Question

If $A=\left[\begin{array}{cc}2& -2\\ 4& 3\end{array}\right]$, then find $A^{-1}$ by adjoint method

Answer 1

We have, $A=\left[\begin{array}{cc}2& -2\\ 4& 3\end{array}\right]$
Now, cofactor of $A{a}_{11}=(-1{)}^{1+1}|3|$
$=3$
$a_{12}=(-1{)}^{1+2}|4|$
$=-4$
$a_{21}=(-1{)}^{2+1}|-2|$
$=2$
$a_{22}=(-1{)}^{2+2}|2|$
$=2$
$\therefore adjA=\left[\begin{array}{cc}3& -4\\ 2& 2\end{array}\right]$
$=\left[\begin{array}{cc}3& 2\\ -4& 2\end{array}\right]$
Now, $|A|=\left[\begin{array}{cc}2& -2\\ 4& 3\end{array}\right]$
$=\left(2\right)\left(3\right)-\left(4\right)(-2)$
$=6+8$
$=14\ne 0$
$A$ is non-singular.
$\therefore A^{-1}$ exists $A^{-1}=\frac{adjA}{|A|}$
$=\frac{1}{14}\left[\begin{array}{cc}3& 2\\ -4& 2\end{array}\right]$
$=\left[\begin{array}{cc}\frac{3}{14}& \frac{2}{14}\\ -\frac{4}{14}& \frac{2}{14}\end{array}\right]=\left[\begin{array}{cc}\frac{3}{14}& \frac{1}{7}\\ -\frac{2}{7}& \frac{1}{7}\end{array}\right]$