If A = begin{bmatrix} 0 & 2 3 & -4 end{bmatr | vedclass

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

Question

(C) Given,$kA = \begin{bmatrix} 0 & 3a \ 2b & 24 \end{bmatrix}$.Substituting $A$,we get $k \begin{bmatrix} 0 & 2 \ 3 & -4 \end{bmatrix} = \begin{bma

Vedclass · Accepted Answer

$ -6, -4, -9$

Vedclass · Answer

(C) Given,$kA = \begin{bmatrix} 0 & 3a \ 2b & 24 \end{bmatrix}$.Substituting $A$,we get $k \begin{bmatrix} 0 & 2 \ 3 & -4 \end{bmatrix} = \begin{bmatrix} 0 & 3a \ 2b & 24 \end{bmatrix}$.Multiplying $k$ inside the matrix: $\begin{bmatrix} 0 & 2k \ 3k & -4k \end{bmatrix} = \begin{bmatrix} 0 & 3a \ 2b & 24 \end{bmatrix}$.Comparing the corresponding elements:$1) -4k = 24 \implies k = -6$.$2) 2k = 3a \implies 2(-6) = 3a \implies -12 = 3a \implies a = -4$.$3) 3k = 2b \implies 3(-6) = 2b \implies -18 = 2b \implies b = -9$.Thus,the values are $k = -6, a = -4, b = -9$.

If $A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$ and $kA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$,then the values of $k, a, b$ are respectively

Similar Questions

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

Let $A$ be a $3 \times 3$ matrix having entries from the set $\{-1, 0, 1\}$. The number of all such matrices $A$ having the sum of all the entries equal to $5$ is:

If $A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$,then the value of $\alpha$ for which $A^2 = B$ is:

$A$ matrix whose elements $a_{ij}$ are defined by $a_{ij} = \frac{1}{3}|i - 5j|$,where $i, j = 1, 2, 3$,is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module