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# difficult matrix question watch

1. A is 2x2 matrix =
1 0
1 3
B is 2x2 matrix =
-2 0
1 x

x does not = 0

b^-1*A*B = 2x2 matrix
y 0
0 z

find x,y,z
I got y =1, z =3 but for x i got every value that's not 0
2. Hi, it looks like what you did was correct. For your last step, how did you try and calculate x once you'd found Z?
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3. (Original post by BDunlop)
Hi, it looks like what you did was correct. For your last step, how did you try and calculate x once you'd found Z?
I got the matrix
1 0
(-6/2x + 6/2x) 3

Then i simplified it to get this

1 0
0/2x 3

then I was stuck
4. (Original post by Balkaran)
I got the matrix
1 0
(-6/2x + 6/2x) 3

Then i simplified it to get this

1 0
0/2x 3

then I was stuck
Did you work out the inverse of B? Is there perhaps an easier way to solve it?

If you were writing out B-1*A*B=x (where x is the matrix (y 0)/0 Z) what could you do to simplify it down so you don't need to start using fractions?
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5. (Original post by BDunlop)
Did you work out the inverse of B? Is there perhaps an easier way to solve it?

If you were writing out B-1*A*B=x (where x is the matrix (y 0)/0 Z) what could you do to simplify it down so you don't need to start using fractions?
x is the value we are trying to work out not a matrix.I'm just a bit confused sorry
6. (Original post by Balkaran)
x is the value we are trying to work out not a matrix.I'm just a bit confused sorry
Sorry the x I wrote isn't anything to do with the other x. I'll give you a hint.

You can rearrange an equation, for argument sake, B-1A=X to A=BX right? Do you understand how I did that? If you do try and apply that principle to your equation at the start and it'll be much easier to solve..
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7. (Original post by Balkaran)
find x,y,z
I got y =1, z =3 but for x i got every value that's not 0
I agree with your answer. y=1, z=3, and x can be anything non-zero.
8. (Original post by BDunlop)
Sorry the x I wrote isn't anything to do with the other x. I'll give you a hint.

You can rearrange an equation, for argument sake, B-1A=X to A=BX right? Do you understand how I did that? If you do try and apply that principle to your equation at the start and it'll be much easier to solve..
I'm sorry i'm lost.
9. (Original post by Balkaran)
I'm sorry i'm lost.
Okay, I would strongly recommend you watching a video on rearranging inverse matrices. Within 10 minutes you'll understand what I'm talking about and you'll find this question much easier do have a look
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10. (Original post by BDunlop)
Okay, I would strongly recommend you watching a video on rearranging inverse matrices. Within 10 minutes you'll understand what I'm talking about and you'll find this question much easier do have a look
could you tell me what you got for x just cause i'm probably gonna go sleep and I'll try working it out tomorrow
11. (Original post by Balkaran)
could you tell me what you got for x just cause i'm probably gonna go sleep and I'll try working it out tomorrow
You already have what I consider to be the correct answer, as I previously posted.

What BDunlop is suggesting, if I understand them correctly, is:

Let

Then premultiplying through by B, we have:

Evaluate both sides, and then since the two resulting matrices are equal, equate corresponding coefficients to get, in general, 4 simultaneous equations, to then solve.

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