JEE Mathematics 2025 Question with Solution

Given: $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta\\ 0 & 1 & 0\\ \sin \theta & 0 & \cos \theta \end{bmatrix}$ and $A^2 = A^T$.

Find: The sum of the diagonal elements of $((A+I)^3+(A-I)^3-6A)$.

Since the given matrix is orthogonal, we have

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

Using $A^2 = A^T$, we get

$$A^2 = A^{-1}$$

Multiplying by $A$ gives

$$A^3 = I$$

Now expand:

$$(A+I)^3 = A^3 + 3A^2 + 3A + I$$

and

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

Adding these,

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

Therefore,

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

Since $A^3 = I$,

$$2A^3 = 2I = \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{bmatrix}$$

Hence, the sum of diagonal elements is

$$2+2+2=6$$

Therefore, the correct option is A.

Given: $A^2 = A^T$ and $A$ is the rotation matrix shown.

Find: $\operatorname{tr} \left((A+I)^3+(A-I)^3-6A\right)$.

Because $A$ is orthogonal,

$$A A^T = I$$

Using $A^2 = A^T$,

$$A \cdot A^2 = A \cdot A^T = I$$

So,

$$A^3 = I$$

Next,

$$\begin{aligned} (A+I)^3+(A-I)^3-6A &= (A^3+3A^2+3A+I) + (A^3-3A^2+3A-I) - 6A \\ &= 2A^3 \end{aligned}$$

Thus the required matrix equals

$$2A^3 = 2I$$

Now,

$$\operatorname{tr}(2I)=2\operatorname{tr}(I)=2\cdot 3=6$$

Therefore, the sum of diagonal elements is $6$, so the correct option is A.