daktielti

2022-07-12

Question: For the value(s) of $k$, if any, will the following system have (a) no solution, (b) a unique solution, (c) infinity many solutons:
$x+y+kz=1,\phantom{\rule{0ex}{0ex}}x+ky+z=1,\phantom{\rule{0ex}{0ex}}xk+y+z=-2.$

lywiau63

Expert

Let's calculate this determinant
$det\left(\begin{array}{ccc}1& 1& k\\ 1& k& 1\\ k& 1& 1\end{array}\right)=det\left(\begin{array}{ccc}k+2& 1& k\\ k+2& k& 1\\ k+2& 1& 1\end{array}\right)=\left(\begin{array}{ccc}k+2& 1& k\\ 0& k-1& 1-k\\ 0& 0& 1-k\end{array}\right)=-\left(k+2\right)\left(k-1{\right)}^{2}$
hence
- If $k\ne 1$ and $k\ne -2$ there's exactly one solution.
- If $k=1$ we see easily that the $3$ equations aren't compatible so there is no solution.
- If $k=-2$ we can see also easily that there's infinitely many solutions.

Frank Day

Expert

Just add. This gives $\left(k+1\right)\left(x+y+z\right)=0$. If $k=-1$, then there are an infinite number of solutions.
Then if $k=0$... and so on.

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