If $A = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}$, then show that $A^2 = \begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$.

> [!TLDR] > This problem uses the fact that the given matrix $A$ is a **rotation matrix** in the plane. Multiplying rotation matrices corresponds to adding their angles, so $A^2$ rotates by $2\theta$, giving the stated result. The matrix $A$ has a beautiful geometric meaning. It represents a rotation of the coordinate axes by an angle $\theta$ in the clockwise direction (or equivalently, a rotation of vectors by $-\theta$). The top row gives the new $x$-axis in terms of the old axes, and the bottom row gives the new $y$-axis. When you multiply two rotation matrices, you get another rotation matrix whose angle is the **sum** of the individual angles. So $A^2$ should rotate by $\theta + \theta = 2\theta$. That’s the core intuition. Let’s verify this algebraically. 1. **Write down the matrix multiplication.** We need $A^2 = A \cdot A$. $$ A^2 = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix} \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix} $$ 2. **Compute the (1,1) entry.** Row 1 × Column 1: $$ (\cos\theta)(\cos\theta) + (\sin\theta)(-\sin\theta) = \cos^2\theta - \sin^2\theta $$ This is exactly $\cos 2\theta$ (double-angle identity). 3. **Compute the (1,2) entry.** Row 1 × Column 2: $$ (\cos\theta)(\sin\theta) + (\sin\theta)(\cos\theta) = \cos\theta\sin\theta + \sin\theta\cos\theta = 2\sin\theta\cos\theta $$ And $2\sin\theta\cos\theta = \sin 2\theta$. 4. **Compute the (2,1) entry.** Row 2 × Column 1: $$ (-\sin\theta)(\cos\theta) + (\cos\theta)(-\sin\theta) = -\sin\theta\cos\theta - \cos\theta\sin\theta = -2\sin\theta\cos\theta $$ That’s $-\sin 2\theta$. 5. **Compute the (2,2) entry.** Row 2 × Column 2: $$ (-\sin\theta)(\sin\theta) + (\cos\theta)(\cos\theta) = -\sin^2\theta + \cos^2\theta = \cos^2\theta - \sin^2\theta = \cos 2\theta $$ 6. **Assemble the result.** …

If A = cosθ sinθ\-sinθ cosθ , then show that A² = cos 2θ sin 2θ\-sin 2θ cos 2θ .

NCERT Exemplar · Q33

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

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Concept understanding — Matrix Rotation Power

Matrix Rotation Power

A rotation matrix turns every vector in the plane through a fixed angle. So what happens when you apply it again and again? Applying a rotation of $\theta$ twice is just a rotation of $2\theta$ ; three times, $3\theta$ ; and so on. Matrix power is exactly this idea written algebraically: $A^n$ means "apply the transformation $A$ a total of $n$ times."

The intuition

Multiplying a vector by a matrix $A$ transforms it once. Multiplying by $A$ again transforms the result once more. Hence

A^n = \underbrace{A \cdot A \cdots A}_{n \text{ times}},

with the conventions $A^1 = A$ and $A^0 = I$ (the identity), just as $x^0 = 1$ for numbers.

Watch out

$A^n$ is not raising each entry to the power $n$ . You must carry out full matrix multiplication. For example, with $B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ , $B^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ — the top-right entry becomes $2$ , not $1^2$ .

The rotation case

The cleanest example is the rotation matrix through angle $\theta$ (counterclockwise):

R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.

Because stacking two rotations adds their angles,

R_\theta^{\,n} = R_{n\theta} = \begin{pmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{pmatrix}.

Proving it by induction

This is a classic exam result, proved by mathematical induction on $n$ .

Base case ( $n = 1$ ): $R_\theta^{\,1} = R_\theta = R_{1\cdot\theta}$ , true.
Inductive step: assume $R_\theta^{\,k} = R_{k\theta}$ . Then

R_\theta^{\,k+1} = R_\theta^{\,k}\, R_\theta = R_{k\theta}\, R_\theta.

Multiplying the two matrices and using the addition formulas

$\cos k\theta\cos\theta - \sin k\theta\sin\theta = \cos(k+1)\theta, \qquad \sin k\theta\cos\theta + \cos k\theta\sin\theta = \sin(k+1)\theta,$ …

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This problem uses the fact that the given matrix $A$ is a rotation matrix in the plane. Multiplying rotation matrices corresponds to adding their angles, so $A^2$ rotates by $2\theta$ , giving the stated result.

The matrix $A$ has a beautiful geometric meaning. It represents a rotation of the coordinate axes by an angle $\theta$ in the clockwise direction (or equivalently, a rotation of vectors by $-\theta$ ). The top row gives the new $x$ -axis in terms of the old axes, and the bottom row gives the new $y$ -axis.

When you multiply two rotation matrices, you get another rotation matrix whose angle is the sum of the individual angles. So $A^2$ should rotate by $\theta + \theta = 2\theta$ . That’s the core intuition.

Let’s verify this algebraically.

Write down the matrix multiplication. We need $A^2 = A \cdot A$ .

A^2 = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix} \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}

Compute the (1,1) entry. Row 1 × Column 1:

(\cos\theta)(\cos\theta) + (\sin\theta)(-\sin\theta) = \cos^2\theta - \sin^2\theta

This is exactly $\cos 2\theta$ (double-angle identity).

Compute the (1,2) entry. Row 1 × Column 2:

(\cos\theta)(\sin\theta) + (\sin\theta)(\cos\theta) = \cos\theta\sin\theta + \sin\theta\cos\theta = 2\sin\theta\cos\theta

And $2\sin\theta\cos\theta = \sin 2\theta$ .

Compute the (2,1) entry. Row 2 × Column 1:

(-\sin\theta)(\cos\theta) + (\cos\theta)(-\sin\theta) = -\sin\theta\cos\theta - \cos\theta\sin\theta = -2\sin\theta\cos\theta

That’s $-\sin 2\theta$ .

Compute the (2,2) entry. Row 2 × Column 2:

(-\sin\theta)(\sin\theta) + (\cos\theta)(\cos\theta) = -\sin^2\theta + \cos^2\theta = \cos^2\theta - \sin^2\theta = \cos 2\theta

Assemble the result. …

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Q.If $A = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}$ , then show that $A^2 = \begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$ .

Concept understanding — Matrix Rotation Power

Matrix Rotation Power

The intuition

The rotation case

Proving it by induction

Method: Powers of a rotation matrix via the double-angle identities

Steps

Common Mistakes

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