Question
If $A=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$, then find $A^{-1}$ by the adjoint method.
✓
Answer
For given matrix $A,$ we get,$
M _{11}=3, A _{11}=(-1)^{1+1}(3)=3$
$M _{12}=4, A _{12}=(-1)^{1+2}(4)=-4$
$M _{21}=-2, A _{21}=(-1)^{2+1}(-2)=2$
$M _{22}=2, A _{22}=(-1)^{2+2}(2)=2 $
$ \therefore \text { adj } A =\left[\begin{array}{cc}3 & 2 \\ -4 & 2 \end{array}\right] $ and $| A |=\left[\begin{array}{cc} 2 & -2 \\ 4 & 3 \end{array}\right]=6+8=14 \neq 0 $
$\therefore $ using $ A^{-1}=\frac{1}{| A |}(\operatorname{adj} A ) $
$ A^{-1}=\frac{1}{14}\left[\begin{array}{cc} 3 & 2 \\ -4 & 2 \end{array}\right]$
M _{11}=3, A _{11}=(-1)^{1+1}(3)=3$
$M _{12}=4, A _{12}=(-1)^{1+2}(4)=-4$
$M _{21}=-2, A _{21}=(-1)^{2+1}(-2)=2$
$M _{22}=2, A _{22}=(-1)^{2+2}(2)=2 $
$ \therefore \text { adj } A =\left[\begin{array}{cc}3 & 2 \\ -4 & 2 \end{array}\right] $ and $| A |=\left[\begin{array}{cc} 2 & -2 \\ 4 & 3 \end{array}\right]=6+8=14 \neq 0 $
$\therefore $ using $ A^{-1}=\frac{1}{| A |}(\operatorname{adj} A ) $
$ A^{-1}=\frac{1}{14}\left[\begin{array}{cc} 3 & 2 \\ -4 & 2 \end{array}\right]$
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