If the system of linear equations \(x-2 y+z=-4\) \(2 \mathrm{x}+\alpha \mathrm{y}+3 \mathrm{z}=5\) \(3 x-y+\beta z=3\) has infinitely many solutions, then \(12 \alpha+13 \beta\) is equal to

Matrices — JEE Main · 2024 · 31 Jan 2024 · Shift 1 Question with Solution

If the system of linear equations

x - 2 y + z = - 4

2 x + α y + 3 z = 5

3 x - y + β z = 3

has infinitely many solutions, then

12 α + 13 β

is equal to

Answer: (D)

x - 2 y + z = - 4 (1), 2 x + α y + 3 z = 5 (2), 3 x - y + β z = 3 (3)

From (1):

x = - 4 + 2 y - z

. Substituting into (2):

2 (- 4 + 2 y - z) + α y + 3 z = 5 ⟹ (4 + α) y + z = 13 (2^{'})

Substituting into (3):

3 (- 4 + 2 y - z) - y + β z = 3 ⟹ 5 y + (β - 3) z = 15 (3^{'})

From

(2^{'})

z = 13 - (4 + α) y

. Substituting into

(3^{'})

5 y + (β - 3) [13 - (4 + α) y] = 15

[5 - (β - 3) (4 + α)] y + 13 (β - 3) - 15 = 0

For this to hold for every

y

(infinitely many solutions), both the coefficient of

y

and the constant term must vanish:

13 (β - 3) = 15 ⟹ β - 3 = \frac{15}{13} ⟹ β = \frac{54}{13}

5 = (β - 3) (4 + α) = \frac{15}{13} (4 + α) ⟹ 4 + α = \frac{13}{3} ⟹ α = \frac{1}{3}

12 α + 13 β = 12 (\frac{1}{3}) + 13 (\frac{54}{13}) = 4 + 54 = 58

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option