Find the value of a, b, c, and d from the equ | vedclass

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

Question

(B) Given the matrix equation: $\begin{bmatrix} a-b & 2a+c \ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \ 0 & 13 \end{bmatrix}$Since the two

Vedclass · Accepted Answer

$a=1, b=2, c=3, d=4$

Vedclass · Answer

(B) Given the matrix equation: $\begin{bmatrix} a-b & 2a+c \ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \ 0 & 13 \end{bmatrix}$Since the two matrices are equal,their corresponding elements must be equal.Comparing the corresponding elements,we get:$a-b = -1$  $(1)$$2a-b = 0$  $(2)$$2a+c = 5$  $(3)$$3c+d = 13$  $(4)$From equation $(2)$,we have $b = 2a$.Substituting $b = 2a$ into equation $(1)$:$a - 2a = -1$$-a = -1 \Rightarrow a = 1$Now,find $b$ using $b = 2a$:$b = 2(1) = 2$Substitute $a = 1$ into equation $(3)$:$2(1) + c = 5$$2 + c = 5 \Rightarrow c = 3$Substitute $c = 3$ into equation $(4)$:$3(3) + d = 13$$9 + d = 13 \Rightarrow d = 4$Therefore,the values are $a=1, b=2, c=3,$ and $d=4$.

Find the value of $a, b, c,$ and $d$ from the equation: $\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$

Similar Questions

If $A = \begin{vmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{vmatrix}$,then $1 + A^2 =$ . . . . . . .

The number of matrices of order $3 \times 3$,whose entries are either $0$ or $1$ and the sum of all the entries is a prime number,is:

If $A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$,then verify that $A^{\prime} A = I$.

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

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -5 & 7 & 1 \\ 1 & -5 & 7 \\ 7 & 1 & -5 \end{bmatrix}$,then $AB$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module