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Math Help - Matrices

  1. #1
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    Matrices

    Hi Thank you all very much for your answers to my previous questions, really appreciate them. Unfortunately, I have one more..

    Find the values of a for which the system of equations has a unique solution.
    3x-ay+2z = 4
    x+2y-3z = 1
    -x-y+z = 12

    In matrix form,
    \\1 & 2 & -3 \\-1 & -1 & 1 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}" alt="\begin{bmatrix} 3 & -a & 2 \\1 & 2 & -3 \\-1 & -1 & 1 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}" />
    = \\1\\12 \end{bmatrix}" alt="\begin{bmatrix} 4 \\1\\12 \end{bmatrix}" />

    I know I am supposed to solve it such that the determinant is not zero to get a unique solution..I know how to get the determinant of 3 by 3 matrices, I got a = -0.5 (or is it 0.5? ( I was confused with the negative signs)) for determinant =0. However, I was not taught the inverse of a 3 by 3 matrix, I was told to key it into the calculator to which I can do, but not here, since there is an unknown a..I tried substituting with a=-0.5 but I goe fractional entries, and I do not think I am doing it the right way anyway..

    Please help me!
    I have done such questions before but they were only asking for the inverse of a two by two matrix, and I could easily find the value of the unknown etc..
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  2. #2
    MHF Contributor

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    Re: Matrices

    the determinant we want is:

    \begin{vmatrix}3&-a&2\\1&2&-3\\-1&-1&1 \end{vmatrix}

    for a 3x3 matrix, the easiest method to employ is the "rule of sarrus" (form 3 triples product by "wrap-around diagonals" from upper-left to lower right, and sum these, form three more triple products from "wrap around diagonals" from upper-right to lower-left, subtract these from the first sum).

    in the matrix at hand that gives us:

    (3)(2)(1) + (-a)(-3)(-1) + (2)(1)(-1) - (2)(2)(-1) - (-a)(1)(1) - (3)(-3)(-1)

    = 6 - 3a - 2 + 4 + a - 9 = -2a - 1

    if the determinant is non-zero, it means the matrix IS invertible, if we call our matrix A, then the unique solution is A-1(4,1,12).

    the determinant will be non-zero whenever:

    2a + 1 \neq 0 that is, when a \neq -\frac{1}{2}

    note that what happens when a = -1/2, is the first column is: twice the second column plus the third. this means that A is of rank 2 (since the 2nd and 3rd columns are clearly LI), and that the image (column space) of A is:

    span({(1/2,2,-1),(2,-3,1)}). in this particular case, when a = -1/2, the system has NO solution, as (4,1,12) is not IN the column space.

    since the problem does not ask you to specifically give a solution when a ≠ -1/2, this is all that needs be done.
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  3. #3
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    Re: Matrices

    Thank you so much! I get why the answer is 'no solutions'!

    Could I trouble you to elaborate more about the part where you identified A is rank 2, I don't remember learning that in school, and we were never taught to look at the relationship of entries within a matrix in such an intricate way!
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