For matrices X and Y,if X+Y = begin{bmatrix} | vedclass

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

Question

(A) Given the system of equations for matrices $X$ and $Y$: $(1)$ $X + Y = \begin{bmatrix} 7 & 0 \ 2 & 5 \end{bmatrix}$ $(2)$ $X - Y = \begin{bmatrix

Vedclass · Accepted Answer

$\begin{bmatrix} 10 & 0 \ 2 & 8 \end{bmatrix}$

Vedclass · Answer

(A) Given the system of equations for matrices $X$ and $Y$: $(1)$ $X + Y = \begin{bmatrix} 7 & 0 \ 2 & 5 \end{bmatrix}$ $(2)$ $X - Y = \begin{bmatrix} 3 & 0 \ 0 & 3 \end{bmatrix}$ To find $2X$,we add equation $(1)$ and equation $(2)$: $(X + Y) + (X - Y) = \begin{bmatrix} 7 & 0 \ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \ 0 & 3 \end{bmatrix}$ $2X = \begin{bmatrix} 7+3 & 0+0 \ 2+0 & 5+3 \end{bmatrix}$ $2X = \begin{bmatrix} 10 & 0 \ 2 & 8 \end{bmatrix}$ Thus,the correct option is $A$.

For matrices $X$ and $Y$,if $X+Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$ and $X-Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$,then $2X =$ . . . . . .

Similar Questions

If $A$ and $B$ are square matrices of the same order such that $AB = BA$,then prove by induction that $AB^{n} = B^{n}A$. Further,prove that $(AB)^{n} = A^{n}B^{n}$ for all $n \in N$.

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

Which of the following statements is not correct?

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$,then $A$ is

Which of the given values of $x$ and $y$ make the following pair of matrices equal?
$\left[\begin{array}{cc}3x+7 & 5 \\ y+1 & 2-3x\end{array}\right]=\left[\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module