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Matrices and Determinants question

2021 · 18 Mar · Shift 1 · Q27
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  5. /2021 · 18 Mar · Shift 1 · Q27

Matrices and Determinants question

2021 · 18 Mar · Shift 1 · Q27

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let α\alphaα, β\betaβ, γ\gammaγ be the real roots of the equation, x3 + ax2 + bx + c = 0, (a, b, c ∈\in∈ R and a, b eee 0). If the system of equations (in u, v, w) given by α\alphaα u + β\betaβ v + γ\gammaγ w = 0, β\betaβ u + γ\gammaγ v + α\alphaα w = 0; γ\gammaγ u + α\alphaα v + β\betaβ w = 0 has non-trivial solution, then the value of a2b{{{a^2}} \over b}ba2​ is
  1. A
    5
  2. B
    3
  3. C
    1
  4. D
    0
View written solutionFree

Correct answer: B

  1. Write the system in matrix form

The given homogeneous system is

{αu+βv+γw=0βu+γv+αw=0γu+αv+βw=0\begin{cases} \alpha u + \beta v + \gamma w = 0 \\ \beta u + \gamma v + \alpha w = 0 \\ \gamma u + \alpha v + \beta w = 0 \end{cases}⎩⎨⎧​αu+βv+γw=0βu+γv+αw=0γu+αv+βw=0​

So the coefficient matrix is

A=(αβγβγαγαβ)A=\begin{pmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{pmatrix}A=​αβγ​βγα​γαβ​​

For a non-trivial solution to exist, since the system is homogeneous, we must have

det⁡(A)=0.\det(A)=0.det(A)=0.
  1. Compute the determinant

For a circulant-type matrix of this form,

det⁡(A)=−(α+β+γ)(α2+β2+γ2−αβ−βγ−γα).\det(A)=-(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha\right).det(A)=−(α+β+γ)(α2+β2+γ2−αβ−βγ−γα).

Equivalently,

det⁡(A)=−(α3+β3+γ3−3αβγ).\det(A)=-(\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma).det(A)=−(α3+β3+γ3−3αβγ).

Thus, for non-trivial solution,

(α+β+γ)(α2+β2+γ2−αβ−βγ−γα)=0.(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha\right)=0.(α+β+γ)(α2+β2+γ2−αβ−βγ−γα)=0.

So either

α+β+γ=0\alpha+\beta+\gamma=0α+β+γ=0

or

α2+β2+γ2−αβ−βγ−γα=0.\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha=0.α2+β2+γ2−αβ−βγ−γα=0.

But

α2+β2+γ2−αβ−βγ−γαn=12[(α−β)2+(β−γ)2+(γ−α)2].\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha n=\frac12\big[(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2\big].α2+β2+γ2−αβ−βγ−γαn=21​[(α−β)2+(β−γ)2+(γ−α)2].

Since α,β,γ\alpha,\beta,\gammaα,β,γ are real roots, this quantity is 000 iff

α=β=γ.\alpha=\beta=\gamma.α=β=γ.

If α=β=γ=r\alpha=\beta=\gamma=rα=β=γ=r, then the cubic is

(x−r)3=x3−3rx2+3r2x−r3.(x-r)^3=x^3-3rx^2+3r^2x-r^3.(x−r)3=x3−3rx2+3r2x−r3.

Hence

a=−3r,b=3r2.a=-3r, \qquad b=3r^2.a=−3r,b=3r2.

Therefore,

a2b=9r23r2=3.\frac{a^2}{b}=\frac{9r^2}{3r^2}=3.ba2​=3r29r2​=3.
  1. Check the other possibility

If

α+β+γ=0,\alpha+\beta+\gamma=0,α+β+γ=0,

then by Vieta,

α+β+γ=−a  ⟹  a=0,\alpha+\beta+\gamma=-a \implies a=0,α+β+γ=−a⟹a=0,

but the question states a≠0a\ne 0a=0. So this case is not allowed.

Thus the only valid possibility is

α=β=γ.\alpha=\beta=\gamma.α=β=γ.

Hence,

a2b=3.\boxed{\frac{a^2}{b}=3}.ba2​=3​.
  1. Option check
  • A: 555 ✗
  • B: 333 ✓
  • C: 111 ✗
  • D: 000 ✗

So the correct option is B.

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