Question

If $A$ is a non-singular square matrix of order $n,$ then $adj\left(adj\right(A\left)\right)$ is equal to

• A. $|A{|}^{n}A$
• B. $|A{|}^{n-1}$
• C. $|A{|}^{n-2}A$
• D. None of these

Answer 1

Answer: (C)

$|A{|}^{n-2}A$

We know that B.adjB$=\left(\left|B\right|I_n\right)$ for every square matrix B of order n

Replacing B by adjA we get

$\left(adj\left(A\right)\right)\left(adj\left(adj\left(A\right)\right)\right)=\left(\left|adj\left(A\right)\right|I_n\right)={{\left|A\right|}^{n+1}I}_n$ $(\because \left|adj\left(A\right)\right|={\left|A\right|}^{n+1})$

Multiply both sides by A to get

$A\left(adj\left(A\right)\right)\left(adj\left(adj\left(A\right)\right)\right)=A\left(\left|adj\left(A\right)\right|I_n\right)={{A\left|A\right|}^{n+1}I}_n$

$A\left(adj\left(A\right)\right)\left(adj\left(adj\left(A\right)\right)\right)=\left(A{I}_n\right){\left|A\right|}^{n-1}$ (by Associativity)

$\Rightarrow \left|A\right|I_n\left(adj\left(adjA\right)\right)=A{\left|A\right|}^{n-1}$

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

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

$(\because \left|A\right|\ne 0)$, dividing both sides by $\left|A\right|$