If P = begin{bmatrix} i & 0 & -i 0 & -i & i | vedclass

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

Question

(B) Given $P = \begin{bmatrix} i & 0 & -i \ 0 & -i & i \ -i & i & 0 \end{bmatrix}$ and $Q = \begin{bmatrix} -i & i \ 0 & 0 \ i & -i \end{bmatrix}$

Vedclass · Accepted Answer

$\begin{bmatrix} 2 & -2 \ -1 & 1 \ -1 & 1 \end{bmatrix}$

Vedclass · Answer

(B) Given $P = \begin{bmatrix} i & 0 & -i \ 0 & -i & i \ -i & i & 0 \end{bmatrix}$ and $Q = \begin{bmatrix} -i & i \ 0 & 0 \ i & -i \end{bmatrix}$.To find $PQ$,we perform matrix multiplication:$PQ = \begin{bmatrix} i & 0 & -i \ 0 & -i & i \ -i & i & 0 \end{bmatrix} \begin{bmatrix} -i & i \ 0 & 0 \ i & -i \end{bmatrix}$$= \begin{bmatrix} i(-i) + 0(0) + (-i)(i) & i(i) + 0(0) + (-i)(-i) \ 0(-i) + (-i)(0) + i(i) & 0(i) + (-i)(0) + i(-i) \ (-i)(-i) + i(0) + 0(i) & (-i)(i) + i(0) + 0(-i) \end{bmatrix}$Since $i^2 = -1$,we have:$PQ = \begin{bmatrix} -i^2 + 0 - i^2 & i^2 + 0 + i^2 \ 0 + 0 + i^2 & 0 + 0 - i^2 \ i^2 + 0 + 0 & -i^2 + 0 + 0 \end{bmatrix} = \begin{bmatrix} -(-1) - (-1) & -1 + (-1) \ -1 & -(-1) \ -1 & -(-1) \end{bmatrix} = \begin{bmatrix} 1+1 & -1-1 \ -1 & 1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 \ -1 & 1 \ -1 & 1 \end{bmatrix}$.

If $P = \begin{bmatrix} i & 0 & -i \\ 0 & -i & i \\ -i & i & 0 \end{bmatrix}$ and $Q = \begin{bmatrix} -i & i \\ 0 & 0 \\ i & -i \end{bmatrix}$,then $PQ$ is equal to

Similar Questions

If $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$,then $(aI + bA)^n$ is (where $I$ is the identity matrix of order $2$)

Find the value of $x, y$ and $z$ from the following equation: $\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$

If $A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 2 & 3 & 0 & 0 \\ 4 & 5 & 6 & 0 \\ 7 & 8 & 9 & 10 \end{bmatrix}$,then $A$ is

If $A$ and $B$ are $n \times n$ square matrices such that $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$,then $A B A B=$

If $f(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}$,then $f\left(\frac{\pi}{6}\right) = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module