Let $\theta \in (0, \frac{\pi}{2})$. If the system of linear equations $(1 + \cos^2 \theta)x + \sin^2 \theta y + 4 \sin 3\theta z = 0$, $\cos^2 \theta x + (1 + \sin^2 \theta)y + 4 \sin 3\theta z = 0$, $\cos^2 \theta x + \sin^2 \theta y + (1 + 4 \sin 3\theta)z = 0$ has a non-trivial solution, then the value of $\theta$ is : Options: (A) $\frac{4\pi}{9}$ (B) $\frac{7\pi}{18}$ (C) $\frac{\pi}{18}$ (D) $\frac{5\pi}{18}$

### Analyzing the Setup For a homogeneous system of linear equations to possess **non-trivial solutions**, the determinant of the coefficient matrix must be equal to zero. The given system is: $$(1 + \cos^2 \theta)x + \sin^2 \theta y + 4 \sin 3\theta z = 0$$ $$\cos^2 \theta x + (1 + \sin^2 \theta)y + 4 \sin 3\theta z = 0$$ $$\cos^2 \theta x + \sin^2 \theta y + (1 + 4 \sin 3\theta)z = 0$$ The condition for the existence of non-trivial solutions is: $$\begin{vmatrix} 1 + \cos^2 \theta & \sin^2 \theta & 4 \sin 3\theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \sin 3\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \sin 3\theta \end{vmatrix} = 0$$ ### The Art of Simplification To simplify the determinant, we apply the column operation $C_1 \to C_1 + C_2$. Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, the first column transforms into a column of constants: $$\begin{vmatrix} 2 & \sin^2 \theta & 4 \sin 3\theta \\ 2 & 1 + \sin^2 \theta & 4 \sin 3\theta \\ 1 & \sin^2 \theta & 1 + 4 \sin 3\theta \end{vmatrix} = 0$$ Next, we perform the row operation $R_1 \to R_1 - R_2$ to introduce zeros into the matrix. This yields: $$\begin{vmatrix} 0 & -1 & 0 \\ 2 & 1 + \sin^2 \theta & 4 \sin 3\theta \\ 1 & \sin^2 \theta & 1 + 4 \sin 3\theta \end{vmatrix} = 0$$ ### The Final Calculation Expanding the determinant along the first row, we obtain: $$-(-1) \cdot \begin{vmatrix} 2 & 4 \sin 3\theta \\ 1 & 1 + 4 \sin 3\theta \end{vmatrix} = 0$$ $$1 \cdot [2(1 + 4 \sin 3\theta) - 4 \sin 3\theta] = 0$$ $$2 + 8 \sin 3\theta - 4 \sin 3\theta = 0$$ $$2 + 4 \sin 3\theta = 0 \Rightarrow \sin 3\theta = -\frac{1}{2}$$ ### Solving for $\theta$ Given the domain $\theta \in (0, \pi/2)$, it follows that $3\theta \in (0, 3\pi/2)$. We seek the value of $3\theta$ in the third quadrant where the sine function equals $-1/2$. The reference angle is $\pi/6$, so the third-quadrant solution is: $$3\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$$ Dividing by 3, we find the final value: $$\theta = \frac{7\pi}{18}$$

I got $\sin 3\theta = -1/2$, but I'm confused about the domain. Why is it $3\theta \in (0, 3\pi/2)$?

This is a classic trap! The problem states $\theta \in (0, \pi/2)$. Since our equation involves $3\theta$, we must transform the domain accordingly. Multiplying the entire inequality by 3 gives $3\theta \in (0, 3\pi/2)$. This means we are looking for solutions in the first, second, and third quadrants. Always transform your domain before solving for the angle!

JEE Main 2021 (26 Aug Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $θ \in (0, \frac{π}{2})$ . If the system of linear equations $(1 + cos^{2} θ) x + sin^{2} θ y + 4 sin 3 θ z = 0$ , $cos^{2} θ x + (1 + sin^{2} θ) y + 4 sin 3 θ z = 0$ , $cos^{2} θ x + sin^{2} θ y + (1 + 4 sin 3 θ) z = 0$ has a non-trivial solution, then the value of $θ$ is :

Q: Wait, why do we set the determinant to zero? Can't we just use substitution or elimination?

That is a fantastic question! For a homogeneous system $AX = 0$, if the determinant of $A$ is non-zero, the matrix $A$ is invertible, meaning the only solution is the trivial one: $X = A^{-1}0 = 0$. To get a non-trivial solution, the matrix must be singular, meaning it cannot be inverted. Thus, $|A| = 0$ is the necessary and sufficient condition for the existence of non-trivial solutions.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. You are dealing with a homogeneous system of linear equations, which is a system where all constants on the right-hand side are zero. For such a system to possess a non-trivial solution (a solution other than

x = y = z = 0

), the determinant of the coefficient matrix must be exactly zero. This is your starting point.

Hint 2

Hint 2: The Arsenal. Do not rush to expand the determinant directly; it will lead to algebraic chaos. Instead, look for patterns. Notice that the sum of the first two columns, specifically

sin^{2} θ + cos^{2} θ

, appears frequently. Use the column operation

C_{1} \to C_{1} + C_{2}

to simplify the first column into constants.

Hint 3

Hint 3: The Execution. After your column operation, you will have a column of

[2, 2, 1]^{T}

. Now, create more zeros by performing row operations like

R_{1} \to R_{1} - R_{2}

. Once you have a row with two zeros, expanding along that row will reduce the problem to a simple trigonometric equation:

2 + 4 sin 3 θ = 0

Visualized Solution

Condition for Non-Trivial Solution

For a homogeneous system $A X = 0$ to have a non-trivial solution, the determinant of the coefficient matrix must be zero.
Condition: $∣ A ∣ = 0$

Setting up the Determinant

Extracting the coefficients of $x$ , $y$ , and $z$ from the given equations.
$1 + cos^{2} θ cos^{2} θ cos^{2} θ sin^{2} θ 1 + sin^{2} θ sin^{2} θ 4 sin 3 θ 4 sin 3 θ 1 + 4 sin 3 θ = 0$

Column Operation $C_{1} \to C_{1} + C_{2}$

To simplify, we apply the column operation: $C_{1} \to C_{1} + C_{2}$
Adding the second column to the first column.
Notice that $cos^{2} θ + sin^{2} θ$ will appear in every element of $C_{1}$ .

Simplifying using $sin^{2} θ + cos^{2} θ = 1$

Using the fundamental identity: $sin^{2} θ + cos^{2} θ = 1$
The new first column becomes:
$C_{1} = 1 + 1 1 + 1 1 = 221$

Row Operation $R_{1} \to R_{1} - R_{2}$

Next, we aim to create zeros to make expansion easier.
Apply row operation: $R_{1} \to R_{1} - R_{2}$
New $R_{1}$ : $[2 - 2, sin^{2} θ - (1 + sin^{2} θ), 4 sin 3 θ - 4 sin 3 θ]$
Simplified $R_{1}$ : $[0, - 1, 0]$

Expanding the Determinant

The determinant is now:
$021 - 1 1 + sin^{2} θ sin^{2} θ 0 4 sin 3 θ 1 + 4 sin 3 θ = 0$
Expanding along $R_{1}$ :
$- (- 1) \cdot [2 (1 + 4 sin 3 θ) - 1 (4 sin 3 θ)] = 0$

Solving the Equation

Simplifying the expanded expression:
$1 \cdot [2 + 8 sin 3 θ - 4 sin 3 θ] = 0$
$2 + 4 sin 3 θ = 0$
$4 sin 3 θ = - 2 ⟹ sin 3 θ = - \frac{1}{2}$

Analyzing the Domain of $3 θ$

We are given that $θ \in (0, \frac{π}{2})$ .
Multiplying by 3, we get the domain for $3 θ$ :
$3 θ \in (0, \frac{3 π}{2})$
This covers the 1st, 2nd, and 3rd quadrants.

Visualizing the Valid Region

The valid region for $3 θ$ is $(0, \frac{3 π}{2})$ .
We need to find where $sin 3 θ = - \frac{1}{2}$ in this region.

Finding the Intersection

Draw the line $y = - \frac{1}{2}$ on the unit circle.
Sine is negative in the 3rd and 4th quadrants.
Within our valid region $(0, \frac{3 π}{2})$ , the only solution is in the 3rd quadrant.

Calculating the Angle

The base angle for $sin α = \frac{1}{2}$ is $\frac{π}{6}$ .
In the 3rd quadrant, the angle is $π + \frac{π}{6}$ .
$3 θ = π + \frac{π}{6} = \frac{7 π}{6}$

Final Value of $θ$

We have $3 θ = \frac{7 π}{6}$ .
Dividing by 3:
$θ = \frac{7 π}{18}$
This matches option (2).

00:00 / 00:00

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Solution Diagram

Analyzing the Setup

For a homogeneous system of linear equations to possess non-trivial solutions, the determinant of the coefficient matrix must be equal to zero. The given system is:

(1 + cos^{2} θ) x + sin^{2} θ y + 4 sin 3 θ z = 0

cos^{2} θ x + (1 + sin^{2} θ) y + 4 sin 3 θ z = 0

cos^{2} θ x + sin^{2} θ y + (1 + 4 sin 3 θ) z = 0

The condition for the existence of non-trivial solutions is:

1 + cos^{2} θ cos^{2} θ cos^{2} θ sin^{2} θ 1 + sin^{2} θ sin^{2} θ 4 sin 3 θ 4 sin 3 θ 1 + 4 sin 3 θ = 0

The Art of Simplification

To simplify the determinant, we apply the column operation

C_{1} \to C_{1} + C_{2}

. Using the identity

sin^{2} θ + cos^{2} θ = 1

, the first column transforms into a column of constants:

221 sin^{2} θ 1 + sin^{2} θ sin^{2} θ 4 sin 3 θ 4 sin 3 θ 1 + 4 sin 3 θ = 0

Next, we perform the row operation

R_{1} \to R_{1} - R_{2}

to introduce zeros into the matrix. This yields:

021 - 1 1 + sin^{2} θ sin^{2} θ 0 4 sin 3 θ 1 + 4 sin 3 θ = 0

The Final Calculation

Expanding the determinant along the first row, we obtain:

- (- 1) \cdot 21 4 sin 3 θ 1 + 4 sin 3 θ = 0

1 \cdot [2 (1 + 4 sin 3 θ) - 4 sin 3 θ] = 0

2 + 8 sin 3 θ - 4 sin 3 θ = 0

2 + 4 sin 3 θ = 0 \Rightarrow sin 3 θ = - \frac{1}{2}

Solving for $θ$

Given the domain

θ \in (0, π /2)

, it follows that

3 θ \in (0, 3 π /2)

. We seek the value of

3 θ

in the third quadrant where the sine function equals

- 1/2

The reference angle is

π /6

, so the third-quadrant solution is:

3 θ = π + \frac{π}{6} = \frac{7 π}{6}

Dividing by 3, we find the final value:

θ = \frac{7 π}{18}

Common Traps & FAQs

Wait, why do we set the determinant to zero? Can't we just use substitution or elimination?

That is a fantastic question! For a homogeneous system

A X = 0

, if the determinant of

A

is non-zero, the matrix

A

is invertible, meaning the only solution is the trivial one:

X = A^{- 1} 0 = 0

. To get a non-trivial solution, the matrix must be singular, meaning it cannot be inverted. Thus,

∣ A ∣ = 0

is the necessary and sufficient condition for the existence of non-trivial solutions.

Why did we choose

C_{1} \to C_{1} + C_{2}

? Is there a specific rule for this?

In determinant problems, we always look for ways to create constants or zeros. When you see

sin^{2} θ

and

cos^{2} θ

in the same row, your brain should immediately scream 'Identity!'. By adding the columns, we force the identity

sin^{2} θ + cos^{2} θ = 1

to appear, which simplifies the expression significantly. It is not a rule, but a strategic choice to reduce complexity.

I got

sin 3 θ = - 1/2

, but I'm confused about the domain. Why is it

3 θ \in (0, 3 π /2)

This is a classic trap! The problem states

θ \in (0, π /2)

. Since our equation involves

3 θ

, we must transform the domain accordingly. Multiplying the entire inequality by 3 gives

3 θ \in (0, 3 π /2)

. This means we are looking for solutions in the first, second, and third quadrants. Always transform your domain before solving for the angle!

Select Answer:

Visualized Solution

Condition for Non-Trivial Solution

Setting up the Determinant

Column Operation $C_{1} \to C_{1} + C_{2}$

Simplifying using $sin^{2} θ + cos^{2} θ = 1$

Row Operation $R_{1} \to R_{1} - R_{2}$

Expanding the Determinant

Solving the Equation

Analyzing the Domain of $3 θ$

Visualizing the Valid Region

Finding the Intersection

Calculating the Angle

Final Value of $θ$

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Art of Simplification

The Final Calculation

Solving for $θ$

Similar Questions

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Select Answer:

Visualized Solution

Condition for Non-Trivial Solution

Setting up the Determinant

Column Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​→C1​+C2​

Simplifying using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2θ+cos2θ=1

Row Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​→R1​−R2​

Expanding the Determinant

Solving the Equation

Analyzing the Domain of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3θ

Visualizing the Valid Region

Finding the Intersection

Calculating the Angle

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Art of Simplification

The Final Calculation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

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Column Operation $C_{1} \to C_{1} + C_{2}$

Simplifying using $sin^{2} θ + cos^{2} θ = 1$

Row Operation $R_{1} \to R_{1} - R_{2}$

Analyzing the Domain of $3 θ$

Final Value of $θ$

Solving for $θ$

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