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In general, if you are given a system with some a and b like

$x + y + z = a $

$x + 2z= b$

$x + ay + z=b$

And asked to find the values a and b such that the system has no solution, infinitely many solutions, or a one solution what would be the constraints when row reducing the augmented matrix (ie you can't divide by something, etc.). For the problem above I did row operations on the augmented until I reached here

($ \begin{matrix} 1 & 0 & 1 & b \\ 0 & 1 & -2 & a-b \\ 0 & 0 & 2a & ab-a^2\\\end{matrix}$)

user278039
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    i got this here $$\left{ x={\frac {2,{a}^{2}-ab-b}{a-1}},y=-{\frac {a-b}{a-1}},z=-{ \frac {a \left( a-b \right) }{a-1}} \right} $$ – Dr. Sonnhard Graubner Nov 20 '15 at 20:35
  • So then there would be no solution if a is one, one solution if a and b were consistent with x, y, z l, but what about infinite solutions – user278039 Nov 20 '15 at 20:48

1 Answers1

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Do the row operation to the augmented matrix: $$ \begin{bmatrix} 1 & 1 & 1 & a \\ 1 & 0 & 2 & b \\ 1 & a & 1 & b \\ \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 1 & a \\ 0 & -1 & 1 & b-a \\ 0 & a-1 & 0 & b-a \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 2 & b \\ 0 & -1 & 1 & b-a \\ 0 & 0 & a-1 & a(b-a) \end{bmatrix} $$

From here, assuming $a-1\neq 0$ you can carry on row operations to get

$$ \begin{bmatrix} 1 & 0 & 0 & b-\frac{2a}{a-1} (b-a)\\ 0 & -1 & 0 & -\frac{1}{a-1}(b-a) \\ 0 & 0 & 1 & \frac{a}{a-1} (b-a) \end{bmatrix} $$

If $a=1$ then you end up with

$$ \begin{bmatrix} 1 & 0 & 2 & b\\ 0 & 1 & -1 & 1-b \\ 0 & 0 & 0 & b-1 \end{bmatrix} $$

So no solution.

user160738
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