# For the system of linear equations:

Question:

For the system of linear equations:

$x-2 y=1, x-y+k z=-2, k y+4 z=6, k \in \mathbf{R}$

consider the following statements:

(A) The system has unique solution if $k \neq 2, k \neq-2$.

(B) The system has unique solution if $k=-2$.

(C) The system has unique solution if $k=2$.

(D) The system has no-solution if $k=2$.

(E) The system has infinite number of solutions if $k \neq-2$.

Which of the following statements are correct?

1. (1) $(B)$ and $(E)$ only

2. (2) $(C)$ and $(D)$ only

3. (3) (A) and (D) only

4. (4) (A) and (E) only

Correct Option: , 3

Solution:

$x-2 y+0 . z=1$

$x-y+k z=-2$

$0 \cdot x+k y+4 z=6$

$\Delta=\left|\begin{array}{ccc}1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4\end{array}\right|=4-k^{2}$

For unique solution $4-\mathrm{k}^{2} \neq 0$

$\mathrm{k} \neq \pm 2$

For $\mathrm{k}=2$

$x-2 y+0 . z=1$

$x-y+2 z=-2$

$0 \cdot x+2 y+4 z=6$

$\Delta \mathrm{x}=\left|\begin{array}{ccc}1 & -2 & 0 \\ -2 & -1 & 2 \\ 6 & 2 & 4\end{array}\right|=(-8)+2[-20]$

$\Delta x=-48 \neq 0$

For $\mathrm{k}=2 \quad \Delta \mathrm{x} \neq 0$

For $\mathrm{K}=2$; The system has no solution