Question

Let $A$ be any $n\times n$ invertible matrix. Then which one of the following is always true

• A. $|Adj\left(AdjA\right)|=|A{|}^{(n-2)}$
• B. $|Adj\left(AdjA\right)|=|A{|}^{(n-1{)}^2}$
• C. $|Adj\left(AdjA\right)|=|A{|}^{(n-1)}$
• D. $|Adj\left(AdjA\right)|=|A{|}^{(n-1{)}^3}$

Answer 1

The correct option is C $|Adj\left(AdjA\right)|=|A{|}^{(n-1)}$

Using the identity,

$adj\left(A\right).A=|A|.I_n$ ………$\left(1\right)$

& $adj\left(AB\right)=adj\left(B\right).adj\left(A\right)$ ………..$\left(2\right)$

By $\left(1\right)$,

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

Using $\left(2\right)$,

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

Multiplying by A,

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

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

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

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

$=|A{|}^{n-1}$.