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JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Determinants

Let

A

and

B

be two square matrices of order 3 such that

∣ A ∣ = 3

and

∣ B ∣ = 2

. Then

A^{T} A (adj (2 A))^{- 1} (adj (4 B)) (adj (A B))^{- 1} A A^{T}

is equal to:

(A)108
(B)81
(C)32
(D)64✓ Correct

Answer: (D)

Solution

Use

adj (M) = ∣ M ∣ M^{- 1}

for a

3 \times 3

matrix, so

∣ adj M ∣ = ∣ M ∣^{2}

, and

∣ adj (k M) ∣ = (k^{3} ∣ M ∣)^{2} = k^{6} ∣ M ∣^{2}

.
Taking determinants of the whole product (with

∣ A^{T} ∣ = ∣ A ∣

):

\frac{∣ A ∣ ^{2} ∣ adj ( 4 B ) ∣ ∣ A ∣ ^{2}}{∣ adj ( 2 A ) ∣ ∣ adj ( A B ) ∣} = \frac{∣ A ∣ ^{4} ( 4 ^{6} ∣ B ∣ ^{2} )}{( 2 ^{6} ∣ A ∣ ^{2} ) ( ∣ A ∣∣ B ∣ ) ^{2}} = \frac{4 ^{6} ∣ A ∣ ^{4} ∣ B ∣ ^{2}}{2 ^{6} ∣ A ∣ ^{4} ∣ B ∣ ^{2}} = \frac{4 ^{6}}{2 ^{6}} = 2^{6} = 64.

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Single correct option

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