array}\right]$
$\text{So, Adj (A)}=\left[Cof\right(A){]}^T=\left[\begin{array}{cc}1& -5\\ -2& -3\end{array}\right]$
$\text{Hence}{A}^{-1}=\frac{adjA}{\left|A\right|}=\frac{-1}{13}\left[\begin{array}{cc}1& -5\\ -2& -3\end{array}\right]$
= $\left[\begin{array}{cc}\frac{-1}{13}& \frac{5}{13}\\ \frac{2}{13}& \frac{3}{13}\end{array}\right]$