Question

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

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

Answer 1

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

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

$\left|A\right|=-2+15=13\ne 0$

Hence $A^{-1}$ exists.

$C_{ij}={(-1)}^{i+j}{M}_{ij}$

$C_{11}=2,C_{12}=3,C_{21}=-5$ and $C_{22}=-1$

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

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

$A^{-1}=\frac{Adj\left(A\right)}{\left|A\right|}=\frac{1}{13}\left[\begin{array}{cc}2& -5\\ 3& -1\end{array}\right]$