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I have a linear equation system: $$\left\{\matrix{2x_1+x_2-x_3=-1\\3x_1-2x_2+x_3=7\\x_1-3x_2+2x_3=8}\right.$$ During the process of solving it with Gauss-Jordan elimination, I get a singular matrix:

$$\left(\matrix{3&-2&1\\0&7/3&-5/3\\0&0&0}\right.\left|\matrix{7\\-17/3\\0}\right)$$ Converting this matrix to a system: $$\left\{\matrix{3x_1-2x_2+1x_3=7\\0x_1+{7\over3}x_2-{5\over3}x_3=-{17\over3}\\0x_1+0x_2+0x_3=0}\right.$$ the third row is true, thus the system has infinite solutions - the planes are cut in a specific line, rather than a dot.

My linear algebra book does not elaborate how to find that specific line. According to it, the final result should be: $$\vec x=\left(\matrix{5/7\\-17/7\\0}\right)+t\left(\matrix{1/7\\-5/7\\1}\right), t\in\mathbb R$$ How do I solve this?


EDIT: If I continue the Gauss-Jordan elimination, I will then get this singular matrix: $$\left(\matrix{1&0&-1/7\\0&1&-5/7\\0&0&0}\right.\left|\matrix{5/7\\-17/7\\0}\right)$$ which paves way to solve $x_1$ and $x_2$ as: $$\left\{\matrix{x_1={1\over7}x_3+{5\over7}\\x_2={5\over7}x_3-{17\over7}}\right.$$ so that it can be plugged into $\vec x$: $\left(\matrix{{1\over7}x_3+{5\over7}\\{5\over7}x_3-{17\over7}\\x_3}\right)$.

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    $\begingroup$ Some care is due here: that the third row becomes all zeros does not necessarily mean there are infinite solutions but merely that there is not a unique solution. Another option is that there may be no solutions at all... $\endgroup$ – DonAntonio Feb 19 at 21:42
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To obtain the parameterisation of the set of solutions, determine the row reduced echelon form of the augmented matrix: \begin{align} \left[\begin{array}{rrr|r} 2&1&-1&-1 \\3&-2&1&7 \\1&-3&2&8 \end{array}\right]&\rightsquigarrow \left[\begin{array}{rrr|r} 1&-3&2&8 \\ 2&1&-1&-1 \\ 3&-2&1&7 \end{array}\right]\rightsquigarrow \left[\begin{array}{rrr|r} 1&-3&2&8 \\ 0&7&-5&-17 \\ 0&7&-5&-17 \end{array}\right]\rightsquigarrow\\[1ex] \rightsquigarrow \left[\begin{array}{rrr|r} 1&-3&2&8 \\ 0&7&-5&-17 \\ 0&7&-5&-17 \end{array}\right]&\rightsquigarrow \left[\begin{array}{rrr|r} 1&-3&2&8 \\ 0&1&-\frac5 7&-\frac{17}7 \\ 0&0&0&0 \end{array}\right]\rightsquigarrow \left[\begin{array}{rrr|r} 1&0&-\frac17&\frac 57 \\ 0&1&-\frac5 7&-\frac{17}7 \\ 0&0&0&0 \end{array}\right] \end{align} This final form says the solutions are \begin{cases} x_1=\frac57+\frac17 x_3, \\ x_2=-\frac{17}7+\frac 57x_3, \end{cases} or, setting $x_3=t$, we have, in vector form: $$\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}\frac57\\-\frac{17}7\\0\end{pmatrix}+t\begin{pmatrix}\frac17\\-\frac{5}7\\1\end{pmatrix}.$$

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  • $\begingroup$ That's the method! Thanks for your help! $\endgroup$ – Marko Ivanović Feb 19 at 22:31
  • $\begingroup$ You're welcome. Always glad to feel helpful! $\endgroup$ – Bernard Feb 19 at 22:33
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I have got $$x_1-3x_2+2x_3=8$$ $$7x_2-5x_3=-17$$ The third equation is the same as second!

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  • $\begingroup$ Looks like it's not yet in row-reduced echelon form. Try dividing the second equation with 7, and get rid of $x_2$ in the first equation: $R_1+{3\over7}R_2$. Then find $x_1$ and $x_2$. $\endgroup$ – Marko Ivanović Feb 19 at 22:50
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Set one Variable as $t\in\mathbb R$, e.g $x_3=t$ and solve the new system.

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