Question

For what real values of k, the system of equations
$x+2y+z=1,x+3y+4z=k;x+5y+10z=k^2$ has solution? Find the solution in each case.

Answer 1

$△=\left|\begin{array}{ccc}1& 2& 1\\ 1& 3& 4\\ 1& 5& 10\end{array}\right|=0$ after making two zeros
$\therefore$ Solution is not unique.
The system will have infinite solutions if ${△}_1=0,{△}_2=0$
${△}_3=0$
${△}_1=0\Rightarrow k^2-3k+2=0\therefore k=1,2$
For these values ${△}_2$ and ${△}_3$ are also zero.
Now take $z=\lambda ,k=2$ then $x=-1+5\lambda ,y=1-3\lambda$