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Math Help - Find exactly one solution

  1. #1
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    Find exactly one solution

    I'm confussed on how to solve this because of the constants a for the given value of s.
    Given:
    x1 + 0x2 + x3 = a
    5x1 + (s-1)x2 + 3x3 = a
    sx1 + 0x2 + 4x3 = 1

    Find what value of s does the system have on solution?
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  2. #2
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    Re: Find exactly one solution

    For the general case, a makes no difference:

    Have MX = B, which is \left( \begin{matrix} 1 & 0 & 1 \\ 5 & (s-1) & 3 \\ s & 0 & 4 \end{matrix} \right)\left( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right) = \left( \begin{matrix} a \\ a \\ 1 \end{matrix} \right),

    so, when M^{-1} exists, have X = M^{-1}B and it's solved. But M^{-1} exists iff det(M) \ne 0.

    Thus, the general condition that will make that system solvable has nothing to do with the B. It's decided by entirely by M, via det(M) \ne 0.

    Using cofactors:

    det(M) = (1)det \left( \begin{matrix} (s-1) & 3 \\ 0 & 4 \end{matrix} \right) - (0)det \left( \begin{matrix} 5 & 3 \\ s & 4 \end{matrix} \right) + (1)det \left( \begin{matrix} 5 & (s-1) \\ s & 0 \end{matrix} \right).

    det(M) = (1)((s-1)(4)-(3)(0)) - (0)(whatever) +(1)((5)(0) - (s)(s-1))

    = (4s-4) + (- s^2 + s) = -s^2 +5s -4

    So det(M) =0 when -s^2 +5s -4 = 0 when s^2 -5s +4= 0

    Factoring gives s^2 -5s +4= (s-4)(s-1), so s^2 -5s +4=0 when  s\in \{1, 4\}.

    Thus this is solveable whenever s \notin \{ 1, 4 \}, no matter what a is.

    Now, when s \in \{ 1, 4 \}, it *might* have solutions, but then the value of a will decide that.

    Check by doing row reduction on: \left( \begin{matrix} 1 & 0 & 1 & | & a \\ 5 & (s-1) & 3 & | & a\\ s & 0 & 4 & | & 1 \end{matrix} \right)

    When s = 1 you get:

    \left( \begin{matrix} 1 & 0 & 1 & | & a \\ 5 & 0 & 3 & | & a\\ 1 & 0 & 4 & | & 1 \end{matrix} \right)

    so \left( \begin{matrix} 1 & 0 & 1 & | & a \\ 0 & 0 & -2 & | & -4a\\ 0 & 0 & 3 & | & 1-a \end{matrix} \right) (used row 1 to clear column 1)

    so \left( \begin{matrix} 1 & 0 & 1 & | & a \\ 0 & 0 & 1 & | & 2a\\ 0 & 0 & 1 & | & \frac{1-a}{3} \end{matrix} \right)

    so \left( \begin{matrix} 1 & 0 & 0 & | & -a \\ 0 & 0 & 1 & | & 2a\\ 0 & 0 & 0 & | & (\frac{1-a}{3}-2a) \end{matrix} \right) (used row 2 to clear column 3)

    Thus x_1 = -a, x_3 = 2a, and 0 = \frac{1-a}{3}-2a.

    The last equation puts a restriction on a: \frac{1-a}{3}-2a = 0 implies 1-a = 6a implies a = 1/7.

    Thus when s = 1, it has a solution, but only if a = 1/7.

    The s = 1 and a = 1/7 solution is: x_1 = -1/7, x_2 = anything, x_3 = 2/7.

    I'll leave it to you to see what happens when s = 4.
    Last edited by johnsomeone; October 1st 2012 at 11:18 AM.
    Thanks from nivek0078
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