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

We are given the adjugate (adjoint) of a $$3 \times 3$$ matrix $$A$$ as

$$\operatorname{adj}A=\begin{bmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{bmatrix}.$$

First, we recall the basic relation between a square matrix and its adjugate:

$$\operatorname{adj}A = |A|\,A^{-1}.$$

For a matrix of order $$n$$, another very useful fact is

$$|\operatorname{adj}A| = |A|^{\,n-1}.$$

Since we are dealing with a $$3 \times 3$$ matrix, $$n=3$$ and therefore

$$|\operatorname{adj}A| = |A|^{\,2}.$$

So, if we compute the determinant of the given $$\operatorname{adj}A$$, call it $$\Delta$$, we can immediately get $$|A|$$ from the equation

$$\Delta = |A|^{\,2}.$$

Let us now calculate $$\Delta$$ step by step. Using the first row expansion for the determinant, we have

$$$ \begin{aligned} \Delta &= 2\begin{vmatrix} 0 & 2 \\ -2 & -1 \end{vmatrix} \;-\; (-1)\begin{vmatrix} -1 & 2 \\ 1 & -1 \end{vmatrix} \;+\; 1\begin{vmatrix} -1 & 0 \\ 1 & -2 \end{vmatrix}\\[6pt] &= 2\Bigl(0\cdot(-1) \;-\; 2\cdot(-2)\Bigr) \;-\; (-1)\Bigl((-1)\cdot(-1) \;-\; 2\cdot1\Bigr) \;+\; 1\Bigl((-1)\cdot(-2) \;-\; 0\cdot1\Bigr)\\[6pt] &= 2\Bigl(0 - (-4)\Bigr) \;-\; (-1)\Bigl(1 - 2\Bigr) \;+\; 1\Bigl(2 - 0\Bigr)\\[6pt] &= 2\cdot 4 \;-\; (-1)\cdot(-1) \;+\; 1\cdot 2\\[6pt] &= 8 \;-\; 1 \;+\; 2\\[6pt] &= 9. \end{aligned} $$$

Thus

$$|\operatorname{adj}A| = 9.$$

Setting this equal to $$|A|^{\,2}$$ we get

$$|A|^{\,2}=9 \quad\Longrightarrow\quad |A|=\pm 3.$$

The problem denotes $$|A|$$ by $$\lambda$$. We shall need only the absolute value, so

$$|\lambda| = |\,|A|\,| = 3.$$

Next, we must evaluate $$\mu = \bigl|\,(B^{-1})^{T}\bigr|$$ where

$$B = \operatorname{adj}(\operatorname{adj}A).$$

There is a standard identity for the double adjugate (valid for all nonsingular $$n \times n$$ matrices):

$$\operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}\,A.$$

With $$n=3$$ this becomes

$$B = \operatorname{adj}(\operatorname{adj}A) = |A|\,A.$$

Let us now find $$B^{-1}$$. Because $$|A|$$ is just a scalar, we have

$$B^{-1} = (|A|\,A)^{-1} = \frac{1}{|A|}\,A^{-1}.$$

Taking the transpose, we get

$$(B^{-1})^{T} = \frac{1}{|A|}\,(A^{-1})^{T}.$$

The determinant of a transpose equals the determinant of the original matrix, and extracting a scalar $$\frac{1}{|A|}$$ from a $$3 \times 3$$ determinant raises it to the third power. Hence

$$$ \begin{aligned} \mu = \bigl|\,(B^{-1})^{T}\bigr| &= \left|\frac{1}{|A|}\,(A^{-1})^{T}\right|\\[6pt] &= \left(\frac{1}{|A|}\right)^{3}\,\bigl|A^{-1}\bigr|. \end{aligned} $$$

But $$|A^{-1}| = \dfrac{1}{|A|}$$, so we obtain

$$$ \mu = \left(\frac{1}{|A|}\right)^{3}\,\frac{1}{|A|} = \left(\frac{1}{|A|}\right)^{4}. $$$

Substituting $$|A| = \pm 3$$ (whose absolute value is $$3$$) gives

$$\mu = \left(\frac{1}{3}\right)^{4} = \frac{1}{81}.$$

Collecting the two required numbers, we have

$$\bigl(|\lambda|,\mu\bigr) = \left(3,\frac{1}{81}\right).$$

Hence, the correct answer is Option A.