Let the system of linear equations $x + y + \alpha z = 2$ $3x + y + z = 4$ $x + 2z = 1$ have a unique solution $(x^{*}, y^{*}, z^{*})$. If $(\alpha, x^{*})$, $(y^{*}, \alpha)$ and $(x^{*}, -y^{*})$ are collinear points, then the sum of absolute values of all possible values of $\alpha$ is:

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Determinants

Let the system of linear equations

x + y + α z = 2

3 x + y + z = 4

x + 2 z = 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:

Answer: (C)

For a unique solution, the determinant

Δ = 131110 α 12 = - (α + 3) \neq = 0

, so

α \neq = - 3

.

Using Cramer's rule, we find

x^{*} = 1

y^{*} = 1

z^{*} = 0

.

For collinearity of

(α, 1)

(1, α)

, and

(1, - 1)

α 11 1 α - 1 111 = 0

Expanding:

α (α + 1) - 1 (0) + 1 (- 1 - α) = 0

α^{2} + α - 1 - α = 0

α^{2} = 1

, so

α = \pm 1

Sum of absolute values:

∣1∣ + ∣ - 1∣ = 2

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option