If P = begin{bmatrix} cos frac{pi}{4} & -sin | vedclass

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

Question

(C) Given $P = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix} = \begin{bmatrix} \fr

Vedclass · Accepted Answer

$\begin{bmatrix} -1 \ 0 \end{bmatrix}$

Vedclass · Answer

(C) Given $P = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$.Note that $P$ is a rotation matrix $R_{	heta}$ where $	heta = \frac{\pi}{4}$.The property of a rotation matrix is $R_{	heta}^n = R_{n	heta}$.Therefore, $P^3 = R_{3 	imes \frac{\pi}{4}} = R_{\frac{3\pi}{4}} = \begin{bmatrix} \cos \frac{3\pi}{4} & -\sin \frac{3\pi}{4} \ \sin \frac{3\pi}{4} & \cos \frac{3\pi}{4} \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$.Now, $P^3 X = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} \end{bmatrix}$.$P^3 X = \begin{bmatrix} (-\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) \ (\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} 	imes \frac{1}{\sqrt{2}}) \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} - \frac{1}{2} \ \frac{1}{2} - \frac{1}{2} \end{bmatrix} = \begin{bmatrix} -1 \ 0 \end{bmatrix}$.

If $P = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix}$ and $X = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$, then $P^3 X$ is equal to:

Similar Questions

For the matrices $A$ and $B$,verify that $(AB)^{\prime} = B^{\prime}A^{\prime}$ where $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$.

If $A = \begin{vmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{vmatrix}$,then $1 + A^2 =$ . . . . . . .

If $A = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix}$ is such that $A^2 = I$,then

Compute the following: $\begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix}$

The number of all $3 \times 3$ matrices $A$,with entries from the set $\{-1, 0, 1\}$ such that the sum of the diagonal elements of $AA^{T}$ is $3$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module