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Please help, do not know what this type of problem is called or how to solve it!

  1. Sep 18, 2012 #1
    Suppose that A is an nxn matrix which satisfies the equation

    A3 - 2A2 + 3A - I = 0

    Show that the nxn matrix B defined by the equation B = A2 - 2A +3I is invertible and find its inverse.

    Does anyone have any idea what this type of problem is called, and what steps you take to solve it?
     
  2. jcsd
  3. Sep 18, 2012 #2

    hotvette

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    It's a matrix algebra (linear algebra) problem. Try factoring it like you would a normal algebra problem. You'll also need to recall the definition of a matrix inverse to solve the problem.
     
  4. Sep 18, 2012 #3
    can you be more specific? for some reason i just can't see how to do this, i've literally spent over 2 hours attempting to solve it.
     
  5. Sep 18, 2012 #4

    hotvette

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    You need to tell us what you've tried and where you are stuck.
     
  6. Sep 18, 2012 #5
    i think what's messing me up is the fact that there are equations involved. i understand how to do the inverse of a matrix, but I can't seem to connect the dots between the idea of a matrix and how it relates to the given equations.
     
  7. Sep 18, 2012 #6

    jbunniii

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    Try to recognize how the two equations are related to each other.
     
  8. Sep 18, 2012 #7

    hotvette

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    OK, so the conceptual block is the fact that matrices can be combined with operators to make expressions. The rules are almost the same as with normal algebra. The following is a matrix algebra equation that says a matrix B added to a matrix A equals matrix C:

    A + B = C

    Matrices can also be multiplied by scalars (meaning each element of the matrix is multiplied by the scalar):

    A + 2B = C

    Here are examples of an expressions using matrix multiplication (and addition):

    AB = C
    AB + C = D
    A2 = AA
    AB = I

    In the last case, what can you say about B?

    Also, matrix mutliplication is distributive:

    A(B + C) = AB + AC

    Is none of the above familiar?
     
  9. Sep 18, 2012 #8
    I recognize all that. In the last case, B would be equal to the inverse of A. I know there is just something so small that I am missing, and I know I'll feel like an idiot when I see it. do i have to solve for A in the first equation so that I have an A= and a B=? idk i'm just frustrated, this stuff usually comes so easily to me, for the first time I understand what it's like to not be good at math.
     
  10. Sep 18, 2012 #9

    hotvette

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    Suggestion: solve the first equation for I and factor the opposite side.
     
  11. Sep 18, 2012 #10
    Let's just abuse some notation for a minute and say [itex]I[/itex] is [itex]1[/itex]. Carrying this out would give you

    [tex]\begin{align*}
    & A^3 - 2A^2 +3A - 1 = 0 \\
    & A^2 - 2A + 3 = B
    \end{align*}
    [/tex]

    Do you think you can make a simple substitution here?
     
  12. Sep 18, 2012 #11
    you can factor out an A from the first equation and be left with A(B) = 1, am I on the right track?
     
  13. Sep 18, 2012 #12

    hotvette

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    Yes. And what does that tell you about B?
     
  14. Sep 18, 2012 #13
    that it's equal to the inverse of A.
     
  15. Sep 18, 2012 #14

    hotvette

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    Yep. Now you should be in a position to answer the original questions.
     
  16. Sep 18, 2012 #15
    okay I wanna say I get it but I think my issue here is I don't know what format answer I am looking for. Is the inverse just going to be in the form of an equation as well?
     
  17. Sep 18, 2012 #16

    Ray Vickson

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    What is wrong with saying that the inverse of B is A? After all, that is what you have just shown, and it does answer the original question completely.

    RGV
     
  18. Sep 18, 2012 #17
    yeah that would work, I guess it's one of those times the simplest answer is the right answer. thanks for all your help.
     
  19. Sep 18, 2012 #18

    HallsofIvy

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    From [itex]A^3 - 2A^2 + 3A - I = 0[/itex] we can get [itex]A^3- 2A^2+ 3A= I[/itex] and then [itex]A(A^2- 2A+ 3)= I[/itex]. Do you see the point?
     
  20. Sep 18, 2012 #19

    Curious3141

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    Yes, you're on the right track, although that should be AB = I.

    Now postmultiply both sides by [itex]B^{-1}[/itex].
     
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