Let $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. If $A^T + A = I$, then

We have a matrix $A$ (called a rotation matrix) and we need to find the value of $\theta$ when $A^T + A = I$.

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Find $A^T$ (Transpose of A)

Transpose means flip the matrix across its diagonal - rows become columns.

Given: $A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$

So: $A^T = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$

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Add $A^T + A$

Add corresponding positions:

$A^T + A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$

$A^T + A = \begin{bmatrix} \cos\theta + \cos\theta & \sin\theta + (-\sin\theta) \\ -\sin\theta + \sin\theta & \cos\theta + \cos\theta \end{bmatrix}$

$A^T + A = \begin{bmatrix} 2\cos\theta & 0 \\ 0 & 2\cos\theta \end{bmatrix}$

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Use the given condition $A^T + A = I$

$\begin{bmatrix} 2\cos\theta & 0 \\ 0 & 2\cos\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

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Compare corresponding elements

For two matrices to be equal, every element must match.

From position $(1,1)$: $2\cos\theta = 1$

From position $(2,2)$: $2\cos\theta = 1$

From positions $(1,2)$ and $(2,1)$: $0 = 0$ (automatically satisfied)

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Solve for $\theta$

$2\cos\theta = 1$

$\cos\theta = \dfrac{1}{2}$

Which angles have cosine equal to $\dfrac{1}{2}$?

$\theta = 60° = \dfrac{\pi}{3}$ radians

OR

$\theta = 300° = \dfrac{5\pi}{3}$ radians (or $-60°$)

General solution: $\theta = 2n\pi \pm \dfrac{\pi}{3}$, where $n$ is any integer

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$\theta = \dfrac{\pi}{3}$ or $60°$

(or $\theta = 2n\pi \pm \dfrac{\pi}{3}$ for the general solution)