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JEE Main 202124 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2021 · 24 February · Evening Shift · Mathematics · Determinants

For the system of linear equations: x - 2y = 1, x - y + kz = - 2,...

For the system of linear equations:

x−2y=1,x−y+kz=−2,ky+4z=6,k∈Rx - 2y = 1, x - y + kz = - 2, ky + 4z = 6, k \in R

Consider the following statements:
(A) The system has unique solution if k≠2,k≠−2.k \ne 2, k \ne - 2 .
(B) The system has unique solution if k=−2.k = - 2 .
(C) The system has unique solution if k=2.k = 2 .
(D) The system has no-solution if k=2.k = 2 .
(E) The system has infinite number of solutions if k≠−2.k \ne - 2 .

Which of the following statements are correct?

Options

  1. A(A)\left(A\right) and (E)\left(E\right) only
  2. B(B)\left(B\right) and (E)\left(E\right) only
  3. C(A)\left(A\right) and (D)\left(D\right) only
  4. D(C)\left(C\right) and (D)\left(D\right) only

Correct answer

C. (A)\left(A\right) and (D)\left(D\right) only

Step-by-step solution

D = \left|\begin{cases}1 & - 2 & 0 \\ 1 & - 1 & k \\ 0 & k & 4\end{cases}\right| = 4 - {k}^{2}

so, AA is correct and B,C,EB, C, E are incorrect. If k=±2k = \pm 2

{D}_{1} = \left|\begin{cases}1 & - 2 & 0 \\ - 2 & - 1 & 2 \\ 6 & 2 & 4\end{cases}\right| = - 48 \ne 0

So, no solution if k=±2k = \pm 2

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