# How is it organized Cramer's rule when determinant is zero

\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

• Cramer’s rule is invalid in such a case. You’ll have to use some other method. – amd Jan 14 '17 at 16:13
• If the determinant is zero you can't use Cramer's rule. In fact there cannot be a unique solution. – Arnaud D. Jan 14 '17 at 16:14
• Your determinant's very first entry is wrong: it should be $\;-1\;$ ... – DonAntonio Jan 14 '17 at 16:19
• Yes, I corrected. – mrsengineer Jan 14 '17 at 16:22
• 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. – DonAntonio Jan 14 '17 at 16:24

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.