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# Stuck on Matrix Algebra Watch

Thanks
2. (Original post by AG27)

Thanks
It's been a while since I've done matrices, so this may be completely wrong, but unlike normal equations, where , this isn't true for matrices. So you've got to be careful about which side of the equation you apply inverses, and make sure you do the same to both.

So I think this is just a case of fiddling around.

...

Warning, I may be completely off
3. (Original post by Noble.)
It's been a while since I've done matrices, so this may be completely wrong, but unlike normal equations, where , this isn't true for matrices. So you've got to be careful about which side of the equation you apply inverses, and make sure you do the same to both.

So I think this is just a case of fiddling around.

...

Warning, I may be completely off
Yeah, I have tried fiddling around, just hasn't worked. The thing that annoys me is that B is in the middle of ABC, so it is difficult to multiply by B^-1.
4. (Original post by AG27)
Yeah, I have tried fiddling around, just hasn't worked. The thing that annoys me is that B is in the middle of ABC, so it is difficult to multiply by B^-1.
Ok, I've done it. I'll give you some more of the working.

5. Abc=i
abca=ia
a(-1)abca=a(-1)ia=i
b(-1)bca=b(-1)
ca=b(-1)
6. (Original post by Chardin)
Abc=i
abca=ia
a(-1)abca=a(-1)ia=i
b(-1)bca=b(-1)
ca=b(-1)
You're not suppose to post full solutions.

I so need that reply on a macro.
7. (Original post by AG27)
I will if you can help me solve my Maths problem.
Check chardins proff its correct
8. (Original post by Noble.)
Ok, I've done it. I'll give you some more of the working.

Thanks, would this finish it off?

CA = CC^(-1)B^(-1)
CA = IB^(-1)
CA = B^-1
9. (Original post by AG27)
Thanks, would this finish it off?

CA = CC^(-1)B^(-1)
CA = IB^(-1)
CA = B^-1
Yep.
10. I think a faster way would be:

11. Thank you, how will I go about finding the matrix of B given that I know that matrices for A and C?

A = (0 1 ) C = (2 1)
(-1 -6 ) (-3 -1)
12. (Original post by IchiCC)
I think a faster way would be:

Can you please explain how you got from the penultimate step to the last step?
13. (Original post by AG27)
Thank you, how will I go about finding the matrix of B given that I know that matrices for A and C?

A = (0 1 ) C = (2 1)
(-1 -6 ) (-3 -1)
Well you know

So you'd just need to multiply C with A, and find the inverse.
14. (Original post by Noble.)
Well you know

So you'd just need to multiply C with A, and find the inverse.
I've done exactly that and I get (3 -1)
(4 -1)

Apparently, this is wrong (according to my textbook) as the answer is (3 4)
(-1 -1)
15. (Original post by AG27)
Can you please explain how you got from the penultimate step to the last step?
If , then:

(Original post by AG27)
I've done exactly that and I get (3 -1)
(4 -1)

Apparently, this is wrong (according to my textbook) as the answer is (3 4)
(-1 -1)
16. (Original post by AG27)
I've done exactly that and I get (3 -1)
(4 -1)

Apparently, this is wrong (according to my textbook) as the answer is (3 4)
(-1 -1)

Did you get up to that?
17. (Original post by Noble.)

Did you get up to that?
Sorry for my late reply, I think what you have done is find AC instead of CA, surely the matrices should be multiplied the other way round?
18. (Original post by AG27)
Sorry for my late reply, I think what you have done is find AC instead of CA, surely the matrices should be multiplied the other way round?
Yep sorry, when I wrote down the matrices I wrote the values for A as C's instead.
19. (Original post by Noble.)
Yep sorry, when I wrote down the matrices I wrote the values for A as C's instead.
That's okay, did you get my outcome or the actual answer?
20. (Original post by AG27)
That's okay, did you get my outcome or the actual answer?

Did you get up to that?

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