Let A = [[cos θ, 0, -sin θ], [0, 1, 0], [sin θ, 0, cos θ]]. If for some θ ∈ (0, π), A² = Aᵀ, then th

Q: Wait, why can we assume $A^T = A^{-1}$? Is that always true?

That is a fantastic question! It is not true for all matrices, but it is true for orthogonal matrices. If you calculate $A \cdot A^T$ for this specific matrix, you will find that the off-diagonal elements vanish and the diagonal elements become $1$. Since $A \cdot A^T = I$, by definition, $A^T$ must be the inverse of $A$.

Q: Can I really use the binomial expansion $(A + I)^3 = A^3 + 3A^2I + 3AI^2 + I^3$? I thought matrix multiplication wasn't commutative.

You are absolutely right to be cautious! Matrix multiplication is generally not commutative. However, the identity matrix $I$ is special; it commutes with every matrix ($AI = IA = A$). Because $I$ commutes with $A$, the binomial theorem holds perfectly here. You are safe to proceed!

Q: Do I need to find the value of $\theta$ to solve this?

Not at all! The beauty of this problem lies in the algebraic structure. The specific value of $\theta$ is irrelevant because the expression simplifies to a form that depends only on $A^3$ and $I$. Trust the algebra; it will lead you to the answer regardless of the angle.

Question

Let $A = \begin{bmatrix} \cos 	heta & 0 & -\sin 	heta \ 0 & 1 & 0 \ \sin 	heta & 0 & \cos 	heta \end{bmatrix}$. If for some $	heta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6 A$ is equal to _____ .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE Advanced landscape. Today, we stand before a matrix that might look like a daunting wall of trigonometric functions, but it is merely a gateway to a beautiful, symmetric truth.

Let us dissect the matrix:

$$A = \begin{bmatrix} \cos 	heta & 0 & -\sin 	heta \ 0 & 1 & 0 \ \sin 	heta & 0 & \cos 	heta \end{bmatrix}$$

If you have spent enough time in the trenches of linear algebra, you might recognize this structure as a rotation matrix. Even if you do not see the rotation immediately, the math will reveal its secrets.

### The Orthogonal Insight

The first step in any great battle is to understand your terrain. We are given the condition $A^2 = A^T$.

If we compute the product $A \cdot A^T$, we find that the off-diagonal elements cancel out, and the diagonal elements collapse into $\cos^2 	heta + \sin^2 	heta$, which is $1$. Thus, $A \cdot A^T = I$.

This confirms that $A$ is an orthogonal matrix, which implies that its transpose is its inverse: $A^T = A^{-1}$. This is our first major breakthrough.

### The Algebraic Dance

Now, let us look at the condition provided: $A^2 = A^T$. Since we established that $A^T = A^{-1}$, we can rewrite this as $A^2 = A^{-1}$.

If we multiply both sides of this equation by $A$, we get $A^3 = A \cdot A^{-1}$, which simplifies to:

$$A^3 = I$$

This is the key that unlocks the entire problem. We have reduced a complex matrix power to the simplest possible form: the identity matrix.

### The Expansion

We are asked to evaluate the sum of the diagonal elements of $(A + I)^3 + (A - I)^3 - 6A$. Because the identity matrix $I$ commutes with $A$, we can use the binomial expansion just as we would with scalar variables.

First, we expand the terms:

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

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

Now, let us add these two expansions together:

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

The terms $3A^2$ and $I$ cancel out with such elegance that it almost feels like the math is doing the work for us. Finally, we subtract the $6A$ term as required by the original expression:

$$2A^3 + 6A - 6A = 2A^3$$

### The Grand Finale

We are left with the expression $2A^3$. Since we already know that $A^3 = I$, our expression simplifies to $2I$.

The matrix $2I$ is simply a diagonal matrix with $2$ on the main diagonal and $0$ everywhere else:

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

The question asks for the sum of the diagonal elements, which is the trace of the matrix. The trace is $2 + 2 + 2 = 6$.

**The final answer is 6.**

Animated Solution for Mathematics - Matrices and Determinants: 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 _____ .

Enter Numerical Value:

Visualized Solution

Analyze Matrix $A$

Identify Orthogonality

Given Condition

Derive $A^{3} = I$

The Target Expression

Expand $(A + I)^{3}$

Expand $(A - I)^{3}$

Combine and Simplify

Final Matrix Substitution

Calculate the Trace

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Orthogonal Insight

The Algebraic Dance

The Expansion

The Grand Finale

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