Question

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

Answer 1

$3x-y+yz=-3------\left(1\right)$

$3x+2y-3z=-3------\left(2\right)$

$-26x+5y+\lambda z=-3-------\left(3\right)$

$\left[\begin{array}{ccc}3& -1& 4\\ 3& 2& -3\\ -26& 5& -5\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-3\\ -3\\ -3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-3\\ -3\\ -3\end{array}\right]\left[\begin{array}{ccc}1/38& 3/38& -1/38\\ 93/190& 89/190& 21/190\\ 67/190& 11/190& 9/190\end{array}\right]\therefore x=\frac{-9}{38},y=\frac{-609}{190},z=\frac{-261}{190}$