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← JEE Main 2022 · 24 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 3 of 90

JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 3

Let the system of linear equations
x+y+αz=2x+y+\alpha z=2
3x+y+z=43x+y+z=4
x+2z=1x+2z=1
have a unique solution (x∗,y∗,z∗)(x^*,y^*,z^*). If (α,x∗)(\alpha,x^*), (y∗,α)(y^*,\alpha) and (x∗,−y∗)(x^*,-y^*) are collinear points, then the sum of absolute values of all possible values of α\alpha is

Choose one option
Answer and worked solution

Official answer: C

Solve. From the third equation x=1−2zx=1-2z; the second gives y=1+5zy=1+5z; the first then gives (3+α)z=0(3+\alpha)z=0. For a unique solution α≠−3\alpha\ne-3, so z=0z=0, x∗=1x^*=1, y∗=1y^*=1.

Collinearity of (α,1)(\alpha,1), (1,α)(1,\alpha), (1,−1)(1,-1):
∣α111α11−11∣=α(α+1)−0+(−1−α)=α2−1=0.\begin{vmatrix}\alpha&1&1\\1&\alpha&1\\1&-1&1\end{vmatrix}=\alpha(\alpha+1)-0+(-1-\alpha)=\alpha^2-1=0.

So α=±1\alpha=\pm1, and the sum of absolute values is 2.

(C)

Topics:Systems of linear equationsDistance and section formulae
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