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MCQMediumJEE 2024Solving Linear Equations (Matrix Method)

JEE Mathematics 2024 Question with Solution

If the system of equations x+(2sin⁡α)y+(2cos⁡α)z=0,x + (\sqrt{2} \sin \alpha) y + (\sqrt{2} \cos \alpha) z = 0, x + (\cos \alpha) y + (\sin \alpha) z = 0, x + (\sin \alpha) y − (\cos \alpha) z = 0 has a non-trivial solution, then α\alpha in the interval (0,π/2)(0, \pi/2) is equal to:

  • A

    3π/43\pi/4

  • B

    7π/247\pi/24

  • C

    5π/245\pi/24

  • D

    11π/2411\pi/24

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The homogeneous system has a non-trivial solution.

Find: The value of α\alpha in (0,π/2)(0, \pi/2).

For a homogeneous system to have a non-trivial solution, the determinant of its coefficient matrix must be zero.

∣12sin⁡α2cos⁡α1cos⁡αsin⁡α1sin⁡α−cos⁡α∣=0\begin{vmatrix} 1 & \sqrt{2}\sin\alpha & \sqrt{2}\cos\alpha \\ 1 & \cos\alpha & \sin\alpha \\ 1 & \sin\alpha & -\cos\alpha \end{vmatrix} = 0

Now expand along the first column:

D=∣cos⁡αsin⁡αsin⁡α−cos⁡α∣−∣2sin⁡α2cos⁡αsin⁡α−cos⁡α∣+∣2sin⁡α2cos⁡αcos⁡αsin⁡α∣\begin{aligned} D &= \begin{vmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} - \begin{vmatrix} \sqrt{2}\sin\alpha & \sqrt{2}\cos\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} + \begin{vmatrix} \sqrt{2}\sin\alpha & \sqrt{2}\cos\alpha \\ \cos\alpha & \sin\alpha \end{vmatrix} \end{aligned}

Evaluate each minor:

∣cos⁡αsin⁡αsin⁡α−cos⁡α∣=−cos⁡2α−sin⁡2α=−1∣2sin⁡α2cos⁡αsin⁡α−cos⁡α∣=−22sin⁡αcos⁡α∣2sin⁡α2cos⁡αcos⁡αsin⁡α∣=2(sin⁡2α−cos⁡2α)\begin{aligned} \begin{vmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} &= -\cos^2\alpha - \sin^2\alpha = -1 \\ \begin{vmatrix} \sqrt{2}\sin\alpha & \sqrt{2}\cos\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} &= -2\sqrt{2}\sin\alpha\cos\alpha \\ \begin{vmatrix} \sqrt{2}\sin\alpha & \sqrt{2}\cos\alpha \\ \cos\alpha & \sin\alpha \end{vmatrix} &= \sqrt{2}(\sin^2\alpha - \cos^2\alpha) \end{aligned}

So,

D=−1+22sin⁡αcos⁡α+2(sin⁡2α−cos⁡2α)=−1+2sin⁡2α−2cos⁡2α\begin{aligned} D &= -1 + 2\sqrt{2}\sin\alpha\cos\alpha + \sqrt{2}(\sin^2\alpha - \cos^2\alpha) \\ &= -1 + \sqrt{2}\sin 2\alpha - \sqrt{2}\cos 2\alpha \end{aligned}

Hence,

2(sin⁡2α−cos⁡2α)=1\sqrt{2}(\sin 2\alpha - \cos 2\alpha) = 1

Using sin⁡θ−cos⁡θ=2sin⁡(θ−π4)\sin\theta - \cos\theta = \sqrt{2}\sin\left(\theta - \frac{\pi}{4}\right),

2sin⁡(2α−π4)=12\sin\left(2\alpha - \frac{\pi}{4}\right) = 1

Therefore,

sin⁡(2α−π4)=12\sin\left(2\alpha - \frac{\pi}{4}\right) = \frac{1}{2}

So,

2α−π4=π62\alpha - \frac{\pi}{4} = \frac{\pi}{6}

or

2α−π4=5π62\alpha - \frac{\pi}{4} = \frac{5\pi}{6}

This gives

α=5π24\alpha = \frac{5\pi}{24}

or

α=13π24\alpha = \frac{13\pi}{24}

Since α∈(0,π/2)\alpha \in (0, \pi/2), the valid value is α=5π24\alpha = \frac{5\pi}{24}.

Therefore, the correct option is C.

The solution appears unrelated to this question, so the answer is derived from the question itself.

Common mistakes

  • Setting each equation equal to zero and trying to solve directly for x,y,zx, y, z without using the determinant condition. For a homogeneous system, a non-trivial solution exists only when the coefficient determinant is zero. First form the coefficient matrix and impose D=0D = 0.

  • Making an error while expanding the 3×33 \times 3 determinant, especially with signs of cofactors. This changes the trigonometric equation completely. Expand systematically along one column and track the signs +,−,++,-,+ carefully.

  • Using the identity for sin⁡2α\sin 2\alpha or cos⁡2α\cos 2\alpha incorrectly. In particular, 2sin⁡αcos⁡α=sin⁡2α2\sin\alpha\cos\alpha = \sin 2\alpha and sin⁡2α−cos⁡2α=−cos⁡2α\sin^2\alpha - \cos^2\alpha = -\cos 2\alpha. Verify these before simplifying.

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