Question

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

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

Answer 1

The correct option is C

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

Explanation for correct answer:

Step-1 : Finding the value of $\lambda$:

$\text{adj}A=\left[\begin{array}{ccc}2& -1& 1\\ -1& 0& 2\\ 1& -2& -1\end{array}\right]$ and $B=\text{adj}(\text{adj}A)$

$$\begin{gathered} \Rightarrow \left|\text{adj}A\right|=2\left(0+4\right)+1(1-2)+1\left(2\right) \\ =8-1+2 \\ =9 \end{gathered}$$

$\Rightarrow$${\left|A\right|}^{n-1}=9\left[\because \left|\text{adj}A\right|={\left|A\right|}^{n-1}\right]$

$$\begin{gathered} {\left|A\right|}^{3-1}=9 \\ {\left|A\right|}^2=9 \\ \left|A\right|=\pm 3 \end{gathered}$$

If $\left|A\right|=\lambda$

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

$\Rightarrow$ $\left|\lambda \right|=3$

Step-2 : Finding the value of$\mu$:

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

$B=\text{adj}(\text{adj}A)$

Taking determinants on both sides, we get

$$\begin{gathered} \text{det}B=\text{det}\left(\text{adj}(\text{adj}A)\right) \\ \left|B\right|=\text{det}\left(\text{adj}(\text{adj}A)\right) \\ =\text{det}\left({\left|\text{adj}A\right|}^{n-1}\right)\left[\because \left|adjA\right|={\left|A\right|}^{n-1}\right] \\ ={\left|\left({\left|A\right|}^{n-1}\right)\right|}^{n-1}\left[\because \left|\text{adj}A\right|={\left|A\right|}^{n-1}\right] \end{gathered}$$

$3\times 3$ matrix so $n=3$

Therefore,

$$\begin{gathered} \left|B\right|={\left|\left({\left|A\right|}^{3-1}\right)\right|}^{3-1} \\ ={\left|A\right|}^4 \\ =3^4\mathbf{}\left[\because \left|A\right|=3\right] \\ =81 \end{gathered}$$

$$\begin{gathered} \mu =\left|{\left(B^{-1}\right)}^T\right| \\ =\frac{1}{\left|B^T\right|} \\ =\frac{1}{81}\left[\because \left|B\right|=\left|B^T\right|=81,\right] \end{gathered}$$

Therefore, the ordered pair, $\left(\right|\lambda |,\mu )$ is $\left(3,\frac{1}{81}\right)$

Hence, option (C) is the correct answer.