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For which values of $a$ the system has one/none/infinite solutions, write the general solution

$$ \begin{cases} x_1+(a-1)x_2-x_3=4\\ ax_1+(a-1)x_2-x_3=a+3\\ x_1+(a-1)x_2+(a-3)x_3=7 \end{cases} $$

$$ \left( \begin{array}{ccc|c} 1 & a-1 & -1 & 4 \\ 0 & -(a-1)^2 & a-1 & -3a+3 \\ 0 & 0& a-2 & 3 \\ \end{array} \right) $$

So for $a=2$ there is no solution, $a\neq2,1$ there is one solution and for $a=1$ there is infinite solutions in this case we have

$$ \left( \begin{array}{ccc|c} 1 & 0 & -1 & 4 \\ 0 & 0 & -1 & 3 \\ 0 & 0& 0 & 0 \\ \end{array} \right) $$

$$ \begin{cases} x_1-x_3=4\rightarrow x_1=1\\ -x_3=3\rightarrow x_3=-3\\ \end{cases} $$

So the general solution is

$$\begin{pmatrix} 1 \\ t \\ -3\\ \end{pmatrix}=t\begin{pmatrix} 0 \\ 1 \\ 0\\ \end{pmatrix}+\begin{pmatrix} 1 \\ 0\\ -3\\ \end{pmatrix}$$

Is that correct?

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  • $\begingroup$ @Moo you must assume $a\neq 0$ right? $\endgroup$ – gbox Nov 22 '17 at 13:44
  • $\begingroup$ @Moo Yes, but for the process of row reduction, you need to assume that $a\neq 0$ else for example you can divide a row with $0$ $\endgroup$ – gbox Nov 22 '17 at 14:05
  • $\begingroup$ @Moo yes it will just delete the (2,1) entry $\endgroup$ – gbox Nov 22 '17 at 14:11
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Yes, we have $\det(A)=(a - 1)^2(a - 2)$ for the system $(A,b)$, so that there is a unique solution for $a\neq 1,2$. For $a=2$, as you said, we obtain $0=3$, which is correct for fields of characteristic $3$, but a contradiction otherwise. For $a=1$ we have a $1$-dimensional kernel $(0,1,0)^T$, and the general solution follows from this. Assuming that the field is infinite, we obtain infinitely many solutions.

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