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 $\Delta \ne 0,$
$\therefore \Delta=\begin {vmatrix}3 & -1 & 4 \\1 & 2 & -3 \\6 & 5 & \lambda \end {vmatrix} \ne 0$
$\Rightarrow 3(2\lambda +15)+1(\lambda+18)+4(5-12) \ne 0$
$\Rightarrow 7(\lambda +5 ) \ne 0 \Rightarrow \lambda \ne -5$
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$