Consider the matrix f(x) = [[cos x, -sin x, 0], [sin x, cos x, 0], [0, 0, 1]].
Given below are two s

Q: Wait, why is the third row and column just 0, 0, 1? Does it matter?

That is a brilliant observation! That structure indicates that this is a 3D rotation matrix where the rotation is occurring strictly around the $z$-axis. The $z$-coordinate remains unchanged, which is why you see the identity-like values in the third row and column. It is essentially a 2D rotation embedded in 3D space.

Q: Can I just use the determinant to check if it is invertible?

You absolutely can! The determinant of this matrix is $\cos^2 x + \sin^2 x = 1$. Since the determinant is non-zero for all $x$, the matrix is indeed invertible. However, Statement I asks specifically if $f(-x)$ is the inverse, so you must perform the multiplication $f(x)f(-x)$ to confirm it results in the identity matrix $I$.

Q: Why does $f(x)f(y) = f(x+y)$ hold true? Is it always true for any matrix?

Not at all! Matrix multiplication is generally non-commutative. This property holds here because these matrices belong to a special group of rotation matrices. They represent the same physical operation—rotation—and performing two successive rotations is equivalent to a single rotation by the sum of the angles.

Q: I keep getting the signs wrong when multiplying the rows and columns. Any tips?

The secret is to be methodical. When you multiply $R_1$ by $C_2$, you are calculating $\cos x(-\sin y) + (-\sin x)(\cos y)$. Factor out the negative sign, and you will immediately see the structure of the sine addition formula: $-(\sin x \cos y + \cos x \sin y)$. Patience is your best tool here!

Q: Is there a shortcut to avoid all this matrix multiplication?

In an exam, speed is key, but understanding is the foundation. Once you recognize that $f(x)$ represents a rotation operator $R(x)$, you can intuitively see that $R(x)R(y) = R(x+y)$ because rotating by $y$ and then by $x$ is physically identical to rotating by $x+y$. Use the algebra to verify your intuition, but let the geometry guide your thinking!

Question

Consider the matrix $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.
Given below are two statements :
Statement I: $f(-x)$ is the inverse of the matrix $f(x)$.
Statement II: $f(x)f(y) = f(x + y)$.
In the light of the above statements, choose the correct answer from the options given below

Options:

(A) Statement I is false but Statement II is true

(B) Both Statement I and Statement II are false

(C) Statement I is true but Statement II is false

(D) Both Statement I and Statement II are true

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a matrix problem; we are peering into the elegant machinery of space itself.

When you look at the matrix $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$, do not see a static array of numbers. See a physical action. This matrix is a rotation operator that takes any vector in the $xy$-plane and pivots it by an angle $x$ around the $z$-axis.

### Statement I: The Inverse as a Return Journey

Statement I asks us to consider $f(-x)$ as the inverse of $f(x)$. Think about this geometrically: if $f(x)$ rotates a point forward by $x$, the inverse operation must be rotating the point back by $x$, which is equivalent to rotating by $-x$.

To prove this algebraically, we substitute $-x$ into our matrix. Using the parity properties of trigonometry, where $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$, we find:

$$f(-x) = \begin{bmatrix} \cos x & \sin x & 0 \ -\sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$$

Now, we verify if $f(x) \cdot f(-x) = I$. When we multiply these, the top-left element becomes $\cos^2 x + \sin^2 x$, which is $1$. The off-diagonal elements become $\cos x \sin x - \sin x \cos x$, which is $0$.

The identity matrix emerges from the chaos like a lighthouse in the fog. **Statement I is confirmed**: $f(-x)$ is indeed the inverse.

### Statement II: The Composition of Rotations

Now, let us tackle Statement II: $f(x)f(y) = f(x+y)$. This is the heart of group theory hidden in plain sight. If you rotate a coordinate system by $y$ and then rotate it again by $x$, the total rotation is simply $x+y$.

Let us perform the multiplication $f(x) \cdot f(y)$:

$$f(x) \cdot f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \ \sin y & \cos y & 0 \ 0 & 0 & 1 \end{bmatrix}$$

As we multiply the first row by the first column, we get $\cos x \cos y - \sin x \sin y$. This is the classic cosine addition formula: $\cos(x+y)$.

Similarly, the first row multiplied by the second column gives us $-(\sin x \cos y + \cos x \sin y)$, which is $-\sin(x+y)$. When you complete the multiplication for the second row, you find the exact same structure.

The resulting matrix is:

$$f(x+y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \ \sin(x+y) & \cos(x+y) & 0 \ 0 & 0 & 1 \end{bmatrix}$$

### The Final Synthesis

We have walked through the geometry and the algebra. We have seen that $f(-x)$ acts as the inverse because it reverses the rotation, and we have seen that the composition of two rotations is simply the sum of their angles.

**Both statements stand firm.**

Remember, in JEE Advanced, the math is just the language; the physics is the story. You have mastered the story today. Keep this clarity of vision, and no matrix will ever intimidate you again. You are ready.

Select Answer:

Visualized Solution

Defining the Matrix $f (x)$

Visualizing the Inverse

Substituting $- x$ into the Matrix

The Inverse Condition $A \cdot A^{- 1} = I$

Matrix Multiplication

Visualizing $f (x) f (y)$

Algebraic Setup for $f (x) \cdot f (y)$

Multiplying Row 1

The Addition Formulas

Finalizing $f (x) \cdot f (y)$

Final Answer

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

Statement I

Statement II

The Final Synthesis

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List-I

List-II