# Solve this

Question:

If $A=\left[\begin{array}{ccc}-1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{array}\right]$, show that adj $A=3 A^{\top}$.

Solution:

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

Now,

$C_{11}=\left|\begin{array}{cc}1 & -2 \\ -2 & 1\end{array}\right|=-3, C_{12}=-\left|\begin{array}{cc}2 & -2 \\ 2 & 1\end{array}\right|=-6$ and $C_{13}=\left|\begin{array}{cc}2 & 1 \\ 2 & -2\end{array}\right|=-6$

$C_{21}=-\left|\begin{array}{cc}-2 & -2 \\ -2 & 1\end{array}\right|=6, C_{22}=\left|\begin{array}{cc}-1 & -2 \\ 2 & 1\end{array}\right|=3$ and $C_{23}=-\left|\begin{array}{cc}-1 & -2 \\ 2 & -2\end{array}\right|=-6$

$C_{31}=\left|\begin{array}{cc}-2 & -2 \\ 1 & -2\end{array}\right|=6, C_{32}=-\left|\begin{array}{cc}-1 & -2 \\ 2 & -2\end{array}\right|=-6$ and $C_{33}=\left|\begin{array}{cc}-1 & -2 \\ 2 & 1\end{array}\right|=3$

$\operatorname{adj} A=\left[\begin{array}{ccc}-3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3\end{array}\right]^{T}=\left[\begin{array}{ccc}-3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3\end{array}\right]$

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

$\Rightarrow 3 A^{T}=\left[\begin{array}{ccc}-3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3\end{array}\right]$

$\Rightarrow 3 A^{T}=\operatorname{adj} A$