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Find when the equations $$\begin{cases}x + y - 2z = 0\\ax + by + cz = 0\\bx + cy + az = d\end{cases}$$ are consistent and solve them completely when they are consistent.

I have tried the conventional way to solve these equations by taking them into the echelon form but I get stuck when it comes to the part where I have to show the consistency of the system. Can someone please guide me as to how I should proceed with the problem?

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  • $\begingroup$ I am not sure but I think that a system is consistent if and only if the inverse of the coefficient matrix exists, right? Does this include the case where two equations are 'parallel'? $\endgroup$ – shardulc Apr 26 '16 at 11:24
  • $\begingroup$ @AyanShah: Well, row-reduced-echelon-form produces $\left( \begin{array}{cccc} 1 & 0 & 0 & \dfrac{(2 b+c) d}{-a^2+b a-2 c a+2 b^2-c^2+b c} \\ 0 & 1 & 0 & \dfrac{(2 a+c) d}{a^2-b a+2 c a-2 b^2+c^2-b c} \\ 0 & 0 & 1 & \dfrac{(a-b) d}{a^2-b a+2 c a-2 b^2+c^2-b c} \\ \end{array} \right)$. When are those consistent? $\endgroup$ – Moo Apr 26 '16 at 12:23
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Cramer's rule is perfectly suited for such a task. We have to solve: $$ \begin{pmatrix} 1 & 1 & -2 \\ a & b & c \\ b & c & a\end{pmatrix}\begin{pmatrix} x \\ y \\ z\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ d\end{pmatrix} $$ where the determinant of the matrix in the LHS is $D=(a+b+c)(a+c-2b)$.

Assuming that $D\neq 0$ and $d\neq 0$, explicit solutions are given by: $$ (x,y,z) = \frac{d}{D}\cdot\left(-(2b+c),(2a+c),(a-b)\right). $$

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