Introduction

This article explains step by step how to solve the matrix angle problem where
\(A = \begin{bmatrix}\sin\theta & \cos\theta \\ \cos\theta & \sin\theta\end{bmatrix}\)
and \(A + A^T – 2I = 0\), and shows how to find the exact value of \(\theta\) in degrees.

The final answer is \(\theta = 90^\circ\).

Step 1: Write A, Aᵀ and I

Given

\(A =
\begin{bmatrix}
\sin\theta & \cos\theta\\
\cos\theta & \sin\theta
\end{bmatrix}\)

This matrix is symmetric, so its transpose is the same:
\(A^T =
\begin{bmatrix}
\sin\theta & \cos\theta\\
\cos\theta & \sin\theta
\end{bmatrix} = A\).

The identity matrix of order 2 is
\(I =
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}\).

Step 2: Use the condition A + Aᵀ − 2I = 0

Because \(A = A^T\),
\(A + A^T – 2I = 0 \Rightarrow 2A – 2I = 0\).
Dividing by 2 gives \(A – I = 0 \Rightarrow A = I\).

So every entry of \(A\) must match the corresponding entry of \(I\), which means
\[
\begin{bmatrix}
\sin\theta & \cos\theta\\
\cos\theta & \sin\theta
\end{bmatrix}
=
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}.
\]

Equating corresponding elements gives:

• \(\sin\theta = 1\) from the (1,1) and (2,2) entries.
• \(\cos\theta = 0\) from the (1,2) and (2,1) entries.

Step 3: Solve for θ

From \(\sin\theta = 1\), we get \(\theta = 90^\circ + 360^\circ k\) where \(k \in \mathbb{Z}\).

From \(\cos\theta = 0\), we get \(\theta = 90^\circ + 180^\circ n\) where \(n \in \mathbb{Z}\).

The common angles satisfying both conditions are
\(\theta = 90^\circ + 360^\circ m\), \(m \in \mathbb{Z}\).
In the principal range \(0^\circ \leq \theta < 360^\circ\), this gives \(\theta = 90^\circ\).

Therefore, the required value of \(\theta\) is \(90^\circ\).

Option-wise explanation

In a typical multiple-choice setting, the options might be \(0^\circ\), \(45^\circ\), \(60^\circ\), and \(90^\circ\).
Here is how each behaves:

• 0°:
  \(\sin 0^\circ = 0\), \(\cos 0^\circ = 1\) so
  \(A = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \neq I\);
  hence \(A + A^T – 2I \neq 0\).
• 45°:
  \(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}\) so all entries of \(A\) are \(\frac{\sqrt{2}}{2}\),
  which does not equal \(I\); the condition fails.
• 60°:
  \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 60^\circ = \frac{1}{2}\);
  off-diagonal elements are nonzero and \(A \neq I\), so the equation fails.
• 90°:
  \(\sin 90^\circ = 1\), \(\cos 90^\circ = 0\), so
  \(A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} = I\),
  giving \(A + A^T – 2I = I + I – 2I = 0\); this satisfies the condition.

Thus, among these options, \(90^\circ\) is the correct choice.