JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 11)
Consider the matrix $f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$.
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 :
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 :
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Both Statement I and Statement II are true
Statement I is true but Statement II is false
Explanation
$$\begin{aligned} & f(-x)=\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(-x)=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I \end{aligned}$$
Hence statement- I is correct
Now, checking statement II
$$\begin{aligned} & f(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow f(x) \cdot f(y)=f(x+y) \end{aligned}$$
Hence statement-II is also correct.
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