Question

If $A=\left[\begin{array}{ccc}3& 2& 4\\ 1& 4& 1\\ 2& 6& 3\end{array}\right]$.

Verify that $A\left(\text{adj.}A\right)=\left(\text{adj.}A\right)A=|A|I$.

Answer 1

$A=\left[\begin{array}{ccc}3& 2& 4\\ 1& 4& 1\\ 2& 6& 3\end{array}\right]$ verity that $A(adj.A)=\left(adjA\right)A=|A|I$

$|A|=3(12-6)-2(3-2)+4(6-8)$

$|A|=18-2-8=8$

$|A|=8$

$adjA=(CFA{)}^T$

Let $c_{13}$ be the cotaelor of $9_{13}$ in $A$

$c_{11}={(-1)}^{1+1}\left|\begin{array}{cc}4& 1\\ 6& 3\end{array}\right|=6$

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

$c_{13}={(-1)}^{1+3}\left|\begin{array}{cc}1& 4\\ 2& 6\end{array}\right|=-1$

$c_{21}={(-1)}^{2+1}\left|\begin{array}{cc}2& 4\\ 6& 3\end{array}\right|=18$

$c_{22}={(-1)}^{2+2}\left|\begin{array}{cc}3& 4\\ 2& 3\end{array}\right|=-1$

$c_{23}={(-1)}^{2+3}\left|\begin{array}{cc}3& 2\\ 2& 6\end{array}\right|=-14$

$c_{31}={(-1)}^{3+1}\left|\begin{array}{cc}2& 4\\ 4& 1\end{array}\right|=-14$

$c_{32}={(-1)}^{3+2}\left|\begin{array}{cc}3& 4\\ 1& 1\end{array}\right|=1$

$c_{33}={(-1)}^{3+3}\left|\begin{array}{cc}3& 2\\ 1& 4\end{array}\right|=10$

$adjA={\left[\begin{array}{ccc}6& -1& -2\\ 18& +1& -14\\ -14& 1& 10\end{array}\right]}^T=\left[\begin{array}{ccc}6& 18& 14\\ -1& +1& 1\\ -2& -14& 10\end{array}\right]$

$A\left(adjA\right)=\left[\begin{array}{ccc}3& 2& 4\\ 14& 4& 1\\ 2& 6& 3\end{array}\right]=\left[\begin{array}{ccc}6& 18& 14\\ -1& +1& 1\\ -2& -14& 10\end{array}\right]$

$=\left[\begin{array}{ccc}18-2-8& 54+2-56& -42+2+40\\ 6-4-2& 18+4-14& -14+4+10\\ 12-6-6& 36+6-42& -28+6+30\end{array}\right]$

$A\left(adjA\right)=\left[\begin{array}{ccc}8& 0& 0\\ 0& 8& 0\\ 0& 0& 8\end{array}\right]$

$A\left(adjA\right)=\left[\begin{array}{ccc}6& 18& -14\\ -1& 1& 1\\ -2& -14& 10\end{array}\right]\left[\begin{array}{ccc}3& 2& 4\\ 1& 4& 1\\ 2& 6& 3\end{array}\right]$

$=\left[\begin{array}{ccc}8& 0& 0\\ 0& 8& 0\\ 0& 0& 8\end{array}\right]$

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