Let \(\mathrm{A}=\left[\begin{array}{ccc}\cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right]\). If for some \(\theta \in(0, \pi)\), \(A^{2}=A^{T}\), then the sum of the diagonal elements of the matrix \((\mathrm{A}+\mathrm{I})^{3}+(\mathrm{A}-\mathrm{I})^{3}-6 \mathrm{~A}\) is equal to

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Matrices

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

Answer: 6

The matrix is a rotation about the

y

-axis, hence orthogonal, so

A^{T} = A^{- 1}

; the condition

A^{2} = A^{T}

then gives

A^{3} = I

. Expanding,

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

, so

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

. The sum of the diagonal elements of

2 I

6

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Integer answer