JEE Main 2025 Apr 4 Shift 1, Mathematics Q22: Algebra Of Matrices

JEE Main2025Apr 4, Shift 1Mathematics

Let $A = cos θ 0 - sin θ 010 sin θ 0 cos θ$ . If for some $θ \in (0, π)$ , $A^{2} = A^{T}$ , then the sum of the diagonal elements of the matrix $(A + I)^{3} + (A - I)^{3} - 6 A$ is equal to ____.

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Solution

$A$ is a rotation matrix about the $y$ -axis, so $A$ is orthogonal: $A^{T} = A^{- 1}$ . The condition $A^{2} = A^{T}$ becomes $A^{2} = A^{- 1}$ , i.e. $A^{3} = I$ .

Expand the cubic expression:

$(A + I)^{3} + (A - I)^{3} = (A^{3} + 3 A^{2} + 3 A + I) + (A^{3} - 3 A^{2} + 3 A - I) = 2 A^{3} + 6 A$ .

Subtracting $6 A$ and using $A^{3} = I$ : $(A + I)^{3} + (A - I)^{3} - 6 A = 2 A^{3} = 2 I$ .

Trace of $2 I$ (a $3 \times 3$ matrix) $= 2 + 2 + 2 = 6$ .

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JEE Main 2025 Apr 4 Shift 1, Mathematics Q22: Algebra Of Matrices

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