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Math Help - Reduce the matrix [A|I]

  1. #1
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    Reduce the matrix [A|I]

    Hi,

    I need guidance with the following question:

    Question: Reduce the matrix [A|I] in order to determine if A is invertible. Compute A^-1 when A is invertible and check that AA^-1 = I. Verify that the (1,1)-entry in the product A^-1 is 1.

    A=
    2 1
    5 3

    For starters,I was going to find the determinant of A like this: |A|= ad-bc, but realized the question is asking me to reduce the matrix [A|I]. So how do I reduce [A|I]? Do I multiply matrix A * Identity matrix, then use reduced row echelon like this:

    2 1|1 0
    5 3|0 1
    Last edited by sparky; February 16th 2011 at 01:00 PM. Reason: Typographical error
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  2. #2
    A Plied Mathematician
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    Perform the necessary row operations on the entire matrix [A|I] necessary to take A to I. If you can do it, you've got an invertible matrix A, and you should end up with

    [A|I]\to[I|A^{-1}].

    Make sense?
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  3. #3
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    Do you know what it means to "reduce" a matrix? You want to go from that matrix to the identity matrix by "row operations" and there are three kinds of row operations:
    1) Divide (or multiply) an entire row by a number
    2) Add (or subtract) a multiple of a row to (or from) another row
    3) Swap two rows.

    Since you want a "1" in the upper left, a good way to start is to divide the entire first row by the number that is in that position- here that is a "2". Then, because you want a "0" below that, subtract the new first row times the number you want to get rid of, here a "5", from the second row.

    (Be careful with your terminology. You are not "multiplying" A by the identity matrix- you are simply writing them next to one another so that you can apply the same row operations to both matrices at the same time.)
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  4. #4
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    Thanks for your replies Ackbeet and HallsofIvy!

    Ok, here is what I've done so far:

    2 1 | 1 0 R1 * 1/2
    5 3 | 0 1

    1 1/2 | 1/2 0
    5 3 | 0 1 R2 -5R1

    1 1/2 | 1/2 0
    0 1/2 | 5/2 1 R2 *2

    1 1/2 | 1/2 0 R1 -1/2R2
    0 1 | 5 2

    1 0 | -2 -1
    0 1 | 5 2

    Now that I've reduced A to 1s and 0s, how do I know that A is invertible? I noticed that my Identity matrix has changed close to what A is (except for the minus 2 and minus 1 and the 2)
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  5. #5
    A Plied Mathematician
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    So your candidate for A^{-1} is now

    A^{-1}\overset{?}{=}\begin{bmatrix}-2 &-1\\5&2\end{bmatrix}.

    So, multiply it out to check that

    AA^{-1}=A^{-1}A=I.

    Does it work? I think you'll find that you made a sign error in there somewhere in your row reduction.
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  6. #6
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    Ackbeet, you were right about the sign:

    2 1 | 1 0 R1 * 1/2
    5 3 | 0 1

    1 1/2 | 1/2 0
    5 3 | 0 1 R2 -5R1

    1 1/2 | 1/2 0
    0 1/2 | -5/2 1 R2 *2

    1 1/2 | 1/2 0 R1 -1/2R2
    0 1 | -5 2

    1 0 | 3 -1
    0 1 | -5 2
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  7. #7
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    You multiply that out, and I think you'll see that you now have the correct inverse.

    Verify that the (1,1)-entry in the product A^-1 is 1.
    This is a quote from the OP. I think you meant that the (1,1) entry in the product AA^(-1) is 1? If so, you can see that that's true.

    Any other questions about this problem?
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  8. #8
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    Yes you are right Ackbeet!

    A^-1 =
    3 -1
    -5 2

    2 1 | 3 -1
    5 3 | -5 2

    6 -5 -2 + 2
    15-15 -5 + 6

    1 0
    0 1

    And yes, I did make a mistake in the question. It should read:
    ...Verify that the (1,1)-entry in the product AA^-1 is 1.
    But how do I verify that the (1,1)-entry in the product AA^-1 is 1? Do I calculate it like this: ad-cb, that is, 1*1 - 0*0 = 1?

    Thanks a lot Ackbeet!
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  9. #9
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    The (1,1) entry is the upper left corner entry of the matrix you calculated as

    \begin{bmatrix}6-5&-2+2\\15-15 &-5+6\end{bmatrix}.

    So what would you say its value is?
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  10. #10
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    Quote Originally Posted by Ackbeet View Post
    The (1,1) entry is the upper left corner entry of the matrix you calculated as

    \begin{bmatrix}6-5&-2+2\\15-15 &-5+6\end{bmatrix}.

    So what would you say its value is?
    I would say the value of the upper left corner is 6-5 = 1 Thanks a lot Ackbeet!
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  11. #11
    A Plied Mathematician
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    Yep, I'd say you've got it. You're welcome, and have a good one!
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