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Linear Algebra; AB=AC

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Let A=
    [-1 4
    3 -12]
    Find two 2x2 matrices B and C such that AB=AC but B does not equal C .

    2. Relevant equations

    3. The attempt at a solution

    I was going through my book, and am a bit confused with this problem. How would I solve this?

    I know it's easy to prove AB = AC if B = C, as you could cancel the A's out by multiplying A^(-1) on each side if the matrix was invertible... but I'm not sure how to solve for the matrices B and C, given A if B cannot equal C. Any suggestions on how to begin this problem would be very helpful. Thanks!
  2. jcsd
  3. Nov 17, 2008 #2


    Staff: Mentor

    The matrix you are given isn't invertible, so you won't be able to solve for B and C by multiplying AB and AC on the left by the inverse of A.

    Let B be a 2 x 2 matrix whose entries are (a, b) (top row) and (c d) (bottom row).
    Then A*B = (-a + 4c, -b + 4d) (top row) and (3a - 12c, 3b - 12d) (bottom row).

    Can you find values for a, b, c, and d, so that AB is the zero matrix?
    Can you find another set of values for a, b, c, and d, getting another matrix C so that AC is also the zero matrix?
  4. Nov 18, 2008 #3
    Thanks for the help!

    This may sound like a stupid question, as I've encountered this problem before in this course... how would I solve for 4 unknowns in this way; I haven't taken Math for 4 years before Linear Algebra, so I'm rusty on how to do these types of equations.
    I would set all equal to zero, and then.....
  5. Nov 18, 2008 #4
    It's a very simple problem.

    AB=AC. It means A(B-C)=0. You can find the matrix B-C as the kernel of A. So we have infinite couple B,C.

    Easy, right?
  6. Nov 18, 2008 #5


    Staff: Mentor

    I wound up with two equations in four unknowns, with one equation involving a and c, and the other involving b and d. That means that I could put in arbitrary values for c and d to get, respectively, a and b.
  7. Nov 18, 2008 #6
    Hi Mark,

    When you set all of them equal to zero do you see any pattern?

    What if you solve the equations that have an 'a' in them for 'a' and solve the equations that have a 'b' in them for 'b'. Do you see a pattern now?
  8. Nov 18, 2008 #7
    Great, thanks guys!

    I set each to equal the zero matrix; so ended up getting a = 4c, and b = 4d. I used arbitrary numbers for C and D, and got two separate matrices B and C such that AB = AC...

    for example, these worked out:
    [4 8
    1 2]
    [12 16
    3 4]

    Thanks for clearing that up for me!
  9. Nov 18, 2008 #8
    Thats it! Good work.

    Also my appologies to Mark44, I see you were helping answering the question, when lubricarret was asking it.
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