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Homework Help: Properties of Inverse Matrix

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Determine which of the formulas hold for all invertible nhttp://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmsy10/alpha/144/char02.png [Broken]n matrices A andB

    A. 7A is invertible
    B. ABA^−1=B
    C. A+B is invertible
    D. (A+B)2=A2+B2+2AB
    E. (A+A^−1)^8=A8+A−8
    F. (ABA^−1)^3=AB3A−1

    2. Relevant equations

    3. The attempt at a solution

    I think 1 should be true and c should be true. Those are for certain. The rest, I don't know how.
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 11, 2014 #2


    Staff: Mentor

    Your efforts just barely qualify as an attempt at a solution...

    For A (not 1), you are correct. Can you figure out what the inverse is?
    For C, consider these matrices:
    $$A = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$
    $$B = \begin{bmatrix} -1 & 0 \\ 0 & -1\end{bmatrix}$$
    Both A and B are invertible (clearly, I hope). Is their sum invertible?

    For the others, try expanding what's on the left side of the given equation, and see if you get what's on the right side.

    Your book should have some properties or theorems of invertible matrices.
    Last edited by a moderator: May 7, 2017
  4. Oct 11, 2014 #3
    Still having trouble. B is true. C is false. D is false after expanding. Don't know how to expand e and f.
  5. Oct 11, 2014 #4


    Staff: Mentor

    For E you have "(A+A^−1)^8=A8+A−8"
    Should the right side be A8 + A-8?
    At the very least, use ^ to indicate exponents.

    For E, how do you expand a binomial?
    For F, what properties do you know of to help with expanding ABA-1 to the third power?
  6. Oct 11, 2014 #5
    So after like 100x, I finally got the answer. Turns out A and F are the only true ones. I don't understand why though.
  7. Oct 11, 2014 #6


    Staff: Mentor

    Several of the problems test your understanding of matrix multiplication. In particular, that multiplication isn't commutative, so in general, AB ##\neq## BA.
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