Question

Verify $A\left(adjA\right)=\left(adjA\right)A=|A|I$

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

Answer 1

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

Here, $|A|=-1(9-2)=-7$ (expanding along the second column)

$|A|I=-7I$ .....(1)

We know that, $adjA=C^T$

So, we will find out co-factors of each element of A.

$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}0& 2\\ 0& 3\end{array}\right|$

$\Rightarrow C_{11}=0$

$C_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}3& 2\\ 1& 3\end{array}\right|$

$\Rightarrow C_{12}=-(9-2)=-7$

$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}3& 0\\ 1& 0\end{array}\right|$

$\Rightarrow C_{13}=0$

$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}1& 2\\ 0& 3\end{array}\right|$

$\Rightarrow C_{21}=-(3-0)=-3$

$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& 2\\ 1& 3\end{array}\right|$

$\Rightarrow C_{22}=3-2=1$

$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& 1\\ 1& 0\end{array}\right|$

$\Rightarrow C_{23}=-(0-1)=1$

$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}1& 2\\ 0& 2\end{array}\right|$

$\Rightarrow C_{31}=2-0=2$

$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& 2\\ 3& 2\end{array}\right|$

$\Rightarrow C_{32}=-(2-6)=4$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& 1\\ 3& 0\end{array}\right|$

$\Rightarrow C_{33}=0-3=-3$

So, the cofactor matrix $C=\left[\begin{array}{ccc}0& -7& 0\\ -3& 1& 1\\ 2& 4& -3\end{array}\right]$

$\Rightarrow adjA=C^T=\left[\begin{array}{ccc}0& -3& 2\\ -7& 1& 4\\ 0& 1& -3\end{array}\right]$

Consider $\left(adjA\right)A=\left[\begin{array}{ccc}0& -3& 2\\ -7& 1& 4\\ 0& 1& -3\end{array}\right]\left[\begin{array}{ccc}1& 1& 2\\ 3& 0& 2\\ 1& 0& 3\end{array}\right]$

$=\left[\begin{array}{ccc}0-9+2& 0+0+0& 0-6+6\\ -7+3+4& -7+0+0& -14+2+12\\ 0+3-3& 0+0+0& 0+2-9\end{array}\right]$

$=\left[\begin{array}{ccc}-7& 0& 0\\ 0& -7& 0\\ 0& 0& -7\end{array}\right]$

$=-7\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow \left(adjA\right)A=-7I$ .....(2)

Now, $A\left(adjA\right)=\left[\begin{array}{ccc}1& 1& 2\\ 3& 0& 2\\ 1& 0& 3\end{array}\right]\left[\begin{array}{ccc}0& -3& 2\\ -7& 1& 4\\ 0& 1& -3\end{array}\right]$

$=\left[\begin{array}{ccc}0-7+0& -3+1+2& 2+4-6\\ 0+0+0& -9+0+2& 6+0-6\\ 0+0+0& -3+0+3& 2+0-9\end{array}\right]$

$=\left[\begin{array}{ccc}-7& 0& 0\\ 0& -7& 0\\ 0& 0& -7\end{array}\right]$

$=-7\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow A\left(adjA\right)=-7I$ ....(3)

From (1), (2) and (3), we get

$A\left(adjA\right)=\left(adjA\right)A=|A|I$