QJEEM21Mae090: JEE Main 2021 Mathematics - Class 12 JEEM21Mae090 Solution: For the system of linear equations: , consider the following…

Question

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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 unique solution if

k \ne 2,k \ne - 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

\ne

-

2.

Which of the following statements are correct?

Accepted Answer

x - 2y + 0.z = 1

x - y + kz = - 2

0.x + ky + 4z = 6

Δ = | \begin{matrix} 1 & - 2 & 0 \\ 1 & - 1 & k \\ 0 & k & 4 \end{matrix} | = 4 - k^{2}

For unique solution

4 - {k^2} \ne 0

\Rightarrow

k

\ne

\pm

2

For k = 2 :

x - 2y + 0.z = 1

x - y + 2z = - 2

0.x + 2y + 4z = 6

Δ x = | \begin{matrix} 1 & - 2 & 0 \\ 2 & - 1 & 2 \\ 6 & 2 & 4 \end{matrix} | = (- 8) + 2 [- 20]

\Delta x = - 48 \ne 0

For k = 2,

\Delta x \ne 0

\therefore

For K = 2; The system has no solution.

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