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Here's my system of equations:

$\begin{cases}y + 2kz = 0\\x + 2y + 6z = 2\\kx + 2z = 1\\ \end{cases}$

So I have $\left[\begin{array}{ccc|c}1&2&6&2\\0&1&2k&0\\k&0&2&1\end{array}\right]$

When I row reduce, I get:

$\left[\begin{array}{ccc|c}1&0&6-4k&2\\0&1&2k&0\\0&0&2-6k-4k^2&1-2k\end{array}\right]$

Not really sure where to go from here...

Any suggestions much appreciated!

Thanks, Mariogs

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3 Answers 3

up vote 3 down vote accepted

While DonAntonio's answer is certainly correct, it is likely that your question comes from a class where determinants have not yet been discussed, so you may need a different perspective.

In that case, recall that your system will be inconsistent if, after row reduction, you have a row of the form $( 0 \ 0 \ 0 \mid 1)$ since this row would correspond to the equation $0x+0y+0z=1$ which clearly has no solutions.

On the other hand, you also don't want a row of of the form $( 0 \ 0 \ 0 \mid 0)$, which would give you a free variable and hence infinitely many solutions.

Thus, you should find the values of $k$ for which $2-6k-4k^2 = 0$. By our discussion, we can see that as long as you avoid those values of $k$, your system will have a solution, and this solution will be unique.

You can find these "bad" values of $k$ by methods from high school algebra, e.g. the quadratic formula.

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The coefficients' matrix of your non-homogeneous linear system is:

$$\begin{pmatrix}0&1&2k\\ 1&2&6\\ k&0&2\end{pmatrix}$$

and the system has (a unique, by the way) solution iff the above matrix is regular, which happens iff its determinant is non-zero ...

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Now the last row could be rewritten as $$\pmatrix{1&0&6-4k&2\\0&1&2k&0\\ 0 & 0&2(2k-1)(k-1)&-(2k-1)}$$ Now we have the following cases

  1. $k=1$: the system has no solution. $$\pmatrix{1&0&2&2\\0&1&2&0\\ 0 & 0&0&-1}$$
  2. $k=1/2$: the system has infinitely many solution. $$\pmatrix{1&0&4&2\\0&1&1&0\\ 0 & 0&0&0}$$
  3. $k\in \mathbb{R}, k\ne 1,1/2$: the system has unique solution.
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