For the system of linear equations:

x – 2y = 1, x – y + kz = -2, ky + 4z = 6, k R ,  

Consider the following statements:

(A) The system has unique solution if k 2 , k 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 2 .  

Which of the following statements are correct?

Option 1 - <p>(B) and (E) only</p>
Option 2 - <p>(A) and (E) only</p>
Option 3 - <p>(C) and (D) only</p>
Option 4 - <p>(A) and (D) only</p>
2 Views|Posted 4 months ago
Asked by Shiksha User
1 Answer
V
4 months ago
Correct Option - 3
Detailed Solution:

x – 2y = 1, x – y + kz = -2, ky + 4z = 6

x – 2y + 0. z – 1 = 0

x – y + kz + 2 = 0

0x + ky + 4z – 6 = 0

0x + ky + 4z – 6 = 0

Δ 1 = | 1 2 0 2 1 k 6 k 4 | = ( k + 1 0 ) ( k + 2 )

For no solution

Δ = 0 , Δ 1 0

k = 2

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