Question

If $A$ is a square matrix of order $n\times n$, then adj(adj A) is equal to

• A. ${\left|A\right|}^{n}A$
• B. ${\left|A\right|}^{n-1}A$
• C. ${\left|A\right|}^{n-2}A$
• D. ${\left|A\right|}^{n-3}A$

Answer 1

The correct option is C ${\left|A\right|}^{n-2}A$

For any square matrix $B$, we have
$B\left(adjB\right)=\left|B\right|I_n$
On taking $B=adjA$, we get
$\left(adjA\right)\left[adj\left(adjA\right)\right]=\left|adjA\right|I_n$
$\Rightarrow adjA\left[adj\left(adjA\right)\right]={\left|A\right|}^{n-1}{I}_n(\because \left|adjA\right|={\left|A\right|}^{n-1})$
$\Rightarrow \left(AadjA\right)\left[adj\left(adjA\right)\right]={\left|A\right|}^{n-1}A$
$\Rightarrow \left(\left|A\right|I_n\right)\left[adj\left(adjA\right)\right]={\left|A\right|}^{n-1}A$
$\Rightarrow adj\left(adjA\right)={\left|A\right|}^{n-2}A$