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Question

For the system of linear equations :
x2y=1, xy+kz=2,ky+4z=6, kR,
consider the following statements :
(A) The system has unique solution if k2,k2.
(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 k2.

Which of the following statements are correct ?

A
(B) and (E) only
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B
(C) and (D) only
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C
(A) and (D) only
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D
(A) and (E) only
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Solution

The correct option is C (A) and (D) only
x2y+0z=1
xy+kz=2
0x+ky+4z=6
Δ=∣ ∣12011k0k4∣ ∣=4k2
For unique solution,
4k20
k±2

For k=2,
x2y+0z=1
xy+2z=2
0x+2y+4z=6
Δx=∣ ∣120212624∣ ∣=(8)+2(20)
Δx=480
For k=2,Δx0
So, for k=2, the system has no solution.

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