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JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Matrices

Let

A

and

B

be two invertible matrices of order

3 \times 3

. If

det (A B A^{⊤}) = 8

and

det (A B^{- 1}) = 8

, then

det (B A^{- 1} B^{⊤})

is equal to:

(A) $\frac{1}{4}$
(B)1
(C) $\frac{1}{16}$ ✓ Correct
(D)16

Answer: (C)

Solution

Taking determinants,

det (A B A^{⊤}) = ∣ A ∣^{2} ∣ B ∣ = 8

and

det (A B^{- 1}) = \frac{∣ A ∣}{∣ B ∣} = 8

, so

∣ A ∣ = 8∣ B ∣

. Substituting,

(8∣ B ∣)^{2} ∣ B ∣ = 64∣ B ∣^{3} = 8

, giving

∣ B ∣^{3} = \frac{1}{8}

, so

∣ B ∣ = \frac{1}{2}

and

∣ A ∣ = 4

. Then

det (B A^{- 1} B^{⊤}) = \frac{∣ B ∣ ^{2}}{∣ A ∣} = \frac{1/4}{4} = \frac{1}{16}

.

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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