Given that $$\begin{array}{} 2x & – & y & + & 2z & = & 2 \\ x & – & 2y & + & z & = & -4 \\ x & + & y & + & \lambda z & = & 4 \end{array}$$
Non-homogeneous eqautions is the form of $AX = B$ and augmented matrix $[A:B]$
Case$1:$ If $\text{rank}(A) \neq \text{rank}([A:B]),$ then no solution.
Case$2:$ If $\text{rank}(A) = \text{rank}([A:B]),$ then
- If $\text{rank}(A) = \text{rank}([A:B]) = \text{number of variables(unknown)},$ then unique solution.
- If $\text{rank}(A) = \text{rank}([A:B]) < \text{number of variables(unknown)},$ then infinite solution.
Now, we can write the above equations into an augmented form
$[A: B]=\begin{bmatrix} 2&-1 &2 &:2 \\ 1&-2 &1 &:-4 \\ 1& 1& \lambda&:4 \end{bmatrix}$
Operation $R_{3}\rightarrow R_{3} + R_{2}$
$[A: B]=\begin{bmatrix} 2&-1 &2 &:2 \\ 1&-2 &1 &:-4 \\ 2& -1& \lambda + 1&:0 \end{bmatrix}$
Operation $R_{3}\rightarrow R_{3} - R_{1}$
$[A: B]=\begin{bmatrix} 2&-1 &2 &:2 \\ 1&-2 &1 &:-4 \\ 0& 0& \lambda - 1 &:-2 \end{bmatrix}$
No solution means $\text{rank}(A) \neq \text{rank}([A:B])$
It is only possible when $\lambda - 1 = 0 \implies \lambda = 1$
Here $,\text{rank}(A) = 2$ and $\text{rank}([A:B]) = 3,$ this is satisfied the no solution condition.
$\therefore \lambda = 1$
So, the correct answer is $(B).$
