Consider the matrix f(x) = begin{bmatrix} cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To check Statement $I$: We find $f(-x)$ by replacing $x$ with $-x$ in $f(x)$.$f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \ \sin(-x) & \cos(

Vedclass · Accepted Answer

Both Statement $I$ and Statement $II$ are true.

Vedclass · Answer

(D) To check Statement $I$: We find $f(-x)$ by replacing $x$ with $-x$ in $f(x)$.$f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \ \sin(-x) & \cos(-x) & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \ -\sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix}$.Now,calculate $f(x) \cdot f(-x)$:$f(x) \cdot f(-x) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \ -\sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos^2 x + \sin^2 x & 0 & 0 \ 0 & \sin^2 x + \cos^2 x & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.Since $f(x) \cdot f(-x) = I$,$f(-x)$ is the inverse of $f(x)$. Thus,Statement $I$ is true.To check Statement $II$: Calculate $f(x) \cdot f(y)$:$f(x) \cdot f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \ \sin x & \cos x & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \ \sin y & \cos y & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x \cos y - \sin x \sin y & -(\cos x \sin y + \sin x \cos y) & 0 \ \sin x \cos y + \cos x \sin y & \cos x \cos y - \sin x \sin y & 0 \ 0 & 0 & 1 \end{bmatrix}$.Using trigonometric identities $\cos(x+y) = \cos x \cos y - \sin x \sin y$ and $\sin(x+y) = \sin x \cos y + \cos x \sin y$,we get:$f(x) \cdot f(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \ \sin(x+y) & \cos(x+y) & 0 \ 0 & 0 & 1 \end{bmatrix} = f(x+y)$.Thus,Statement $II$ is true.

Similar Questions

If $a, b, c$ and $d$ are complex numbers,then the determinant $\Delta = \begin{vmatrix} 2 & a+b+c+d & ab+cd \\ a+b+c+d & 2(a+b)(c+d) & ab(c+d)+cd(a+b) \\ ab+cd & ab(c+d)+cd(a+b) & 2abcd \end{vmatrix}$ is

If $\omega$ is an imaginary cube root of unity,then the value of $\left[\begin{array}{ccc}1 & \omega^{2} & 1-\omega^{4} \\ \omega & 1 & 1+\omega^{5} \\ 1 & \omega & \omega^{2}\end{array}\right]$ is

If $P$ and $Q$ are square matrices such that $P^{2006} = O$ and $PQ = P + Q$,then $\det(Q)$ will be

If $A+B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2\end{array}\right]$ and $AB=\left[\begin{array}{lll}1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1\end{array}\right]$, then $A^2+B(A+B)=$

Let $A = \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix}$. If $B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} A \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}$,then the sum of all the elements of the matrix $\sum_{n=1}^{50} B^n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module