Find AB, if A = begin{bmatrix} 0 & -1 0 & 2 | vedclass

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

Question

(A) Given $A = \begin{bmatrix} 0 & -1 \ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 5 \ 0 & 0 \end{bmatrix}$.To find $AB$,we perform matrix mu

Vedclass · Accepted Answer

$\begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$

Vedclass · Answer

(A) Given $A = \begin{bmatrix} 0 & -1 \ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 5 \ 0 & 0 \end{bmatrix}$.To find $AB$,we perform matrix multiplication:$AB = \begin{bmatrix} 0 & -1 \ 0 & 2 \end{bmatrix} \begin{bmatrix} 3 & 5 \ 0 & 0 \end{bmatrix}$$AB = \begin{bmatrix} (0 	imes 3) + (-1 	imes 0) & (0 	imes 5) + (-1 	imes 0) \ (0 	imes 3) + (2 	imes 0) & (0 	imes 5) + (2 	imes 0) \end{bmatrix}$$AB = \begin{bmatrix} 0 + 0 & 0 + 0 \ 0 + 0 & 0 + 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Thus,the product $AB$ is the zero matrix $\begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.

Find $AB,$ if $A = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$.

Similar Questions

If $A$ is an idempotent matrix,then $(I + A)^4$ is (where $I$ is the identity matrix of the same order as $A$)

The value of $\cos \alpha$ is . . . . . . if $A+A^{\prime}=I$,where $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.

In a third-order matrix $A$, $a_{ij}$ denotes the element in the $i$-th row and $j$-th column. If $a_{ij} = 0$ for $i = j$, $1$ for $i > j$, and $-1$ for $i < j$, then the matrix is:

If $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 0 \\ 1 & -2 \\ 0 & 3 \end{bmatrix}$,then $AB =$ . . . . . . .

If for $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$,$A^2 = I$,then . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module