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Let the system of linear equations x + y + az = 2; 3x + y + z = 4 ; x + 2z = 1 have a unique solution (x*, y*, z*). If (α, x*), (y*, α) and (x*, –y*) are collinear points -

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Question

Let the system of linear equations x + y + az = 2; 3x + y + z = 4; x + 2z = 1 have a unique solution (x*, y*, z*). If (α, x*), (y*, α) and (x*, –y*) are collinear points, then the sum of absolute values of all possible values of α is ______.

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  • 3

  • 2

  • 1

MCQ
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Solution

Let the system of linear equations x + y + az = 2; 3x + y + z = 4; x + 2z = 1 have a unique solution (x*, y*, z*). If (α, x*), (y*, α) and (x*, –y*) are collinear points, then the sum of absolute values of all possible values of α is 1.

Explanation:

Given a system of linear equations and apply the determinant rule.

Δ = `|(1, 1, α),(3, 1, 1),(1, 0, 2)|` = – (α + 3);

Δ1 = `|(2, 1, α),(4, 1, 1),(1, 0, 2)|` = – (3 + α)

Δ2 = `|(1, 2, α),(3, 4, 1),(1, 1, 2)|` = – (α + 3);

Δ3 = `|(1, 1, 2),(3, 1, 4),(1, 0, 1)|` = 0

x = `Δ_1/Δ = (-(α + 3))/(-(3 + α))` = 1;

y = `Δ_2/Δ = (-(α + 3))/(-(α + 3))` = 1

z = `Δ_3/Δ = 0/(-(α + 3))` = 0

So, x = 1, y = 1, z = 0. Here, α cannot be –3.

Points (α, 1), (1, α) and (1, –1) are collinear, then

`|(α, 1, 1),(1, α, 1),(1, -1, 1)|` = 0

`\implies` α(α + 1) – 1(1 – 1) + 1(–1 – α) = 0

`\implies` α2 + α – 1 – α = 0

`\implies` α = ±1

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