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Thread: Matrice Equation

  1. #1
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    Matrice Equation

    Hello, I need to solve the following matrice equation. The known matrices are A & B.

    AX + 3IX = B

    If anyone could help me here I would appreciate it.

    I figured it would look something like this, though I am not sure:

    $\displaystyle AX + 3IX = B$

    $\displaystyle inv(A)AX + 3IX = inv(A)B$

    $\displaystyle X + 3IX = inv(A)B$

    $\displaystyle X + inv(3I)(3I)X = inv(3I)inv(A)B$

    $\displaystyle 2X = inv(3I)inv(A)B$

    IF this is correct how would I count the inverse of 3I ? I would be the identity matrix of course, it is all 3 in a diagonal instead of 1?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Goldberg Variation View Post
    Hello, I need to solve the following matrice equation. The known matrices are A & B.

    AX + 3IX = B

    If anyone could help me here I would appreciate it.

    I figured it would look something like this, though I am not sure:

    $\displaystyle AX + 3IX = B$

    $\displaystyle inv(A)AX + 3IX = inv(A)B$
    You are incorrect at this point. You are multiplyin both sides by $\displaystyle A^{-1}$ so it should be
    $\displaystyle A^{-1}(AX+ 3X)= A^{-1}Ax+ A^{-1}3Ix= A^{-1}B$

    But then you will have trouble with $\displaystyle A^{-1}3Ix= 3A^{-1}x$ (since Ix= x).
    Instead, write the equation as $\displaystyle Ax+ 3Ix= (A+ 3I)x= Bx$ and multiply by the inverse of $\displaystyle A+ 3I$, not A.

    Of course, A+ 3I is the matrix A with "3" added to each number on the main diagonal.

    By the way, the English singular of "matrices" is "matrix", not "matrice".

    $\displaystyle X + 3IX = inv(A)B$

    $\displaystyle X + inv(3I)(3I)X = inv(3I)inv(A)B$

    $\displaystyle 2X = inv(3I)inv(A)B$

    IF this is correct how would I count the inverse of 3I ? I would be the identity matrix of course, it is all 3 in a diagonal instead of 1?

    Thanks in advance.
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  3. #3
    A Plied Mathematician
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    ...multiply by the inverse of $\displaystyle A+3I$...
    Of course, this only works if -3 is not an eigenvalue of A. Otherwise, said inverse does not exist! Do we know what A is?
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  4. #4
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    Hello and thank you both for your replies.

    HallsofIvy I am not sure I follow your steps you posted. On your last line of the equation do you break out the matrix X? If so why does it appear next to the matrix B on the right side of the equation?

    Or if it was a typo I can see the answer of the equation I think. Then it would be:

    $\displaystyle X = B inv(A + 3I)$ Correct?

    Oh and Ackbeet the known matrices are the ones I mentioned in my first post, matrix A and B. I just need to figure out the equation so I can solve matrix X.

    Thanks in advance!

    P.S Where can I find the MATH codes for specific functions and such like the inverse (power of -1). Tried search but nothing.
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  5. #5
    A Plied Mathematician
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    Type $\displaystyle (A+3I)^{-1}.$ You can double-click the code to see how I did it. Use the caret symbol ^ to get exponents, and the underscore _ to get subscripts.

    I think HallsofIvy assumed that X was a vector, instead of a matrix. The general method was fine, however. That is, you have

    $\displaystyle AX+3IX=B$

    $\displaystyle (A+3I)X=B$

    $\displaystyle X=(A+3I)^{-1}B,$

    assuming -3 is not an eigenvalue of A, as I said before.
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  6. #6
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    Thank you both HallsofIvy and Ackbeet for your answers. I get it now and can solve the X matrix now on my own.

    P.S I didn't know you could double-click the MATH code to actually view the code. Really neat, thanks.
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  7. #7
    A Plied Mathematician
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    You're welcome. Have a good one!

    Like your username, by the way. I'm partial to the Glenn Gould recording myself.
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