JEE Main 2025 — 4 April, Morning Shift — Question 44

Finding $\theta$

Given $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$.

$A^2$ is the rotation by $2\theta$:

$$A^2 = \begin{bmatrix} \cos 2\theta & 0 & -\sin 2\theta \\ 0 & 1 & 0 \\ \sin 2\theta & 0 & \cos 2\theta \end{bmatrix}, \quad A^T = \begin{bmatrix} \cos \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta \end{bmatrix}$$

From $A^2 = A^T$, comparing $(1,3)$ elements:

$$-\sin 2\theta = \sin \theta \implies 2\sin\theta\cos\theta + \sin\theta = 0$$

Since $\theta \in (0, \pi)$, $\sin\theta \neq 0$, so $2\cos\theta + 1 = 0 \implies \cos\theta = -1/2$.

Simplifying the Matrix Expression

Let $S = (A+I)^3 + (A-I)^3 - 6A$.

Using $(A+I)^3 = A^3 + 3A^2 + 3A + I$ and $(A-I)^3 = A^3 - 3A^2 + 3A - I$:

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

Since $A^2 = A^T$ and $A$ is orthogonal ($A^T = A^{-1}$), we have $A^2 = A^{-1} \implies A^3 = I$.

Thus, $S = 2I$.

Trace Calculation

The sum of diagonal elements (Trace) of $2I$ for a $3 \times 3$ matrix is:

$$\text{Tr}(2I) = 2 + 2 + 2 = 6$$