Q.If $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$, and $A + A' = I$, then the value of $\alpha$ is (A) $\frac{\pi}{6}$ (B) $\frac{\pi}{3}$ (C) $\pi$ (D) $\frac{3\pi}{2}$

Concept: Rotation Matrix Composition — $A$ is a rotation matrix, and $A'$ is its transpose (which equals its inverse for an orthogonal matrix). The condition $A + A' = I$ forces a specific angle.

Step 1: Write $A$ and $A'$:

Step 2: Add them:

Step 3: Set equal to $I$: …

The matrix $A$ is a classic 2D rotation matrix: it rotates any vector by angle $\alpha$ anticlockwise. Its transpose $A'$ is the inverse rotation (by $-\alpha$), because for rotation matrices, $A' = A^{-1}$. The condition $A + A' = I$ is therefore a neat equation linking a rotation and its inverse.

Why does this approach work? Instead of blindly substituting and solving, we recognise that the sum of a rotation and its inverse must equal the identity. That gives us a direct trigonometric equation for $\alpha$, with no messy algebra.

Let’s work through it step by step.

1. Write down $A$ and $A'$ explicitly.

2. Add them entry by entry.

The off-diagonal terms cancel perfectly — that’s the symmetry of the rotation matrix at work.

3. Set this equal to the identity matrix $I$.

Comparing entries gives a single condition:

4. Find $\alpha$ in the usual range $[0, 2\pi)$. …

Method: Solving for an angle from a matrix condition

When a condition such as $A+A'=I$ is imposed on a trigonometric matrix, form $A+A'$ explicitly, equate it entry-by-entry with the target matrix, and solve the resulting trigonometric equation.

Steps

Step 1: Write $A'$ and add

For a rotation matrix the off-diagonal terms cancel in $A+A'$, leaving $\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}=2\cos\alpha\,I$. …

Common Mistakes

Mistake 1: Confusing the transpose $A'$ with the inverse when forming $A+A'$

Why it's wrong: $A'$ is the transpose (which happens to equal $A^{-1}$ for a rotation), but the condition needs the transpose added entry-wise, giving $2\cos\alpha$ on the diagonal. Correct approach: write $A'$ by swapping the off-diagonal signs, then add.

Mistake 2: Solving $\cos\alpha=\tfrac12$ but picking the wrong angle …