Question

For the system of linear equations: $x-2y=1,x-y+kz=-2,ky+4z=6,k\in R$

Consider the following statements:

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

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

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

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

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

Which of the following statements are correct?

• A. (B) and (E) only
• B. (C) and (D) only
• C. (A) and (D) only
• D. (A) and (E) only

Answer 1

The correct option is C

(A) and (D) only

The explanation for the correct options:

Given data

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

$\Rightarrow$ $x–y+kz=-2$

$\Rightarrow$$0.x+ky+4z=6$

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

$=4-k^2$

For $k=2,$

The system of linear equations:

$$\begin{gathered} x–2y+0.z=1 \\ x–y+2z=-2 \\ 0.x+2y+4z=6 \end{gathered}$$

Now, calculate ${∆}_x$and ${∆}_y$

${∆}_x=\left|\begin{array}{ccc}1& -2& 0\\ -2& -1& 2\\ 6& 2& 4\end{array}\right|$ and ${∆}_y=\left|\begin{array}{ccc}1& -2& 1\\ -2& -1& -2\\ 6& 2& 6\end{array}\right|$

$\Rightarrow {∆}_x=-8+2(-20)$ and ${∆}_y=-2+0+2$

$\Rightarrow {\Delta }_x=-48\ne 0$ and ${∆}_y=0$

For no solution ${∆}_x\ne {∆}_y$

for $k=2,{\Delta }_x\ne {∆}_y$

So, for $k=2$, the system has no solution.

Thus the statement (A) and (D) is correct.

Hence the correct option is C.

The explanation for the incorrect options:

For unique solutions,

$4–k^2\ne 0$

$\Rightarrow$ $k\ne \pm 2$

Thus statement (B) is incorrect.

For an infinite number of solution ${∆}_x={∆}_y$

for $k\ne 2,{\Delta }_x\ne {∆}_y$

So, for $k\ne 2$, the system has no infinite number of solutions.

Thus the statement (E) is incorrect.

Hence, option(C) is the correct .