Question

Show that the system of equations, $3x-y+4z=3,x+2y-3z=-2$ and $6x+5y+\lambda z=-3$ has atleast one solution for any real number $\lambda \ne -5$. Find the set of solutions, if $\lambda =-5$.

Answer 1

The given system of equatios
$3x-y+4z=3$
$x+2y-3z=-2$
$6x+5y+\lambda z=-3$
has atleast one solution ,if $\Delta \ne 0,$
$\therefore \Delta =\left|\begin{array}{ccc}3& -1& 4\\ 1& 2& -3\\ 6& 5& \lambda \end{array}\right|\ne 0$
$\Rightarrow 3(2\lambda +15)+1(\lambda +18)+4(5-12)\ne 0$
$\Rightarrow 7(\lambda +5)\ne 0\Rightarrow \lambda \ne -5$
Let $z=-k$, then equation become
$3x-y=3-4k$ and $x+2y=3k-2$
on solving, we get
$x=\frac{4-5k}{7},y=\frac{13k-9}{7},z=k$