NIMCET 2020 — Mathematics PYQ

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Question

NIMCET 2020 — Mathematics PYQ

NIMCET | Mathematics | 2020

The number of values of k for which the linear equations 4x + ky + z = 0 lx + 4y + z = 0 2x + 2y + z = 0 possess a non-zero solution is:

Choose the correct answer:

Explanation

Concept:
Cramer's rule for Linear Equations of Three Variables:
$a_{1} x + b_{1} y + c_{1} z = d_{1}, a_{2} x + b_{2} y + c_{2} z = d_{2}, a_{3} x + b_{3} y + c_{3} z = d_{3}$
$D = a_{1} b_{1} c_{1} am p; a_{2} am p; b_{2} am p; c_{2} am p; a_{3} am p; b_{3} am p; c_{3}$
$D_{x} = d_{1} b_{1} c_{1} am p; a_{2} am p; b_{2} am p; c_{2} am p; a_{3} am p; b_{3} am p; c_{3}$ $D_{y} = a_{1} b_{1} c_{1} am p; d_{2} am p; b_{2} am p; c_{2} am p; a_{3} am p; b_{3} am p; c_{3}$ $D_{z} = a_{1} d_{1} d_{1} am p; a_{2} am p; b_{2} am p; c_{2} am p; a_{3} am p; b_{3} am p; c_{3}$
- If $D \neq = 0$ : a unique solution (consistent).
The solution is: $x = \frac{D _{x}}{D}$ , $y = \frac{D _{y}}{D}$ , $z = \frac{D _{z}}{D}$ .
- If $D = 0$ : either infinitely many solutions (consistent and dependent) or no solution (inconsistent). To find out if the system is dependent or inconsistent, another method, such as elimination or Rouché–Capelli theorem, will have to be used.

Calculation:

For the given set of equations:
4x + ky + z = 0
kx + 4y + z = 0
2x + 2y + z = 0

It can be observed that x = y = z = 0 is one solution of the system.

In order to have infinitely many (including non-zero) solutions, D must be zero.

(10)

$D = 4 k 1 am p; k am p; 4 am p; 1 am p; 2 am p; 2 am p; 1 = 0$

$4 (4 - 2) + k (2 - k) + 2 (k - 4) = 0$

$8 + 2 k - k^{2} + 2 k - 8 = 0$

$k^{2} + 4 k = 0$

$k (k + 4) = 0$

$k = 0 OR k + 4 = 0$

$k = 0 OR k = - 4$

Hence, there are 2 possible values of k but non-zero solution is one.