System of Linear Equations, when does it have infnitely many solutions?

Question

$$ \left\{ \begin{array}{rrrrrc} x& +& 2y& -& z& =& -3 \\ & & y& -& (k-3)z& =& -5 \\ && && (k^2-2k)z& =& 5k + 11 \\ \end{array} \right. $$ I have row reduced the system of equations to the above, and am trying to find values of k for which the system has infinitely many solutions.

I have found that when

$k= 3.5 + \sqrt{93}/2$ and $3.5 - \sqrt{93}/2$,

there are infinitely many solutions. However have been advised that there are more cases of infinitely many solutions and cannot figure out which values of k these are. Could someone please point me in the right direction?

By the way I have no done determinants fully and am needing to try and use gaussian elimination to solve this.

Thank you

Answer 1

By using the first two equations, every value of $z$ gives exactly one value for $y$ and one value for $x$. So the only way to get infinitely many solutions is to get infinitely many $z$. The only way that will happen is if your third equation reduces to $0z=0$. So for the system to have infinitely many solutions, $k$ must satisfy $$k^2-2k=0\quad\hbox{and}\quad 5k+11=0\ .$$ There is no such $k$, so your system never has infinitely many solutions.

Comment. If you are definitely being told that it is possible to have infinitely many solutions, then you must have made a mistake before the equations you posted. Please check them carefully and post another question if you can't find an error.

Answer 2

The first two rows have pivot elements.

We have to make the third row a zero row.

Hence we need $k^2-2k=0$ and $5k+11=0$. Check that there is no such solution.

Remark:

It seems that you set $k^2-2k=5k+11$ and solve for $k$, this doesn't guarantee that they are both equal to zero.

Answer 3

The system is already in echelon form and you don't need any additional Gaussian elimination.

The third equation is a candidate for an infinite number of solutions, if it can be of the form $0z=0$. But this never occurs because $k^2-2k$ and $5k+11$ have no common root.

So when $k^2-2k=0$, the system is impossible.

And when $k^2-2k\ne0$ you can solve for $z$ and get a reduced system on $x,y$ by moving $z$ to the RHS. This reduced system is also in echelon form, with constant coefficients, and has a unique solution.

The answer is "never".