## anonymous one year ago Medal and Fan! Which product represents the solution to the system?

1. anonymous

2. anonymous

Do you know how to multiple matrices?

3. anonymous

4. anonymous

yes that is correct, however, thats a lil complicated how you've done it.

5. anonymous

oh wait, you have just used the inverse formula, i see.

6. anonymous

i wouldn't leave it in this form , i would simplify this further

7. anonymous

because a solution to the system involves values for x and y, here you have just simply solved the inverse.

8. anonymous

$AX=B$ $A ^{-1}AX=A ^{-1} B$ $IX=A ^{-1}B$ $X=A ^{1}B$

9. anonymous

that should be A^(-1) in the last part sorry.

10. anonymous

you have solved A^(-1)|dw:1443662042771:dw|

11. anonymous

12. anonymous

i would simplify it further $X=\left[\begin{matrix}2 & -1/2 \\ -1 & 1/2\end{matrix}\right]\left(\begin{matrix}3 \\ 2\end{matrix}\right)=\left(\begin{matrix}5 \\ -2\end{matrix}\right)$

13. anonymous

here, this simplification says that the solution to this system is x=5, y=-2

14. anonymous

so you don't get confused, $X=\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}5 \\ 2\end{matrix}\right)$

15. anonymous

im not sure if you could see the first part of it but theres a 1/2 in fornt of it

16. anonymous

@chris00

17. anonymous

i did, i just simplified it into the matrix

18. anonymous

$\left[\begin{matrix}2 & -1/2 \\ -1& 1/2\end{matrix}\right]=\frac{ 1 }{ 2 }\left[\begin{matrix}4 & -1 \\ -2 & 1\end{matrix}\right]$