# For what values of $k$ will these equations have no solution/infinite solutions/unique solution

Here are the 3 linear equations:

$$x+y-z=-1$$

$$2x-4y-6z=-1$$

$$x-y+(k^2-1)z=k$$

I understand a $4\times3$ matrix must be set up in order to solve this particular problem.The part which I get stuck at is putting the equation into row reduced form.

And also how to determine if for which values $k$ has no solution/infinite solutions and unique solution

• Commented Sep 20, 2014 at 15:18

## 1 Answer

Apply the Gauss-elimination algorithm:

$\begin{pmatrix} 1&1&-1&-1\\ 2&-4&-6&-1\\ 1&-1&k^2-1&k \end{pmatrix} \implies\begin{pmatrix} 1&1&-1&-1\\ 0&-6&-4&1\\ 0&-2&k^2&k+1 \end{pmatrix}\implies \begin{pmatrix} 1&1&-1&-1\\ 0&-6&-4&1\\ 0&0&k^2+4/3&k+2/3 \end{pmatrix}$

Now, since $\ k^2+\frac{4}{3}>0 \text{ } \forall k$, you can find a solution, and it is always unique.

You can get the same result if you compute the determinant of the coefficient matrix, which cannot be equal to zero:

det$\begin{pmatrix} 1&1&-1\\ 2&-4&-6\\ 1&-1&k^2-1 \end{pmatrix}=[-4(k^2-1)-6]-[2(k^2-1)+6]-[-2+4]=$

$\ =-4k^2+4-6-2k^2+2-6+2-4=-6k^2-8<0 \forall k$