If$\begin{bmatrix}x 3\\ 0 2y-x\end{bmatrix} +\begin{bmatrix}1 2\\ 0 -2\end{bmatrix} =\begin{bmatrix}4 5\\ 0 3\end{bmatrix}$.

Find the value of x and y.


Given:

$\begin{bmatrix} x & 3\\ 0 & 2y-x \end{bmatrix} +\begin{bmatrix} 1 & 2\\ 0 & -2 \end{bmatrix} =\begin{bmatrix} 4 & 5\\ 0 & 3 \end{bmatrix}$

To do:

We have to find the values of x and y.

Solution:

$\begin{bmatrix} x & 3\\ 0 & 2y-x \end{bmatrix} +\begin{bmatrix} 1 & 2\\ 0 & -2 \end{bmatrix} =\begin{bmatrix} 4 & 5\\ 0 & 3 \end{bmatrix}$

This implies,

$\begin{bmatrix} x+1 & 3+2\\ 0+0 & 2y-x+( -2) \end{bmatrix} =\begin{bmatrix} 4 & 5\\ 0 & 3 \end{bmatrix}$

$\begin{bmatrix} x+1 & 5\\ 0 & 2y-x-2 \end{bmatrix} =\begin{bmatrix} 4 & 5\\ 0 & 3 \end{bmatrix}$

Therefore,

$x+1=4$

$x=4-1$

$x=3$

And,

$2y-x-2=3$

$2y-( 3) -2=3$

$2y=3+5$

$2y=8$

$y=\frac{8}{2}$

$y=4$.

The values of x and y are 3 and 4 respectively.

Tutorialspoint
Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

35 Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements