Question

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

Answer 1

$\left|\begin{array}{ccc}1& 1& 1\\ 1& 3& -2\\ 3& \lambda +2& -3\end{array}\right|\left|\begin{array}{c}1\\ \lambda \\ 2\lambda +1\end{array}\right|$

$△=1(-9+2\lambda +4)-1(-3+6)+1(\lambda +2-9)$

$△=2\lambda -5-3+\lambda -7$

$△=3\lambda -15$

The given equations are

$x+y+z=1⟶1$

$x+3y-27=\lambda ⟶2$

$3x+(\lambda +2)y-3z=2\lambda +1⟶3$

Irrespective of values of $\lambda ,x,y,z$

From equation $1$ we see that $x+y+z=1$

Answer:$1$

For all values of $\alpha$, we find that $x+y+z=1$.