# 3x3 system of equations

I have the following system of equations: $$$$x+y+kz = 1$$$$ $$$$x+ky+z = 1$$$$ $$$$kx+y+z =1$$$$

where $$x,y,z \in \mathbf{R}$$.

For what values of $$k \in \mathbf{R}$$ i) the system has a single solution ii) the system has multiple solutions iii) the system has no solution

Can anyone help with that please?

I wrote the above system in a matrix form:

\begin{equation*}

$$\begin{array}{rrr|r} 1 & 1 & k & 1 \\ 1 & k & 1 & 1\\ k & 1 & 1 & 1 \end{array}$$

\end{equation*}

I am not sure how to proceed.. If I subtract the second row from the first (1)-(2) I have the following:

\begin{equation*}

$$\begin{array}{rrr|r} 1 & 1 & k & 1 \\ 0 & 1-k & k-1 & 0\\ k & 1 & 1 & 1 \end{array}$$

\end{equation*}

Then if I subtract the third row from the first row (1)-(3): \begin{equation*}

$$\begin{array}{rrr|r} 1 & 1 & k & 1 \\ 0 & 1-k & k-1 & 0\\ 1-k & 0 & k-1 & 0 \end{array}$$

\end{equation*}

• Yes, of course! Show please your attempts. Commented Apr 11, 2020 at 18:11
• I edited my first post to see how I started to work out the above system.
– Gina
Commented Apr 11, 2020 at 18:22
• I think, it's better to use the Cramer's rule. en.wikipedia.org/wiki/Cramer%27s_rule See now. Commented Apr 11, 2020 at 18:24
• In principle, this is duplicate of math.stackexchange.com/q/2921226/265466 and many others on this site. Look through the handy list of related questions at right and you’ll find several examples of similar problems.
– amd
Commented Apr 11, 2020 at 19:17
• @MichaelRozenberg I mostly agree, with the caveat that when all of the determinants vanish, you have to investigate further. Unlike the $2\times2$ case, for a $3\times3$ system you can’t immediately conclude that the system is indeterminate in that situation. See en.wikipedia.org/wiki/… for an example of an incompatible system for which all of the determinants vanish.
– amd
Commented Apr 11, 2020 at 19:27

I think Michael Rozenberg’s suggestion to use Cramer’s rule is a good one because the determinants are all very easy to compute. However, if you must proceed via row-reduction, you need to proceed systematically: your goal should be to end up with all zeros below each pivot. You also need to be a bit careful about the operations that you perform. The operation $$R_n\to cR_n+R_m$$, which is what you did for your second step, isn’t an elementary row operation, and doing thing like this can cause problems for you in other calculations, such as computing the determinant of a matrix. This is really a combination of two elementary operations: multiply a row by a constant and add one row to another.

Anyway, with the goal of zeroing out everything below the $$1$$ in the upper-left corner, you should add $$-k$$ times the first row the the third to obtain $$\left[\begin{array}{ccc|c}1&1&k&1\\0&k-1&1-k&0\\0&1-k&1-k^2&1-k\end{array}\right].$$ At this point, it should be obvious that when $$k=1$$, the last two rows are both zero, so there’s an infinite number of solutions in that case. We continue with the row reduction by adding the second row to the third: $$\left[\begin{array}{ccc|c}1&1&k&1\\0&k-1&1-k&0\\0&0&2-k-k^2&1-k\end{array}\right].$$ Now examine the last row to determine which values of $$k$$ result in none, one or an infinite number of solutions.

$$\Delta=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ k & 1 & 1 \end{array}\right)\right|=-k^3+3k-2=-(k+2)(k-1)^2,$$ $$\Delta_x=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ 1 & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$

$$\Delta_y=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & 1 & 1\\ k & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$ $$\Delta_z=\left|\left(\begin{array}{cc}1 & 1 & 1\\1 & k & 1\\ k & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$ which gives the answer:

i) $$k\neq-2$$ and $$k\neq1$$;

ii) $$k=1$$;

iii) $$k=-2$$.

• Using Cramer's rule when I calculated the determinant I found: $3k -k^3-2$. Also what $\Delta_x, \Delta_y, \Delta_z$ mean?
– Gina
Commented Apr 11, 2020 at 18:34
• @Gina Wait please. I'll explain this point, Commented Apr 11, 2020 at 18:49
• Oki, I was trying to convert the initial matrix to identity matrix. Because this is what we've been taught..
– Gina
Commented Apr 11, 2020 at 18:55
• How did you decide that there’s an infinite number of solutions when $k=1$? For a $3\times3$ system, you can’t conclude this from only the fact that all of the determinants vanish. For example, all of the determinants also vanish in the system $x+y+z=1$, $x+y+z=2$, $x+y+z=3$, but that system obviously doesn’t have an infinite number of solutions.
– amd
Commented Apr 11, 2020 at 19:33
• @amd In our problem for $k=1$ we got $\Delta_x=\Delta_y=\Delta_z=\Delta=0$. Did you get the same result for your system? If so, we can substitute $k=1$ in the original system. Commented Apr 11, 2020 at 19:38