Question

Let $A$ be a $3\times 3$ matrix such that $adjA=\left[\begin{array}{ccc}2& -1& 1\\ -1& 0& 2\\ 1& -2& -1\end{array}\right]$ and $B=adj\left(adjA\right)$.
If $\left|A\right|=\lambda$ and $\left|{\left(B^{-1}\right)}^T\right|=\mu$, then the ordered pair, $(\left|\lambda \right|,\mu )$ is equal to:

• A. $(9,\frac{1}{81})$
• B. $(9,\frac{1}{9})$
• C. $(3,\frac{1}{81})$
• D. $(3,81)$

Answer 1

The correct option is C $(3,\frac{1}{81})$

$\left|adjA\right|=9$
$\Rightarrow {\left|A\right|}^{(n-1)}=9$
$\Rightarrow {\left|A\right|}^2=9$
$\Rightarrow \left|A\right|=\pm 3=\lambda$
$\therefore |\lambda |=3$

$|adj\left(adjA\right)|={\left(\left|A\right|\right)}^{{(n-1)}^2}$
$\Rightarrow |B|={\left(\left|A\right|\right)}^{{(3-1)}^2}$
$\Rightarrow |B|={\left(\left|A\right|\right)}^4=81$
$\Rightarrow \left|{\left(B^{-1}\right)}^T\right|=|(B^T{)}^{-1}|=\frac{1}{\left|B^T\right|}=\frac{1}{\left|B\right|}=\frac{1}{81}$
$\therefore \mu =\frac{1}{81}$
$\therefore (\left|\lambda \right|,\mu )=(3,\frac{1}{81})$