# For the matrix

Question:

For the matrix $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{array}\right]$, show that $A(\operatorname{adj} A)=0$.

Solution:

$A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{cc}3 & 0 \\ 2 & 10\end{array}\right|=30 \quad C_{12}=-\left|\begin{array}{cc}2 & 0 \\ 18 & 10\end{array}\right|=-20 \quad C_{13}=\left|\begin{array}{cc}2 & 3 \\ 18 & 2\end{array}\right|=-50$

$C_{21}=-\left|\begin{array}{cc}-1 & 1 \\ 2 & 10\end{array}\right|=12 \quad C_{22}=\left|\begin{array}{cc}1 & 1 \\ 18 & 10\end{array}\right|=-8 \quad C_{23}=-\left|\begin{array}{cc}1 & -1 \\ 18 & 2\end{array}\right|=-20$

$C_{31}=\left|\begin{array}{cc}-1 & 1 \\ 3 & 0\end{array}\right|=-3 \quad C_{32}=-\left|\begin{array}{ll}1 & 1 \\ 2 & 0\end{array}\right|=2 \quad C_{33}=\left|\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right|=5$

$\operatorname{adj} A=\left[\begin{array}{ccc}30 & -20 & -50 \\ 12 & -8 & -20 \\ -3 & 2 & 5\end{array}\right]^{T}=\left[\begin{array}{ccc}30 & 12 & -3 \\ -20 & -8 & 2 \\ -50 & -20 & 5\end{array}\right]$

$\therefore A(\operatorname{adj} A)=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{array}\right]\left[\begin{array}{ccc}30 & 12 & -3 \\ -20 & -8 & 2 \\ -50 & -20 & 5\end{array}\right]=\left[\begin{array}{ccc}30+20-50 & 12+18-20 & -3-2+5 \\ 60-60-0 & 24-24-0 & -6+6+0 \\ 540-40-500 & 216-16-200 & -54+4+50\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$