Question

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

Answer 1

$D=\left|\begin{array}{ccc}3& -1& 4\\ 1& 2& -3\\ 6& 5& \lambda \end{array}\right|\ne 0$ for atleast one real solution

$3(2\lambda ÷15)+(\lambda +18)+4(5-12)\ne 0$

$6\lambda +45+\lambda +18-28\ne 0$

$7\lambda \ne -35$

$\lambda \ne -5$ for atleast are real solution

If $\lambda =-5$ then

these are infinites solutions

so for particles $x=K$

$3k-y+4z=3$

$K+2y-3z=-2$

$y=3k-3+4z$

$k+6k-6+8z-3z=-2$

$5z=4-7k=z\frac{4-7k}{5}$

$y=3k-3+\frac{16-23k}{5}=-\frac{13k+1}{5}$

solution is $(k,\frac{-13k+1}{5},\frac{4-7k}{5})k\in R$