Question

If $A=\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]$, then

• A. $adj\left(adjA\right)=A$
• B. $|adj\left(adjA\right)|=1$
• C. $|adjA|=1$
• D. none of these

Answer 1

Answer: (A), (B), (C)

$adj\left(adjA\right)=A$
$|adj\left(adjA\right)|=1$
$|adjA|=1$

$A=\left|\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right|$

$\left|A\right|=3(-3+4)+3\left(2\right)+4(-2)$

$=3+6-8=1$

$\left|adjA\right|={\left|A\right|}^{n-1}$

$={\left(1\right)}^{3-1}=1$

$\left|adj\right(AdjA\left)\right|={\left|A\right|}^{{(n-1)}^2}$

$={\left(1\right)}^{{(3-1)}^2}=1$

$adj\left|adjA\right|={\left|A\right|}^{n-2}\times A$

$=1\times A=A$

Option A, B C are true