UPSEE Solved Model Paper 1

When Δ |

1	2	3
4	6	2λ
1	λ	3

| = 0,
there can be two possibilities
(i) No solution
(ii) Infinite solution
But for Infinite solution
Δ1 is the determinant obtained by replacing the third column of Δ by the elements of the constant matrix.
Δ1 = |

1	2	5
4	6	21
1	λ	3

|
Should also be zero
So, for no solution condition will be,
Δ = 0 and Δ1 = 0
Δ = 0 ⇒ |

1	2	3
4	6	2λ
1	λ	3

| = 0
⇒ 1(18 − 2𝜆2) − 2(12 − 2𝑖) + 3(4𝜆 − 6) = 0
⇒ 𝜆2 − 8𝜆 + 12 = 0
𝜆 = 2,6
Δ1 = 0 ⇒ |

1	2	5
4	6	21
1	λ	3

| = 0
⇒ 1(18 − 21) − 2(12 − 21) + 5(4𝑖 − 6) = 0
⇒ −𝜆 + 6 = 0
𝜆 = 6
So, 𝜆 can have only one value