1.

If system of equationx + y + z = 6x + 2y + 3z = 103x + 2y + λz = μhas more than two solutions. Find (μ – λ2)

Answer»

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


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