JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Matrices

The system of equation

k x + y + z = 1

x + k y + z = k

and

x + y + z k = k^{2}

has no solution if

k

is equal to:

Answer: (C)

Solution

The coefficient determinant is

k 11 1 k 1 11 k = k^{3} - 3 k + 2 = (k - 1)^{2} (k + 2),

which vanishes only at

k = 1

and

k = - 2

. For

k = 1

all three equations reduce to

x + y + z = 1

, so the system has infinitely many solutions. Adding the three equations in general gives

(k + 2) (x + y + z) = 1 + k + k^{2}

; at

k = - 2

this reads

0 = 1 - 2 + 4 = 3

, which no triple

(x, y, z)

can satisfy. The system has no solution when

k = - 2

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option