-1
$\begingroup$

\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

  • 0
    Cramer’s rule is invalid in such a case. You’ll have to use some other method. – 2017-01-14
  • 0
    If the determinant is zero you **can't** use Cramer's rule. In fact there cannot be a unique solution. – 2017-01-14
  • 0
    Your determinant's very first entry is wrong: it should be $\;-1\;$ ... – 2017-01-14
  • 0
    Yes, I corrected. – 2017-01-14
  • 0
    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. – 2017-01-14

1 Answers 1

3

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.

  • 0
    Thank you. This is the solution I want. – 2017-01-14