# Compute the adjoint of each of the following matrices:

Question:

Compute the adjoint of each of the following matrices:

(i) $\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$

(ii) $\left[\begin{array}{ccc}1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1\end{array}\right]$

(iii) $\left[\begin{array}{ccc}2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1\end{array}\right]$

(iv) $\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3\end{array}\right]$

Verify that $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$ for the above matrices.

Solution:

$(\mathrm{i}) A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$

Now,

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

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

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

$\therefore \operatorname{adj} A=\left[\begin{array}{ccc}-3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3\end{array}\right]^{T}=\left[\begin{array}{ccc}-3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3\end{array}\right]$

and $A(\operatorname{adj} A)=\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]$

Thus, $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$

$($ ii $) B=\left[\begin{array}{ccc}1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{ll}3 & 1 \\ 1 & 1\end{array}\right|=2, C_{12}=-\left|\begin{array}{cc}2 & 1 \\ -1 & 1\end{array}\right|=-3$ and $C_{13}=\left|\begin{array}{cc}2 & 3 \\ -1 & 1\end{array}\right|=5$

$C_{21}=-\left|\begin{array}{ll}2 & 5 \\ 1 & 1\end{array}\right|=3, C_{22}=\left|\begin{array}{cc}1 & 5 \\ -1 & 1\end{array}\right|=6$ and $C_{23}=-\left|\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right|=-3$

$C_{31}=\left|\begin{array}{ll}2 & 5 \\ 3 & 1\end{array}\right|=-13, C_{32}=-\left|\begin{array}{ll}1 & 5 \\ 2 & 1\end{array}\right|=9$ and $C_{33}=\left|\begin{array}{cc}1 & 2 \\ 2 & 3\end{array}\right|=-1$

$\therefore \operatorname{adj} B=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 6 & -3 \\ -13 & 9 & -1\end{array}\right]^{T}=\left[\begin{array}{ccc}2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1\end{array}\right]$

$(\operatorname{adj} B) B=\left[\begin{array}{ccc}21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21\end{array}\right]$

and $|B|=21$

$\therefore|B| I=\left[\begin{array}{ccc}21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21\end{array}\right]$

and $B(\operatorname{adj} B)=\left[\begin{array}{ccc}21 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 21\end{array}\right]$

Thus, $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$

$($ iii $) C=\left[\begin{array}{ccc}2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{cc}2 & 5 \\ 4 & -1\end{array}\right|=-22, C_{12}=-\left|\begin{array}{cc}4 & 5 \\ 0 & -1\end{array}\right|=4$ and $C_{13}=\left|\begin{array}{cc}4 & 2 \\ 0 & 4\end{array}\right|=16$

$C_{21}=-\left|\begin{array}{cc}-1 & 3 \\ 4 & -1\end{array}\right|=11, C_{22}=\left|\begin{array}{cc}2 & 3 \\ 0 & -1\end{array}\right|=-2$ and $C_{23}=-\left|\begin{array}{cc}2 & -1 \\ 0 & 4\end{array}\right|=-8$

$C_{31}=\left|\begin{array}{cc}-1 & 3 \\ 2 & 5\end{array}\right|=-11, C_{32}=-\left|\begin{array}{cc}2 & 3 \\ 4 & 5\end{array}\right|=2$ and $C_{33}=\left|\begin{array}{cc}2 & -1 \\ 4 & 2\end{array}\right|=8$

$\therefore \operatorname{adj} C=\left[\begin{array}{ccc}-22 & 4 & 16 \\ 11 & -2 & -8 \\ -11 & 2 & 8\end{array}\right]^{T}=\left[\begin{array}{ccc}-22 & 11 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8\end{array}\right]$

$(\operatorname{adj} C) C=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

and $|C|=0$

$\therefore|C| I=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$

Thus, $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$

$($ iv $) D=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{ll}1 & 0 \\ 1 & 3\end{array}\right|=3, C_{12}=-\left|\begin{array}{ll}5 & 0 \\ 1 & 3\end{array}\right|=-15$ and $C_{13}=\left|\begin{array}{ll}5 & 1 \\ 1 & 1\end{array}\right|=4$

$C_{21}=-\left|\begin{array}{cc}0 & -1 \\ 1 & 3\end{array}\right|=-1, C_{22}=\left|\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right|=7$ and $C_{23}=-\left|\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right|=-2$

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

$\therefore \operatorname{adj} D=\left[\begin{array}{ccc}3 & -15 & 4 \\ -1 & 7 & -2 \\ 1 & -5 & 2\end{array}\right]^{T}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 7 & -5 \\ 4 & -2 & 2\end{array}\right]$

$(\operatorname{adj} D) D=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$

and $|D|=2$

$\therefore|D| I=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$

and $D(\operatorname{adj} D)=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$

Thus, $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$