0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ @Moo you must assume $a\neq 0$ right? $\endgroup$
    – gbox
    Nov 22, 2017 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, 2017 at 14:05
  • $\begingroup$ @Moo yes it will just delete the (2,1) entry $\endgroup$
    – gbox
    Nov 22, 2017 at 14:11

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.