For any 3 times 3 matrix M,let | M | denote t | vedclass

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

Question

(C) First,calculate $PEP$:$PEP = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 4 \ 8 & 13 &

Vedclass · Accepted Answer

$A, B, D$

Vedclass · Answer

(C) First,calculate $PEP$:$PEP = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 4 \ 8 & 13 & 18 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \ 8 & 13 & 18 \ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 2 \ 8 & 18 & 13 \ 2 & 4 & 3 \end{bmatrix} = F$.Also,$P^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$. Thus,$(A)$ is $TRUE$.For $(B)$,note that $|E| = 0$ and $|F| = 0$. Since $|EQ| = |E||Q| = 0$ and $|PFQ^{-1}| = |P||F||Q|^{-1} = 0$,the equation $|EQ + PFQ^{-1}| = 0 + 0 = 0$ holds. Thus,$(B)$ is $TRUE$.For $(C)$,$|EF| = |E||F| = 0 	imes 0 = 0$. Thus,$|(EF)^3| = 0$ and $|EF|^2 = 0$. The inequality $0 > 0$ is $FALSE$.For $(D)$,since $P^2 = I$,$P^{-1} = P$. Then $P^{-1}FP = PFP = P(PEP)P = P^2EP^2 = I E I = E$. Thus,$E + P^{-1}FP = 2E$. Also,$P^{-1}EP + F = PEP + F = F + F = 2F$. The sum of diagonal entries (Trace) of $2E$ is $2(1+3+18) = 44$,while the trace of $2F$ is $2(1+18+3) = 44$. Thus,$(D)$ is $TRUE$.

Similar Questions

Let $S = \left\{ \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix} : a, b \in \{1, 2, 3, \ldots, 100\} \right\}$ and let $T_n = \{A \in S : A^{n(n+1)} = I\}$. Then the number of elements in $\bigcap_{n=1}^{100} T_n$ is

Let $A$ be a square matrix such that $AA^T = I$. Then $\frac{1}{2} A[(A+A^T)^2 + (A-A^T)^2]$ is equal to

If a $2^{nd}$ order determinant with elements $0$ or $1$ is chosen from the set of all such determinants,find the probability that the determinant chosen is non-zero.

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix}$. Then the sum of the diagonal elements of the matrix $(A + I)^{11}$ is equal to:

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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module