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I need to find all the answers of this linear equation system. $$\begin{vmatrix} 1 & 2 & 1 & 3 & 0 & =\beta \\ 2&3&2&5&1&=0\\ 2&1&1&4&1&=0\\ 3&3&2&7&1&=1\\ \end{vmatrix} $$ I tried to solve it like that:

Row 4- row 3->

$$\begin{vmatrix} 1 & 2 & 1 & 3 & 0 & =\beta \\ 2&3&2&5&1&=0\\ 2&1&1&4&1&=0\\ 1&2&1&3&0&=1\\ \end{vmatrix} $$

Row 1- row 4 ->

$$\begin{vmatrix} 0 & 0 & 0 & 0 & 0 & =\beta-1 \\ 2&3&2&5&1&=0\\ 2&1&1&4&1&=0\\ 1&2&1&3&0&=1\\ \end{vmatrix} $$

and if I understand correctly, if a row is in the form of the first row here, all the variable are zero and equals something, the linear equation system don't have an answer. but I checked that LES in online calculator, and apparently it does have an answer. Am I getting something wrong here?

MeepMeep
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  • The first line means $0\cdot x_1 + ... + 0\cdot x_5 = \beta - 1$. In other words $0 = \beta -1$ and thus $\beta = 1$. – Philipp Nov 22 '21 at 17:18

1 Answers1

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As I wrote in the comment. From your work we conclude that the linear system might only possess a solution for $\beta = 1$. As it turns out the solutions for this case are of the form $$\left(\begin{matrix} 1-2\cdot x_4 \\ 2-x_4+x_5 \\ -4+x_4-2\cdot x_5 \\ x_4 \\ x_5 \end{matrix}\right)$$ for $x_4, x_5 \in \mathbb{R}$.

Philipp
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