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    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
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    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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    (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
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    (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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    (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
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    (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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    (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.
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    (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.
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    (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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    (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
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    (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 Y=B^{-1}AB

    Then premultiplying through by B, we have:

    BY=AB

    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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