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I am struggling with solving systems of linear equations with unknown constants.

I can solve simple ones, such as:

In the following system, for which values of k would produce:

  1. Infinitely many solutions?
  2. No solution?
  3. A unique solution? $$ x + ky = 1\\ kx + y = 1 $$
    (where $k$ is some constant)

$$ \left[ \begin{array}{cc|c} 1&k&1\\ k&1&1 \end{array} \right] $$

$$ \text{Operation: } R2 \leftarrow R2 - R1 $$

$$ = \left[ \begin{array}{cc|c} 1&k&1\\ 0&1-k^2&1-k \end{array} \right] $$

Therefore:

a) If k = 1, there are infinitely many solutions:

$$ = \left[ \begin{array}{cc|c} 1&1&1\\ 0&0&0 \end{array} \right] $$

b) if k = -1, there are no solutions:

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

c) for any other values of k, we can do:

$$ \text{Operation: } R2 = \frac1{(1-k^2)}R2 $$

$$ = \left[ \begin{array}{cc|c} 1&k&1\\ 0&1&\frac1{1+k} \end{array} \right] $$

But I can't figure out how to do this:

$$ = \left[ \begin{array}{ccc|c} 1&-2&3&2\\ 1&1&1&k\\ 2&-1&4&k^2 \end{array} \right] $$

I try to put it into echelon form, but I end up with something that suggests there aren't any possible solutions:

$$ = \left[ \begin{array}{ccc|c} 1&-2&3&2\\ 0&3&-2&k-2\\ 0&0&0&k^2-k-6 \end{array} \right] $$

Which was done by:

$$ R2 \leftarrow R2 - R1 $$ $$ R3 \leftarrow R3 - 2R1 $$ $$ R3 \leftarrow R3 - R2 $$

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Assuming that you've row-reduced correctly:

We conclude that the system will be consistent if $k^2 - k - 6 = 0$ (so $k = 3,-2$) and therefore, for these $k$, the system have infinitely many solutions. In all other situations, the equation $$ 0 = k^2 - k - 6 $$ is a contradiction, so there are no solutions.

The system will never have a unique solution.

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  • $\begingroup$ Thank you, that makes sense. I suppose the sensible thing I should have tried to do myself was to see the obvious (duh!) and to factorise the expression which would have given me the solutions for k = 3, -2. I miss so many of the obvious things... Math is practice for people like me who aren't talented, I always need more. $\endgroup$ – Benjamin R Mar 9 '15 at 11:01
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    $\begingroup$ Being a mathematician isn't about "never making silly mistakes"; far more often, it consists of making the silly mistake and feeling dumb enough afterwards to not make it the eighth time. Math is practice for everyone, and everyone feels dumb sometimes. $\endgroup$ – Omnomnomnom Mar 9 '15 at 11:16
  • $\begingroup$ thank you that is very encouraging! $\endgroup$ – Benjamin R Mar 9 '15 at 22:29

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