-1
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\begin{cases} x + 2y - z = 0 \\ 2x +3y – 2z = -1 \\ -x + y + z = 3 \end{cases}

$$ \left | \begin{matrix} 1&2&-1 \\ 2&3&-2 \\ -1&1&1 \end{matrix} \right|=0. $$

How should I organize equations to solve with Cramer's Rule

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  • $\begingroup$ Cramer’s rule is invalid in such a case. You’ll have to use some other method. $\endgroup$ – amd Jan 14 '17 at 16:13
  • $\begingroup$ If the determinant is zero you can't use Cramer's rule. In fact there cannot be a unique solution. $\endgroup$ – Arnaud D. Jan 14 '17 at 16:14
  • $\begingroup$ Your determinant's very first entry is wrong: it should be $\;-1\;$ ... $\endgroup$ – DonAntonio Jan 14 '17 at 16:19
  • $\begingroup$ Yes, I corrected. $\endgroup$ – mrsengineer Jan 14 '17 at 16:22
  • $\begingroup$ With the last edition of your question your system ins incongruent = it has no solution,. as you can easily check by reducing by rows the system's matrix. $\endgroup$ – DonAntonio Jan 14 '17 at 16:24
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Since the determinant is $0$, the system either has no solution or it has infinitely many.

Since $\det\begin{bmatrix}1&2\\2&3\end{bmatrix}\ne0$, you can consider $$ \begin{cases} -x+2y=z\\ 2x+3y=2z-1 \end{cases} $$ Solve it with Cramer's rule and substitute in the last equation to verify whether it holds or not.

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  • $\begingroup$ Thank you. This is the solution I want. $\endgroup$ – mrsengineer Jan 14 '17 at 16:29

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