Question

Discuss the solution of the following system of equation for all values of $\lambda$.
$x+y+z=2,2x+y-2z=2,\lambda x+y+4z=2$

Answer 1

The matrix equation is

$\left[\begin{array}{ccc}1& 1& 1\\ 2& 1& -2\\ \lambda & 1& 4\end{array}\right]\begin{array}{c}x\\ \begin{array}{c}y\\ z\end{array}\end{array}\begin{array}{c}2\\ \begin{array}{c}2\\ 2\end{array}\end{array}$

The augmented matrix

$[A,B]=\left[\begin{array}{cccc}1& 1& 1& 1\\ 2& 1& -2& 2\\ \lambda & 1& 4& 2\end{array}\right]$

$\sim \left[\begin{array}{cccc}1& 1& 1& 2\\ 0& -1& -4& -2\\ \lambda & 1-\lambda & 4-\lambda & 2\lambda \end{array}\right]R_2$

$\to R_2-2R_1$

$R_3\to R_3-\lambda R_1$

$\sim \left[\begin{array}{cccc}1& 1& 1& 2\\ 0& -1& -4& -2\\ 0& -\lambda & -\lambda & -2\lambda \end{array}\right]R_3\to R_3$

$+R_2$

When $\lambda =0$ the last equivalent matrix

becomes $\left[\begin{array}{cccc}1& 1& 1& 2\\ 0& -1& -4& -2\\ 0& 0& 0& 0\end{array}\right]$ which is in

echelon form.

There are two nonzero rows and $\rho \left(A\right)=\rho (A,B)=2$

The system is consistent with infinitely

many solutions.

When $\lambda \ne 0$, there are $3$ nonzero rows

and $\rho \left(A\right)=\rho (A,B)=2$. The system has a unique solution.