I was helping my comrade answering some questions taken from review classes when I stumbled upon this question. It looks like this:

The number of values of $k$ for which the linear equations $$\begin{align}4x + ky + 2z &= 0 \\kx + 4y + z &= 0 \\ 2x + 2y + kz &= 0\end{align}$$ possess a nonzero solution is.....

My work

I honestly don't know how to answer this question. I don't understand what the statement "possess a nonzero solution is....." means. I thought that it would mean $x$ and $y$ and $z$ musn't equal to zero. Moreover, I'm dealing in $4$ unknowns and $3$ equations, making matters worse.

How to answer the above question?

  • $\begingroup$ It means for which values of $k$ do the equations above, have at least one solution in $(x,y,z)$ which is not $(0,0,0)$ ? $\endgroup$ – zwim Sep 27 '17 at 8:32
  • $\begingroup$ Let's link to this important justification of the determinant method : math.stackexchange.com/questions/1455226/… $\endgroup$ – zwim Sep 27 '17 at 10:02

We have the matrix:$$\begin{bmatrix} 4&k&2\\k&4&1\\2&2&k\\\end{bmatrix}$$ Since the determinant of the matrix equals $0$, $$\begin{align}4(4k-2)-k(k^2-2)+2(2k-8)&=0 \\ 16k-8-k^3+2k+4k-16&=0 \\ -k^3+22k-24&=0 \\ (k-4)(k^2+4k-6)&=0 \end{align}$$ So either $k^2+4k-6=0$ or $k-4=0$

If the former applies, then $$\begin{align}k^2+4k&=6\\k^2+4k+4&=10\\(k+2)^2&=10 \\k+2&=\pm\sqrt{10}\end{align}$$ so $k=\sqrt{10}-2,$ or $-\sqrt{10}-2$.

If the latter applies, then $k$ simply equals $4$.

Thus, our solutions are $4,\sqrt{10}-2,$ or $-\sqrt{10}-2$, and so there are three solutions for $k$.

  • $\begingroup$ typo $A_{3,3}=k$ $\endgroup$ – zwim Sep 27 '17 at 8:47
  • $\begingroup$ @zwim fixed, thanks $\endgroup$ – user472341 Sep 27 '17 at 8:48
  • $\begingroup$ determinant should be $4(4k-2)-k(k^2-2)+2(2k-8)$ $\endgroup$ – zwim Sep 27 '17 at 8:57
  • $\begingroup$ @zwim My internet went out, sorry. edited the entire solution $\endgroup$ – user472341 Sep 27 '17 at 9:20

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