Let quad P_1=I=left[begin{array}{lll}1 & 0 & | vedclass

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

Question

(B) Let $Q = \left[\begin{array}{lll}2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1\end{array}ight]$.$X = \sum_{k=1}^6 (P_k Q P_k^T)$.$X^T = \sum_{k=1}^6 (P_k

Vedclass · Accepted Answer

$2, 3, 4$

Vedclass · Answer

(B) Let $Q = \left[\begin{array}{lll}2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1\end{array}ight]$.$X = \sum_{k=1}^6 (P_k Q P_k^T)$.$X^T = \sum_{k=1}^6 (P_k Q P_k^T)^T = \sum_{k=1}^6 (P_k Q^T P_k^T) = \sum_{k=1}^6 (P_k Q P_k^T) = X$ (since $Q$ is symmetric).Thus,$X$ is a symmetric matrix. Option $(4)$ is correct.Let $R = \left[\begin{array}{l}1 \ 1 \ 1\end{array}ight]$. Note that for all permutation matrices $P_k$,$P_k R = R$ and $P_k^T R = R$.$X R = \sum_{k=1}^6 P_k Q P_k^T R = \sum_{k=1}^6 P_k Q R = (\sum_{k=1}^6 P_k) Q R$.$sum_{k=1}^6 P_k = \left[\begin{array}{lll}2 & 2 & 2 \ 2 & 2 & 2 \ 2 & 2 & 2\end{array}ight]$ and $Q R = \left[\begin{array}{lll}2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1\end{array}ight] \left[\begin{array}{l}1 \ 1 \ 1\end{array}ight] = \left[\begin{array}{l}6 \ 3 \ 6\end{array}ight]$.$X R = \left[\begin{array}{lll}2 & 2 & 2 \ 2 & 2 & 2 \ 2 & 2 & 2\end{array}ight] \left[\begin{array}{l}6 \ 3 \ 6\end{array}ight] = \left[\begin{array}{l}30 \ 30 \ 30\end{array}ight] = 30 R$. Thus,$\alpha = 30$. Option $(3)$ is correct.Since $X R = 30 R$,$(X - 30I) R = 0$,which implies $X - 30I$ is non-invertible. Option $(1)$ is incorrect.$	ext{Trace}(X) = \sum_{k=1}^6 	ext{Trace}(P_k Q P_k^T) = \sum_{k=1}^6 	ext{Trace}(P_k^T P_k Q) = \sum_{k=1}^6 	ext{Trace}(I Q) = 6 	imes 	ext{Trace}(Q) = 6 	imes (2+0+1) = 18$. Option $(2)$ is correct.Therefore,options $(2), (3), (4)$ are correct.

Similar Questions

Let $M = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $N = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$. Then $N M^{10} N^{-1} =$

Let the matrices $A$ and $B$ be defined as $A = \begin{bmatrix} 3 & 2 \\ 2 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 1 \\ 7 & 3 \end{bmatrix}$. Then the value of $\det(2A^9B^{-1})$ is:

Let $A$ and $B$ be real matrices of the form $\begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $\begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$,respectively.
Statement $1$: $AB - BA$ is always an invertible matrix.
Statement $2$: $AB - BA$ is never an identity matrix.

Let $A = \begin{bmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \text{adj}(A)$. If $\det(B) = 66$,then $\det(\text{adj}(A))$ equals:

If $S = \{x \in [0, 2\pi] : \begin{vmatrix} 0 & \cos x & -\sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \end{vmatrix} = 0\}$,then $\sum_{x \in S} \tan \left( \frac{\pi}{3} + x \right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module