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 $\left|\left(B^{-1}\right)^{\top}\right|=\mu$, then the ordered pair, $(|\lambda| \cdot \mu)$ is equal to:

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

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

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

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Matrices

Let

A

be a

3 \times 3

matrix such that

adj A = 2 - 1 1 - 1 0 - 2 12 - 1

and

B = adj (adj A)

. If

∣ A ∣ = λ

and

(B^{- 1})^{⊤} = μ

, then the ordered pair,

(∣ λ ∣ \cdot μ)

is equal to:

Answer: (D)

For a

3 \times 3

matrix

∣ adj A ∣ = ∣ A ∣^{2}

. The given adjoint has determinant

9

, so

∣ A ∣^{2} = 9

and

∣ A ∣ = \pm 3

, giving

∣ λ ∣ = 3

. Next,

B = adj (adj A)

has

∣ B ∣ = ∣ adj A ∣^{2} = 9^{2} = 81

. Finally

(B^{- 1})^{⊤} = ∣ B^{- 1} ∣ = \frac{1}{∣ B ∣} = \frac{1}{81} = μ

. The ordered pair is

(3, \frac{1}{81})

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option