Question

If the system of linear equations,

x + y + z = 6

x + 2y + 3z = 10

3x + 2y + λz = µ

has more two solutions, then µ - λ² is equal to ....

Answer 1

x + y + z = 6 ....(1)

x + 2y + 3z = 10 ...(2)

3x + 2y + λz = µ ...(3)

from (1) and (2)

if z = 0 ⇒ x + y = 6 and x + 2y = 10

⇒ y = 4, x = 2

(2,4,0)

if y = 0 ⇒ x + z = 6 and x + 3z = 10

⇒ z = 2 and x = 4

(4,0,2)

So, 3x + 2y + λz = µ

must pass through (2,4,0) and (4,0,2)

so 6 + 8 = µ ⇒ µ = 14

and 12 + 2λ = µ

12 + 2λ = 14 ⇒ λ = 1

so µ – λ² = 14 – 1

= 13

Answer 2

System has intfinitely many solution

µ = 14

µ - λ² = 13