JEE Main 2024 — 27 January, Shift 1 — Question 17

Given the matrix:

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

Evaluating Statement I: $f(-x)$ is the inverse of the matrix $f(x)$. First, we find $f(-x)$ by substituting $-x$ for $x$ in $f(x)$. Using the trigonometric identities $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$:

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

Now, we multiply $f(x)$ by $f(-x)$:

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

Performing the matrix multiplication:

$$\begin{aligned} (f(x)f(-x))_{11} &= (\cos x)(\cos x) + (-\sin x)(-\sin x) + (0)(0) = \cos^2 x + \sin^2 x = 1 \\ (f(x)f(-x))_{12} &= (\cos x)(\sin x) + (-\sin x)(\cos x) + (0)(0) = \cos x \sin x - \sin x \cos x = 0 \\ (f(x)f(-x))_{13} &= (\cos x)(0) + (-\sin x)(0) + (0)(1) = 0 \\ (f(x)f(-x))_{21} &= (\sin x)(\cos x) + (\cos x)(-\sin x) + (0)(0) = \sin x \cos x - \cos x \sin x = 0 \\ (f(x)f(-x))_{22} &= (\sin x)(\sin x) + (\cos x)(\cos x) + (0)(0) = \sin^2 x + \cos^2 x = 1 \\ (f(x)f(-x))_{23} &= (\sin x)(0) + (\cos x)(0) + (0)(1) = 0 \\ (f(x)f(-x))_{31} &= (0)(\cos x) + (0)(-\sin x) + (1)(0) = 0 \\ (f(x)f(-x))_{32} &= (0)(\sin x) + (0)(\cos x) + (1)(0) = 0 \\ (f(x)f(-x))_{33} &= (0)(0) + (0)(0) + (1)(1) = 1 \end{aligned}$$

So,

$$f(x)f(-x) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I$$

Thus, Statement I is true.

Evaluating Statement II: $f(x)f(y) = f(x+y)$. First, we write $f(y)$:

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

Now, we multiply $f(x)$ by $f(y)$:

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

Performing the matrix multiplication:

$$\begin{aligned} (f(x)f(y))_{11} &= (\cos x)(\cos y) + (-\sin x)(\sin y) + (0)(0) = \cos x \cos y - \sin x \sin y = \cos(x+y) \\ (f(x)f(y))_{12} &= (\cos x)(-\sin y) + (-\sin x)(\cos y) + (0)(0) = -(\cos x \sin y + \sin x \cos y) = -\sin(x+y) \\ (f(x)f(y))_{13} &= (\cos x)(0) + (-\sin x)(0) + (0)(1) = 0 \\ (f(x)f(y))_{21} &= (\sin x)(\cos y) + (\cos x)(\sin y) + (0)(0) = \sin x \cos y + \cos x \sin y = \sin(x+y) \\ (f(x)f(y))_{22} &= (\sin x)(-\sin y) + (\cos x)(\cos y) + (0)(0) = -\sin x \sin y + \cos x \cos y = \cos(x+y) \\ (f(x)f(y))_{23} &= (\sin x)(0) + (\cos x)(0) + (0)(1) = 0 \\ (f(x)f(y))_{31} &= (0)(\cos y) + (0)(\sin y) + (1)(0) = 0 \\ (f(x)f(y))_{32} &= (0)(-\sin y) + (0)(\cos y) + (1)(0) = 0 \\ (f(x)f(y))_{33} &= (0)(0) + (0)(0) + (1)(1) = 1 \end{aligned}$$

So,

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

By definition, $f(x+y)$ is:

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

Thus, $f(x)f(y) = f(x+y)$. Statement II is true.

Since both Statement I and Statement II are true.

The final answer is Both Statement I and Statement II are true.