Let A be a 3×3 matrix such that adj A=
Question:

Let $A$ be a $3 \times 3$ matrix such that $\operatorname{adj} A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1\end{array}\right]$ and

$B=\operatorname{adj}(\operatorname{adj} A)$. If $|A|=\lambda$ and $\left|\left(B^{-1}\right)^{T}\right|=\mu$, then the ordered pair, $(|\lambda|, \mu)$ is equal to :

1. (1) $\left(3, \frac{1}{81}\right)$

2. (2) $\left(9, \frac{1}{9}\right)$

3. (3) $(3,81)$

4. (4) $\left(9, \frac{1}{81}\right)$

Correct Option: 1

Solution:

$|\operatorname{adj} A|=|A|^{2}=9 \quad\left[\because|\operatorname{adj} A|=|A|^{n-1}\right]$

$\Rightarrow|A|=\pm 3=\lambda \Rightarrow|\lambda|=3$

$\Rightarrow|B|=|\operatorname{adj} A|^{2}=81$

$\mu=\left|\left(B^{-1}\right)^{T}\right|=\left|B^{-1}\right|=|B|^{-1}=\frac{1}{|B|}=\frac{1}{81}$