## anonymous 4 years ago Solve the matrix equation for A?

1. anonymous

|dw:1327007851191:dw|

2. anonymous

Oh yay my friends are here

3. TuringTest

you need the inverse of the matrix on the left, do you know how to find that?

4. anonymous

I know how to do that but they want me to do a simpler method

5. TuringTest

...like what? maybe it would help if I could actually see the matrix on the right. try to use the equation feature to type it, you can make 2X2 matrices with it.

6. anonymous

|dw:1327008078673:dw|

7. TuringTest

is the right side the identity matrix?

8. anonymous

ya but we didnt really learn that yet. We only learnt multiplying and addition and subtraction

9. TuringTest

I can't see the matrix on the RHS is my problem what is it, the identity matrix?

10. anonymous

Like maybe a is a 2 by 2 matrix

11. anonymous

ohhh i see it okk i will redraw it

12. anonymous

|dw:1327008221822:dw|

13. amistre64

since BA=I wouldnt A=B^-1 I ?

14. anonymous

I know but we didnt really use inverse matrices yet But if i cant solve it any other way i will do it that way

15. TuringTest

...so that is the identity matrix so we could solve this by multiplying out our matrix A with the one on the LHS and get a system of equations:$2a_{11}-a_{21}=1$$3a_{11}-2a_{21}=0$$2a_{12}-a_{22}=0$$3a_{12}-2a_{22}=1$

16. amistre64

$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{det}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$

17. TuringTest

I only did it this way amistre because they said not to use the inverse... but yes, the inverse seems the smarter/easier way to do it.

18. amistre64

youre way is fine, I just did that for the practice really :)

19. anonymous

ok Thanks

20. amistre64

also, i think the GaussJordan comes up with an inverse as well

21. anonymous

Ya it does :D Thanks guys