# Thread: Need help solving for 'x' and 'y' in matrices...

1. ## Need help solving for 'x' and 'y' in matrices...

Here is the equation:

[3 2 -1]
[-2 0 2]

multiplied by

[2 x]
[-1 4]
[3 y]

equals

[1 0]
[2 4]

instructions call to solve for x and y but i can seem to be able to find them, please help me. . .

2. Originally Posted by runner940
Here is the equation:

[3 2 -1]
[-2 0 2]

multiplied by

[2 x]
[-1 4]
[3 y]

equals

[1 0]
[2 4]

instructions call to solve for x and y but i can seem to be able to find them, please help me. . .
Once you multiply the matrices together, compare the values of the elements...

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

Multiplying the two matrices together yields the equation $\displaystyle \begin{bmatrix}1&3x+8-y\\2&-2x+2y\end{bmatrix}=\begin{bmatrix}1&0\\2&4\end{bma trix}$

In comparing the elements, the only system that we need to solve is $\displaystyle \left\{\begin{array}{rcrcr}3x&-&y&=&-8\\-x&+&y&=&2\end{array}\right.$

Do you think that you can find x and y now?

--Chris

3. when you multiply those matrices,you get 2*2 as dimensions of the answer which is
1 3x-y+8 = 1 0
2 -2x+2y = 2 4

which means that 3x-y=-8 and -2x+2y=4
-x+y=2 y=x+2
3x-x-2=2

2x=4

x=2

y=2+2=4

S=(2,4)

4. Originally Posted by franckherve1
when you multiply those matrices,you get 2*2 as dimensions of the answer which is
1 3x-y+8 = 1 0
2 -2x+2y = 2 4

which means that 3x-y=-8 and -2x+2y=4
-x+y=2 y=x+2
3x-x-2=2

2x=4

x=2

y=2+2=4

S=(2,4)
This solution is difficult to follow. The answer given is incorrect.