Question

Inverse of $\left[\begin{array}{cc}3& 1\\ 5& 2\end{array}\right]$ is:

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

Answer 1

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

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

$\left|A\right|=6-5=1\ne 0$

$\therefore A^{-1}$ exists.

$C_{11}={(-1)}^{1+1}{M}_{11}={(-1)}^{2}2=2$

$C_{12}={(-1)}^{1+2}{M}_{12}={(-1)}^{3}5=-5$

$C_{13}={(-1)}^{1+3}{M}_{13}={(-1)}^{4}1=1$

$C_{14}={(-1)}^{1+4}{M}_{14}={(-1)}^{5}3=-3$

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

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

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

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

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