Question

If $A$ is a $3 \times 3$ matrix such that $|A|=27, \text{Adj} A= kA ^{ T }$, then $k^2-3 k+5=$

• A. 5
• B. 3
• C. 0
• D. 2

Answer 1

Answer: (A)

$|A|=27$
$ \operatorname{adj} A=k A^T $
$ \Rightarrow|\operatorname{adj} A|=k^3\left|A^T\right|$
$ \Rightarrow|A|^2=k^3|A| $
$ \Rightarrow|A|\left(|A|-k^3\right)=0$
Since, $|A| \neq 0$
$\Rightarrow k^3=|A|=27$
$k=3$
$k^2-3 k+5=9-9+5=5$