Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    Can someone please help me prove that B^-1 = CA given that ABC = I

    Thanks
    Offline

    17
    ReputationRep:
    (Original post by AG27)
    Can someone please help me prove that B^-1 = CA given that ABC = I

    Thanks
    It's been a while since I've done matrices, so this may be completely wrong, but unlike normal equations, where ab=ba, 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.

    ABC = I
    ABCC^{-1} = IC^{-1}
    AB = C^{-1}
    ...

    Warning, I may be completely off
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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 ab=ba, 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.

    ABC = I
    ABCC^{-1} = IC^{-1}
    AB = C^{-1}
    ...

    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.
    Offline

    17
    ReputationRep:
    (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.

    ABC = I
    ABCC^{-1} = IC^{-1}
    AB = C^{-1}
    ABB^{-1} = C^{-1}B^{-1}
    A = C^{-1}B^{-1}
    Offline

    0
    ReputationRep:
    Abc=i
    abca=ia
    a(-1)abca=a(-1)ia=i
    b(-1)bca=b(-1)
    ca=b(-1)
    Offline

    17
    ReputationRep:
    (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. :facepalm:


    I so need that reply on a macro.
    Offline

    14
    ReputationRep:
    (Original post by AG27)
    I will if you can help me solve my Maths problem.
    Check chardins proff its correct
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Noble.)
    Ok, I've done it. I'll give you some more of the working.

    ABC = I
    ABCC^{-1} = IC^{-1}
    AB = C^{-1}
    ABB^{-1} = C^{-1}B^{-1}
    A = C^{-1}B^{-1}
    Thanks, would this finish it off?

    CA = CC^(-1)B^(-1)
    CA = IB^(-1)
    CA = B^-1
    Offline

    17
    ReputationRep:
    (Original post by AG27)
    Thanks, would this finish it off?

    CA = CC^(-1)B^(-1)
    CA = IB^(-1)
    CA = B^-1
    Yep.
    Offline

    2
    ReputationRep:
    I think a faster way would be:


    ABC = I

    ABCC^{-1} = IC^{-1}

    AB = C^{-1}

    A^{-1}AB=A^{-1}C^{-1}

    B=(CA)^{-1}

    B^{-1}=CA
    • Thread Starter
    Offline

    0
    ReputationRep:
    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)
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by IchiCC)
    I think a faster way would be:


    ABC = I

    ABCC^{-1} = IC^{-1}

    AB = C^{-1}

    A^{-1}AB=A^{-1}C^{-1}

    B=(CA)^{-1}

    B^{-1}=CA
    Can you please explain how you got from the penultimate step to the last step?
    Offline

    17
    ReputationRep:
    (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

    B=(CA)^{-1}

    So you'd just need to multiply C with A, and find the inverse.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Noble.)
    Well you know

    B=(CA)^{-1}

    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)
    Offline

    2
    ReputationRep:
    (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)
    Check your working. I just tried it and got the same answer as your textbook.
    Offline

    17
    ReputationRep:
    (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)

    B=(CA)^{-1}
     B = [\begin{pmatrix} 0 & 1 \\-1 & -6 \end{pmatrix} \times \begin{pmatrix} 2 & 1 \\-3 & -1 \end{pmatrix}]^{-1}

     B = [\begin{pmatrix} 0(2) + 1(-3) & 0(1) + 1(-1) \\-1(2) + -6(-3) & -1(1) + -6(-1) \end{pmatrix}]^{-1}

     B = [\begin{pmatrix} -3 & -1 \\16 & 5 \end{pmatrix}]^{-1}

    Did you get up to that?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Noble.)
    B=(CA)^{-1}
     B = [\begin{pmatrix} 0 & 1 \\-1 & -6 \end{pmatrix} \times \begin{pmatrix} 2 & 1 \\-3 & -1 \end{pmatrix}]^{-1}

     B = [\begin{pmatrix} 0(2) + 1(-3) & 0(1) + 1(-1) \\-1(2) + -6(-3) & -1(1) + -6(-1) \end{pmatrix}]^{-1}

     B = [\begin{pmatrix} -3 & -1 \\16 & 5 \end{pmatrix}]^{-1}

    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?
    Offline

    17
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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?
    Offline

    17
    ReputationRep:
    (Original post by AG27)
    That's okay, did you get my outcome or the actual answer?
    B=(CA)^{-1}
     B = [\begin{pmatrix} 2 & 1 \\-3 & -1 \end{pmatrix} \times \begin{pmatrix} 0 & 1 \\-1 & -6 \end{pmatrix}]^{-1}

     B = [\begin{pmatrix} -1 & -4 \\1 & 3 \end{pmatrix}]^{-1}

    Did you get up to that?

    I get the actual answer, so you've made a mistake somewhere
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.