If $P(x) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}$, then show that $P(x) \cdot P(y) = P(x + y) = P(y) \cdot P(x)$.

> [!TLDR] > The matrix $P(x)$ is a rotation matrix in the plane. Multiplying two such matrices corresponds to adding the rotation angles, and because addition is commutative, the product is commutative — giving $P(x)P(y)=P(x+y)=P(y)P(x)$. The key insight here is that $P(x)$ is not just any $2\times 2$ matrix — it is the standard **rotation matrix** that rotates a vector by angle $x$ in the clockwise direction (or anticlockwise, depending on sign convention). When you multiply two rotation matrices, you are composing two rotations: rotating by $x$ and then by $y$ is the same as rotating by $x+y$ all at once. And since rotating by $x$ then $y$ is the same as rotating by $y$ then $x$, the product commutes. Let’s verify this algebraically. 1. **Write down the product $P(x) \cdot P(y)$.** We have $$P(x) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}, \quad P(y) = \begin{bmatrix}\cos y & \sin y\\ -\sin y & \cos y\end{bmatrix}.$$ Multiply them in the order $P(x)P(y)$: $$P(x)P(y) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix} \begin{bmatrix}\cos y & \sin y\\ -\sin y & \cos y\end{bmatrix}.$$ The entry in row 1, column 1 is $\cos x \cos y + \sin x (-\sin y) = \cos x \cos y - \sin x \sin y$. Row 1, column 2: $\cos x \sin y + \sin x \cos y = \sin x \cos y + \cos x \sin y$. Row 2, column 1: $(-\sin x)\cos y + \cos x (-\sin y) = -(\sin x \cos y + \cos x \sin y)$. Row 2, column 2: $(-\sin x)\sin y + \cos x \cos y = \cos x \cos y - \sin x \sin y$. 2. **Recognise the trigonometric identities.** The expressions we just got are exactly the angle-sum formulas: $$\cos(x+y) = \cos x \cos y - \sin x \sin y,$$ $$\sin(x+y) = \sin x \cos y + \cos x \sin y.$$ So the product becomes: $$P(x)P(y) = \begin{bmatrix}\cos(x+y) & \sin(x+y)\\ -\sin(x+y) & \cos(x+y)\end{bmatrix} = P(x+y).$$ 3. **Now multiply in the reverse order $P(y)P(x)$.** …

If P(x) = cos x sin x\-sin x cos x , then show tha – Class 12 Mathematics

NCERT Exemplar · Q45

P(x) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}

Sikkim CbseShort· 3mImportance★★★★★est

69% · 126/182 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Rotation Matrix Composition

Rotation Matrix Composition

Hold a book, rotate it so the cover faces you, then tilt it to look down at the spine. The final orientation depends on which turn you do first. That is the whole idea: composing rotations means applying one after another, and in 3D the order matters.

Each rotation is a matrix, and doing two in sequence corresponds to multiplying their matrices. The subtlety is that matrix multiplication is generally not commutative: $AB \neq BA$ .

The Rule

R_1

Read right to left: a rotation acts on a column vector as $R\mathbf{v}$ , so applying $R_1$ then $R_2$ gives $R_2(R_1\mathbf{v}) = (R_2 R_1)\mathbf{v}$ . The rotation done first sits on the right.

A 2D Example

In the plane, a counter-clockwise rotation by angle $\theta$ is

R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.

Here $R(60^\circ)\,R(30^\circ) = R(90^\circ)$ : the angles simply add. In 2D rotations do commute, because they all share the same axis (the axis pointing out of the plane).

Why 3D Is Different

In 3D, rotations are about different axes, and swapping the order changes the result. Use the standard right-handed matrices

R_x(90^\circ) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}, \qquad R_y(90^\circ) = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix},

and track the vector $\mathbf{v} = (0,0,1)$ .

Order 1 — $R_x$ first, then $R_y$ (product $R_y R_x$ ):

$R_x(90^\circ)\mathbf{v} = (0,-1,0)$ — the point moves onto the $y$ -axis.
$R_y$ leaves the $y$ -axis fixed, so the result stays $(0,-1,0)$ .

Order 2 — $R_y$ first, then $R_x$ (product $R_x R_y$ ):

$R_y(90^\circ)\mathbf{v} = (1,0,0)$ — the point moves onto the $x$ -axis.
$R_x$ leaves the $x$ -axis fixed, so the result stays $(1,0,0)$ .

Watch out

Same two rotations, opposite order, different final point: $(0,-1,0)$ versus $(1,0,0)$ . So $R_x R_y \neq R_y R_x$ in general. (Notice each rotation leaves its own axis unmoved — a rotation about the $y$ -axis can never shift a vector that already lies on the $y$ -axis.) …

Start your 14-day free trial to unlock the full solution →

The matrix $P(x)$ is a rotation matrix in the plane. Multiplying two such matrices corresponds to adding the rotation angles, and because addition is commutative, the product is commutative — giving $P(x)P(y)=P(x+y)=P(y)P(x)$ .

The key insight here is that $P(x)$ is not just any $2\times 2$ matrix — it is the standard rotation matrix that rotates a vector by angle $x$ in the clockwise direction (or anticlockwise, depending on sign convention). When you multiply two rotation matrices, you are composing two rotations: rotating by $x$ and then by $y$ is the same as rotating by $x+y$ all at once. And since rotating by $x$ then $y$ is the same as rotating by $y$ then $x$ , the product commutes.

Let’s verify this algebraically.

Write down the product $P(x) \cdot P(y)$ . We have

P(x) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}, \quad P(y) = \begin{bmatrix}\cos y & \sin y\\ -\sin y & \cos y\end{bmatrix}.

Multiply them in the order $P(x)P(y)$ :

P(x)P(y) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix} \begin{bmatrix}\cos y & \sin y\\ -\sin y & \cos y\end{bmatrix}.

The entry in row 1, column 1 is $\cos x \cos y + \sin x (-\sin y) = \cos x \cos y - \sin x \sin y$ .

Row 1, column 2: $\cos x \sin y + \sin x \cos y = \sin x \cos y + \cos x \sin y$ .

Row 2, column 1: $(-\sin x)\cos y + \cos x (-\sin y) = -(\sin x \cos y + \cos x \sin y)$ .

Row 2, column 2: $(-\sin x)\sin y + \cos x \cos y = \cos x \cos y - \sin x \sin y$ .

Recognise the trigonometric identities. The expressions we just got are exactly the angle-sum formulas:

\cos(x+y) = \cos x \cos y - \sin x \sin y,

\sin(x+y) = \sin x \cos y + \cos x \sin y.

So the product becomes:

P(x)P(y) = \begin{bmatrix}\cos(x+y) & \sin(x+y)\\ -\sin(x+y) & \cos(x+y)\end{bmatrix} = P(x+y).

Now multiply in the reverse order $P(y)P(x)$ . …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ44 — If the matrix 0 a 3\2 b -1c 1 0 is a skew symmetric matrix, find the v…→Next questionQ46 — If A is square matrix such that A² = A, show that (I + A)³ = 7A + I.

Q.If $P(x) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}$ , then show that $P(x) \cdot P(y) = P(x + y) = P(y) \cdot P(x)$ .

Concept understanding — Rotation Matrix Composition

Rotation Matrix Composition

The Rule

A 2D Example

Why 3D Is Different

Method: Multiplying rotation matrices — angle addition and commutativity

Steps

Common Mistakes

Q.If P(x)=[cos⁡xsin⁡x−sin⁡xcos⁡x]P(x) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}P(x)=[cosx−sinx​sinxcosx​], then show that P(x)⋅P(y)=P(x+y)=P(y)⋅P(x)P(x) \cdot P(y) = P(x + y) = P(y) \cdot P(x)P(x)⋅P(y)=P(x+y)=P(y)⋅P(x).

🎓Unlock everything free for 14 days

Q.If $P(x) = \begin{bmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{bmatrix}$ , then show that $P(x) \cdot P(y) = P(x + y) = P(y) \cdot P(x)$ .

Unlock everything free for 14 days