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JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Matrices

Let A be a square matrix such that

AA^{T} = I

. Then

\frac{1}{2} A [(A + A^{T})^{2} + (A - A^{T})^{2}]

is equal to

(A) $A^{2} + I$
(B) $A^{3} + I$
(C) $A^{2} + A^{T}$
(D) $A^{3} + A^{T}$ ✓ Correct

Answer: (D)

Solution

Given

A A^{T} = A^{T} A = I

, so

A

and

A^{T}

commute.

(A + A^{T})^{2} = A^{2} + A A^{T} + A^{T} A + (A^{T})^{2} = A^{2} + (A^{T})^{2} + 2 I

(A - A^{T})^{2} = A^{2} - A A^{T} - A^{T} A + (A^{T})^{2} = A^{2} + (A^{T})^{2} - 2 I

Adding:

(A + A^{T})^{2} + (A - A^{T})^{2} = 2 A^{2} + 2 (A^{T})^{2}

.

\frac{1}{2} A [2 A^{2} + 2 (A^{T})^{2}] = A^{3} + A (A^{T})^{2} = A^{3} + (A A^{T}) A^{T} = A^{3} + A^{T}

Key confirmed: D.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option

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