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My class was able to produce solutions using Substitution on the following System: $$ \left\{ \begin{array}{c} x+y+z=0 \\ 2x+3y+2z=-1\\ x-y+z=2 \end{array} \right. $$ The solution was: x = 1, y = -1, z = 0

However, when I tried solving that system using matrices in the following format, I received a Singular Matrix Error in my calculator.

$$ \left[ \begin{array}{ccc|c} 1&1&1&0\\ 2&3&2&-1\\ 1&-1&1&2 \end{array} \right] $$

I solved the matrix by hand, and it is true. The det = 0.

So, why are there solutions to the system using substitution, but no solutions when using matrices?

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  • $\begingroup$ Are parameterized solutions numbers that can also fit in the solution such as: 0.5,-1,0.5? $\endgroup$ – Rhendz Oct 18 '14 at 21:25
  • $\begingroup$ Where did you get x = 1-t, and z = t from? $\endgroup$ – Rhendz Oct 18 '14 at 21:33
  • $\begingroup$ Yes that makes sense thank you :D $\endgroup$ – Rhendz Oct 18 '14 at 21:37

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