JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Matrices

Let the system of linear equations

x + y + k z = 2

2 x + 3 y - z = 1

3 x + 4 y + 2 z = k

have infinitely many solutions. Then the system

(k + 1) x + (2 k - 1) y = 7

(2 k + 1) x + (k + 5) y = 10

has:

Answer: (D)

Solution

Infinitely many solutions of the first system require its coefficient determinant to vanish:

123134 k - 1 2 = 10 - 7 - k = 3 - k = 0 \Rightarrow k = 3,

and

k = 3

makes the third equation the sum of the first two, so the system is consistent. Substituting

k = 3

into the second pair gives

4 x + 5 y = 7

and

7 x + 8 y = 10

, whose determinant

32 - 35 = - 3 \neq = 0

yields the unique solution

x = - 2, y = 3

. Since

x + y = - 2 + 3 = 1

, the second system has a unique solution satisfying

x + y = 1.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option