Question

If $A=\left[\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right],$ then the value of $\text{det(adj(adj}A\left)\right)$ is

• A. ${14}^4$
• B. ${14}^2$
• C. ${14}^3$
• D. ${14}^1$

Answer 1

The correct option is A ${14}^4$

We know that $\text{adj(adj}A)=|A{|}^{n-2}A,\text{if}|A|\ne 0;$
provided order of $A$ is $n$
$\therefore \text{adj(adj}A)=|A|A(asn=3);$
$\therefore \text{det(adj(adj}A\left)\right)=|A{|}^3\text{det}A=|A{|}^4$
$A=\left[\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right]$
$\Rightarrow |A|=14$
Hence, $\text{det(adj(adj}A\left)\right)=|A{|}^4={14}^4$