Matrices — JEE Main · 2025 · 23 Jan 2025 · Shift 1 Question with Solution

A, B

and

(adj (A^{- 1}) + adj (B^{- 1}))

are non-singular matrices of same order, then the inverse of

A (adj (A^{- 1}) + adj (B^{- 1}))^{- 1} B

, is equal to

Answer: (C)

Solution

Let

M = A (adj (A^{- 1}) + adj (B^{- 1}))^{- 1} B

, so we need

M^{- 1}

M^{- 1} = B^{- 1} (adj (A^{- 1}) + adj (B^{- 1})) A^{- 1}

For an invertible matrix

X

adj (X^{- 1}) = ∣ X^{- 1} ∣ X = \frac{X}{∣ X ∣}

. So

adj (A^{- 1}) = \frac{A}{∣ A ∣}, adj (B^{- 1}) = \frac{B}{∣ B ∣}

M^{- 1} = B^{- 1} (\frac{A}{∣ A ∣} + \frac{B}{∣ B ∣}) A^{- 1} = \frac{B ^{- 1} A A ^{- 1}}{∣ A ∣} + \frac{B ^{- 1} B A ^{- 1}}{∣ B ∣} = \frac{B ^{- 1}}{∣ A ∣} + \frac{A ^{- 1}}{∣ B ∣}

Writing

B^{- 1} = \frac{adj B}{∣ B ∣}

and

A^{- 1} = \frac{adj A}{∣ A ∣}

M^{- 1} = \frac{adj B}{∣ A ∣∣ B ∣} + \frac{adj A}{∣ A ∣∣ B ∣} = \frac{1}{∣ A B ∣} (adj A + adj B)

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option