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Matrices and Determinants question

2021 · 31 Aug · Shift 2 · Q22
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Matrices and Determinants question

2021 · 31 Aug · Shift 2 · Q22

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If α\alphaα+β\betaβ+γ\gammaγ = 2 π\piπ, then the system of equations x + (cos γ\gammaγ)y + (cos β\betaβ)z = 0 (cos γ\gammaγ)x + y + (cos α\alphaα)z = 0 (cos β\betaβ)x + (cos α\alphaα)y + z = 0 has :
  1. A
    no solution
  2. B
    infinitely many solution
  3. C
    exactly two solutions
  4. D
    a unique solution
View written solutionFree

Correct answer: B

  1. Write the system in matrix form:
[1cos⁡γcos⁡βcos⁡γ1cos⁡αcos⁡βcos⁡α1][xyz]=[000]\begin{bmatrix} 1 & \cos\gamma & \cos\beta \\ \cos\gamma & 1 & \cos\alpha \\ \cos\beta & \cos\alpha & 1 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}​1cosγcosβ​cosγ1cosα​cosβcosα1​​​xyz​​=​000​​

This is a homogeneous linear system. Such a system has:

  • a unique solution iff determinant ≠0\neq 0=0,
  • infinitely many solutions iff determinant =0=0=0.

So we only need to compute the determinant.

  1. Let

a=cos⁡α,b=cos⁡β,c=cos⁡γ.a=\cos\alpha,\quad b=\cos\beta,\quad c=\cos\gamma.a=cosα,b=cosβ,c=cosγ.

Then the coefficient matrix is

A=[1cbc1aba1]A=\begin{bmatrix} 1 & c & b\\ c & 1 & a\\ b & a & 1 \end{bmatrix}A=​1cb​c1a​ba1​​

and

det⁡(A)=1(1−a2)−c(c−ab)+b(ca−b).\det(A)=1(1-a^2)-c(c-ab)+b(ca-b).det(A)=1(1−a2)−c(c−ab)+b(ca−b).

Simplifying,

det⁡(A)=1−a2−c2+abc+abc−b2\det(A)=1-a^2-c^2+abc+abc-b^2det(A)=1−a2−c2+abc+abc−b2

so

det⁡(A)=1−a2−b2−c2+2abc.\det(A)=1-a^2-b^2-c^2+2abc.det(A)=1−a2−b2−c2+2abc.

Hence,

det⁡(A)=1−cos⁡2α−cos⁡2β−cos⁡2γ+2cos⁡αcos⁡βcos⁡γ.\det(A)=1-\cos^2\alpha-\cos^2\beta-\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma.det(A)=1−cos2α−cos2β−cos2γ+2cosαcosβcosγ.
  1. Use the identity when α+β+γ=2π\alpha+\beta+\gamma=2\piα+β+γ=2π.

Since α+β+γ=2π\alpha+\beta+\gamma=2\piα+β+γ=2π, we have

γ=2π−(α+β)\gamma=2\pi-(\alpha+\beta)γ=2π−(α+β)

and therefore

cos⁡γ=cos⁡(α+β).\cos\gamma=\cos(\alpha+\beta).cosγ=cos(α+β).

Now use the standard identity:

cos⁡2α+cos⁡2β+cos⁡2γ=1+2cos⁡αcos⁡βcos⁡γ\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1+2\cos\alpha\cos\beta\cos\gammacos2α+cos2β+cos2γ=1+2cosαcosβcosγ

whenever α+β+γ=2π\alpha+\beta+\gamma=2\piα+β+γ=2π (equivalently, modulo 2π2\pi2π).

Therefore,

1−cos⁡2α−cos⁡2β−cos⁡2γ+2cos⁡αcos⁡βcos⁡γ=0.1-\cos^2\alpha-\cos^2\beta-\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=0.1−cos2α−cos2β−cos2γ+2cosαcosβcosγ=0.

So,

det⁡(A)=0.\det(A)=0.det(A)=0.

  1. Since the system is homogeneous and the determinant is zero, the coefficient matrix is singular. Therefore the system has non-trivial solutions, and hence infinitely many solutions.

  2. Check options:

  • A: no solution ❌ (homogeneous system always has at least trivial solution)
  • B: infinitely many solution ✅
  • C: exactly two solutions ❌
  • D: a unique solution ❌ (would require determinant nonzero)

Therefore the correct option is B.

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