Let . If for some , then the sum of the diagonal elements of the matrix is equal to __________. [2025]

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Solution

Q.
Let $A = [\begin{matrix} \cos θ & 0 & - \sin θ \\ 0 & 1 & 0 \\ \sin θ & 0 & \cos θ \end{matrix}]$ . 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 __________. [2025]

Ans.
(6)

We have, $A = [\begin{matrix} \cos θ & 0 & - \sin θ \\ 0 & 1 & 0 \\ \sin θ & 0 & \cos θ \end{matrix}]$

Since, A is orthogonal.

$\Rightarrow A A^{T} = A^{T} A = I and A^{T} = A^{- 1}$

Given, $A^{2} = A^{T}$

$\Rightarrow A^{3} = I$

Let $B = {(A + I)}^{3} + {(A - I)}^{3} - 6 A$

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

So, sum of diagonal elements of B = 2(1 + 1 + 1) = 6.

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