Question

Let $A$ be a $3\times 3$ square matrix. If $B=adj\left(A\right),$ $C=adj\left(adj\right(A\left)\right)$ and $D=adj\left(adj\right(adj\left(A\right)\left)\right),$ then $|adj\left(adj\right(adj\left(adj\right(ABCD\left)\right)\left)\right)|,$ in terms of $|A|$ is

• A. $|A^{60}|$
• B. $|A^{120}|$
• C. $|A^{180}|$
• D. $|A^{240}|$

Answer 1

The correct option is D $|A^{240}|$

We know that $|adjA|=|A{|}^{n-1}\cdots \left(1\right)$
$|adj\left(adjA\right)|=|adjA{|}^{n-1}$
$=(|A{|}^{n-1}{)}^{n-1}$ [From $\left(1\right)$]
=$|A{|}^{(n-1{)}^2}$

Similarly, $|adj\left(adj\right(adj\left(A\right)\left)\right)|=|A{|}^{(n-1{)}^3}$
and $|adj\left(adj\right(adj\left(adj\right(A\left)\right)\left)\right)|=|A{|}^{(n-1{)}^4}$
So, $|adj\left(adj\right(adj\left(adj\right(ABCD\left)\right)\left)\right)|$
$=|ABCD{|}^{(n-1{)}^4}$
$=|ABCD{|}^{(3-1{)}^4}$
$=|ABCD{|}^{16}$
$=(|A|\cdot |B|\cdot |C|\cdot |D|{)}^{16}$
$=(|A|\cdot |A{|}^{3-1}\cdot |A{|}^{(3-1{)}^2}\cdot |A{|}^{(3-1{)}^3}{)}^{16}$
$=|A{|}^{240}=|A^{240}|$