# Find the adjoint of the matrix

Question:

Find the adjoint of the matrix $A=\left[\begin{array}{ccc}-1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{array}\right]$ and hence show that $A(\operatorname{adj} A)=|A| I_{3}$.

Solution:

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

Now, to find Adj. $A$

$A_{11}=(-1)^{1+1}(-3)=-3 \quad A_{21}=(-1)^{2+1}(-6)=6 \quad A_{31}=(-1)^{3+1}(6)=6$

$A_{12}=(-1)^{1+2}(6)=-6 \quad A_{22}=(-1)^{2+2}(3)=3 \quad A_{32}=(-1)^{3+2}(6)=-6$

$A_{13}=(-1)^{1+3}(-6)=-6 \quad A_{23}=(-1)^{2+3}(6)=-6 \quad A_{33}=(-1)^{3+3}(3)=3$

Therefore,

Adj. $A=\left[\begin{array}{ccc}-3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3\end{array}\right]$

Now,

$|A|=\left|\begin{array}{ccc}-1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{array}\right|$

$=-1(1-4)-2(-2-4)+2(4+2)$

$=3+12+12$

$=27$

To show : $A(\operatorname{adj} A)=|A| I_{3}$

$\mathrm{LHS}=$

$\left[\begin{array}{ccc}-1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{array}\right]\left[\begin{array}{ccc}-3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3\end{array}\right]$

$=\left[\begin{array}{ccc}3+12+12 & -6-6+12 & -6+12-6 \\ -6-6+12 & 12+3+12 & 12-6-6 \\ -6+12-6 & 12-6-6 & 12+12+3\end{array}\right]$

$=\left[\begin{array}{ccc}27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27\end{array}\right]$

$=27\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$

$=|A| I_{3}$

$=\mathrm{RHS}$

Hence, $A(\operatorname{adj} A)=|A| I_{3}$.