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Thread: For what values of "a" is there one, none, or infinitely many solutions?

  1. #1
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    For what values of "a" is there one, none, or infinitely many solutions?

    I normally wouldn't have too much trouble with something like this, but my skills regarding it are somewhat rusty.

    For what values of $\displaystyle a$ does the system of equations
    $\displaystyle x+ay+2z=0$
    $\displaystyle -2x-ay+z=4$
    $\displaystyle 2ax+3a^2y+9z=4$
    has one, none, or infinitely many solutions? Find all the possible solutions.

    Right now, it's mainly that $\displaystyle a^2$ in the third equation that has me a tad confused on how to handle it. This shouldn't be very hard, but, as I said, my memory on this sort of thing is a bit rusty (and muddled by other topics).
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  2. #2
    Senior Member roninpro's Avatar
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    You can put this into a matrix:

    $\displaystyle \left(
    \begin{array}{ccc|c}
    1 & a & 2 & 0 \\
    -2 & -a & 1 & 4 \\
    2 a & 3 a^2 & 9 & 4
    \end{array}
    \right)$

    Try row-reducing this and see what kinds of assumptions you need for $\displaystyle a$ to get each case.
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  3. #3
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    Or, more "unsophisticatedly" (my favorite kind- although it usually involves more work) solve the equations directly.

    For example, if we add twice the first equation, 2x+ 2y+ 4z= 0, to the second equation, -2x- ay+ z= 4, we eliminate x: (2-a)y+ 5z= 4.
    We can also eliminate x by adding -2a times the first equation, -2ax- 2ay- 4az= 0, to the third equation, $\displaystyle 2ax+ 3a^2y+ 9z= 4: (3a^2- 2a)y+ (9- 4a)z= 4$.

    Now that you have two equations in y and z only, you could combine them to eliminate, say, z, leaving an equation for y only (and a, of course). Solving that equation for y will involve dividing one expression in a by another. There will be no solution if and only if the denominator is 0 but the numerator is not.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Or, more "unsophisticatedly" (my favorite kind- although it usually involves more work) solve the equations directly.

    For example, if we add twice the first equation, 2x+ 2y+ 4z= 0, to the second equation, -2x- ay+ z= 4, we eliminate x: (2-a)y+ 5z= 4.
    We can also eliminate x by adding -2a times the first equation, -2ax- 2ay- 4az= 0, to the third equation, $\displaystyle 2ax+ 3a^2y+ 9z= 4: (3a^2- 2a)y+ (9- 4a)z= 4$.

    Now that you have two equations in y and z only, you could combine them to eliminate, say, z, leaving an equation for y only (and a, of course). Solving that equation for y will involve dividing one expression in a by another. There will be no solution if and only if the denominator is 0 but the numerator is not.
    That looks alright, but you made a couple of mistakes: instead of $\displaystyle (2-a)y+5z=4$, it should be $\displaystyle ay+5z=4$, since the first equation is $\displaystyle x+ay+2=0$.

    Also, instead of $\displaystyle (3a^2- 2a)y+ (9- 4a)z= 4$, it's supposed to be $\displaystyle a^2y+(9-4a)z=4$.

    You forgot that the first equation had an $\displaystyle a$ in it.
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