Question

If $A=\left[\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right]$ ,then $\det (adj(adjA))$ is equal to

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

Answer 1

Answer: (A)

We have, $A=\left[\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right]$
$\Rightarrow$ $|A|=1(1+2)-2(-1-4)-1(1-2)$
$=3+10+1=14$
We know that, for a square matrix of order n,
$adj(adjA)=|A{|}^{n-2}A,$ if $|A|\ne 0$
$\Rightarrow$ $\det (adj(adjA))=||A{|}^{n-2}A|$
$\Rightarrow$ $\det (adj(adjA))={(|A{|}^{n-2})}^n|A|$
$\Rightarrow$ $\det (adj(adjA))=|A{|}^{n^2-2n+1}$
Here, $n=3$ and $|A|=14$
$\therefore$ $\det (adj(adjA))={(14)}^{3^2-2\times 3+1}={14}^4$