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$$\begin{cases} &x &- &y &+ &2z &= 4 \\ &3x &- &2y &+ &9z &= 14 \\ &2x &- &4y &+ &az &= b \end{cases} $$ I know that $a$ and $b$ has to either equal to something or not in order to satisfy the $4$ conditions stated above.

my matrices looked like $$ \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & a+12 \\ \end{bmatrix} \begin{bmatrix} x & \\ y &\\ z & \\ \end{bmatrix}= \begin{bmatrix} 6 & \\ 2 & \\ b+8 & \\ \end{bmatrix} $$

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  • $\begingroup$ Exactly $3$ solutions isn't possible. $\endgroup$
    – user63181
    Commented Nov 14, 2013 at 19:55
  • $\begingroup$ does the matrix have to be in RREF or just REF? $\endgroup$
    – user108864
    Commented Nov 14, 2013 at 19:58
  • $\begingroup$ user99680 - yes $\endgroup$
    – user108864
    Commented Nov 14, 2013 at 20:00
  • 1
    $\begingroup$ @SamiBenRomdhane, exactly 3 solutions isn't possible if the problem is over the reals; it is however possible over the finite field with 3 elements. Most likely the former situation is what is meant. $\endgroup$
    – vadim123
    Commented Nov 14, 2013 at 20:04

1 Answer 1

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Apply Gaussian Elimination to the reduced matrix. $$ \pmatrix{1 & -1 & 2 & 4\\3 & -2 & 9 & 14\\2 & -4 & a & b} \to \pmatrix{1 & -1 & 2 & 4\\0 & 1 & 3 & 2\\0 & -2 & a-4 & b-8} \to \pmatrix{1 & -1 & 2 & 4\\0 & 1 & 3 & 2\\0 & 0 & a+2 & b+4} $$ So

  • if you want a unique solution, $\frac{b+4}{a+2}$ must be defined, i.e. $a \neq -2$.
  • if you want infinite number of solutions, $a=-2$ and $b = -4$ removes the last equation.
  • if you want no solutions, $a=-2$ and $b \neq -4$.
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  • $\begingroup$ Shouldn't $a=-2$ in the case for infinite number of solutions? $\endgroup$
    – K. Rmth
    Commented Nov 14, 2013 at 20:18
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    $\begingroup$ @K.Rmth Yes, of course, thanks, fixed the typo $\endgroup$
    – gt6989b
    Commented Nov 14, 2013 at 20:18

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