Question

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

Answer 1

We have,

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

Now, the determinant of the matrix $A$

$|A|=(6+1)=7$

Now, the Adjoint of the matrix $A$

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

We know that

$A^{-1}=\frac{1}{|A|}AdjA$

Therefore,

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

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

Hence, this is the answer.