Question

Test the consistency and solve them when consistent the following system of equations for all values of $\lambda$:
$x+y+z=1,x+3y-2z=\lambda ,3x+(\lambda +z)y-3z=2\lambda +1$.

Answer 1

After making two zeros and expanding, we get

$△=\left|\begin{array}{ccc}1& 1& 1\\ 1& 3& -2\\ 3& \lambda +2& -3\end{array}\right|=3(\lambda -5)$

${△}_x=\left|\begin{array}{ccc}1& 1& 1\\ \lambda & 3& -2\\ 2\lambda +1& 2\lambda +2& -3\end{array}\right|=(\lambda -5)(\lambda +2)$

${△}_y=\left|\begin{array}{ccc}1& 1& 1\\ 1& \lambda & -2\\ 3& 2\lambda +1& -3\end{array}\right|=0$

${△}_z=\left|\begin{array}{ccc}1& 1& 1\\ 1& 3& \lambda \\ 3& \lambda +2& 2\lambda +1\end{array}\right|=-(\lambda -1)(\lambda -5)$

$△=0$ i.e, $\lambda =5$ the system has unique solution given by

$\frac{x}{△x}=\frac{y}{{△}_y}=\frac{z}{{△}_z}=\frac{1}{△}$

Hence the system has infinite solutions.

Putting $\lambda =5$ and eliminating z, we have $3x+5y=7$ i.e only one equation in two variables.

Putting $x=c,y=\frac{7-3c}{5}$ and hence from any $z=-\frac{2(1+c)}{5}$