If [x] denotes the greatest integer ≤ x, then the system of linear equations [sinθ]x + [-cosθ]y = 0,

Q: Wait, why do we only care about the determinant? Can't we just solve for $x$ and $y$ directly?

In a homogeneous system of the form $AX = 0$, you cannot 'solve' for $x$ and $y$ in the traditional sense because they are always zero if the determinant is non-zero. The determinant is the gatekeeper; it tells us if there's a non-trivial solution (infinitely many) or just the trivial one. It is the most elegant and efficient path to the solution.

Q: I always mess up the GIF with negative numbers. Is $[-0.5]$ equal to $0$ or $-1$?

This is the classic JEE trap! Remember, $[x]$ is the greatest integer *less than or equal to* $x$. On the number line, $-0.5$ lies between $-1$ and $0$. The greatest integer *less than* $-0.5$ is $-1$. Always visualize the number line to avoid this common pitfall.

Q: In the second case, I didn't know the exact value of $[\cot\theta]$. Does that make the problem unsolvable?

That is a brilliant observation! In the second case, the determinant expression becomes $D = -1 - (0) \cdot [\cot\theta]$. Since the coefficient of $[\cot\theta]$ is zero, the actual value of $[\cot\theta]$ becomes irrelevant. It is a beautiful mathematical 'cancellation' that simplifies the problem. Trust the algebra!

Question

If $[x]$ denotes the greatest integer $\leq x$, then the system of linear equations $[\sin	heta]x + [-\cos	heta]y = 0$, $[\cot	heta]x + y = 0$

Options:

(A) have infinitely many solutions if $	heta \in (\pi/2, 2\pi/3) \cup (\pi, 7\pi/6)$

(B) have infinitely many solutions if $	heta \in (\pi/2, 2\pi/3)$ and has a unique solution if $	heta \in (\pi, 7\pi/6)$

(C) has a unique solution if $	heta \in (\pi/2, 2\pi/3)$ and have infinitely many solutions if $	heta \in (\pi, 7\pi/6)$

(D) has a unique solution if $	heta \in (\pi/2, 2\pi/3) \cup (\pi, 7\pi/6)$

Sigma Percentile · Accepted Answer

### The Symphony of Algebra and Trigonometry

Welcome, future engineers! Today, we are not just solving a system of equations; we are embarking on a journey through the unit circle, the rigid steps of the Greatest Integer Function, and the elegant logic of linear algebra.

This problem is a classic JEE Advanced challenge because it forces you to synthesize three distinct mathematical domains into one cohesive solution. Let us break it down, step by step.

### Phase 1: The Gatekeeper of Solutions

We are presented with a homogeneous system of linear equations:

$$[\sin	heta]x + [-\cos	heta]y = 0$$
$$[\cot	heta]x + y = 0$$

When you see a system where the right-hand side is zero, your mind should immediately jump to the concept of the determinant. For a homogeneous system $AX = 0$, the determinant of the coefficient matrix $D$ is the ultimate arbiter of fate.

If $D 
eq 0$, the system is 'well-behaved' and possesses only the trivial solution, $x=0$ and $y=0$. However, if $D = 0$, the system collapses into a state of dependency, yielding infinitely many solutions. Our goal is to find when this collapse happens.

Let us construct our determinant $D$:

$$D = \begin{vmatrix} [\sin	heta] & [-\cos	heta] \ [\cot	heta] & 1 \end{vmatrix}$$

Expanding this $2 	imes 2$ matrix is straightforward. We multiply the main diagonal and subtract the product of the off-diagonal elements:

$$D = ([\sin	heta] \cdot 1) - ([-\cos	heta] \cdot [\cot	heta])$$
$$D = [\sin	heta] - [-\cos	heta][\cot	heta]$$

This expression, $D = [\sin	heta] - [-\cos	heta][\cot	heta]$, is our master equation. Everything now depends on the values of these trigonometric functions within the given intervals.

### Phase 2: Navigating the Second Quadrant

Let us examine the first interval: $	heta \in (\frac{\pi}{2}, \frac{2\pi}{3})$.

Visualize the unit circle. This interval sits comfortably in the second quadrant. In this region, $\sin	heta$ is positive, decreasing from $1$ to $\frac{\sqrt{3}}{2}$ (approximately $0.866$).

Since $\sin	heta$ is strictly between $0.866$ and $1$, its greatest integer value is clearly $[\sin	heta] = 0$.

Now, consider $-\cos	heta$. In the second quadrant, $\cos	heta$ is negative, ranging from $0$ to $-\frac{1}{2}$. Therefore, $-\cos	heta$ is positive, ranging from $0$ to $0.5$.

The greatest integer of any value between $0$ and $0.5$ is $0$. Thus, $[-\cos	heta] = 0$.

Finally, look at $\cot	heta$. In the second quadrant, $\cot	heta$ is negative, ranging from $-\frac{1}{\sqrt{3}}$ (approx $-0.577$) to $0$. The greatest integer of a number slightly less than zero is $-1$. So, $[\cot	heta] = -1$.

Substituting these into our determinant $D$:

$$D = 0 - (0)(-1) = 0$$

Because $D = 0$, the system has infinitely many solutions in this interval. The logic holds firm!

### Phase 3: The Third Quadrant Challenge

Now, let us turn our attention to the second interval: $	heta \in (\pi, \frac{7\pi}{6})$.

This interval lies in the third quadrant. Here, $\sin	heta$ is negative, ranging from $0$ to $-\frac{1}{2}$. The greatest integer of a value between $-0.5$ and $0$ is $-1$. So, $[\sin	heta] = -1$.

Next, consider $-\cos	heta$. In the third quadrant, $\cos	heta$ ranges from $-1$ to $-\frac{\sqrt{3}}{2}$. Thus, $-\cos	heta$ ranges from $\frac{\sqrt{3}}{2}$ (approx $0.866$) to $1$.

The greatest integer of a value between $0.866$ and $1$ is $0$. So, $[-\cos	heta] = 0$.

Finally, $\cot	heta$ in the third quadrant is positive and greater than $1$. Specifically, it ranges from $\sqrt{3}$ (approx $1.732$) to $\infty$. Therefore, $[\cot	heta] \geq 1$.

Let us substitute these into our determinant $D$ again:

$$D = [\sin	heta] - [-\cos	heta][\cot	heta]$$
$$D = -1 - (0) \cdot [\cot	heta]$$
$$D = -1$$

Since $D = -1$, which is clearly not zero, the system has a unique solution in this interval. The math has spoken!

### Conclusion: The Elegance of Logic

We have successfully navigated the treacherous waters of the Greatest Integer Function and trigonometric ranges.

We found that in the interval $(\frac{\pi}{2}, \frac{2\pi}{3})$, the determinant vanishes, leading to **infinitely many solutions**. In the interval $(\pi, \frac{7\pi}{6})$, the determinant stands firm at $-1$, leading to a **unique solution**.

This confirms that the system behaves differently depending on the quadrant, a beautiful reminder that in mathematics, context is everything. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process!

Animated Solution for Mathematics - Matrices and Determinants: If $[x]$ denotes the greatest integer $\leq x$ , then the system of linear equations $[sin θ] x + [- cos θ] y = 0$ , $[cot θ] x + y = 0$

Select Answer:

Visualized Solution

System of Equations

Condition for Solutions

The Determinant $D$

Expanding $D$

Case 1: $θ \in (\frac{π}{2}, \frac{2 π}{3})$

Trig Ranges in Case 1

GIF Values in Case 1

Determinant in Case 1

Case 2: $θ \in (π, \frac{7 π}{6})$

Trig Ranges in Case 2

GIF Values in Case 2

Determinant in Case 2

Final Conclusion

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

The Symphony of Algebra and Trigonometry

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

Let for some real numbers $α$ and $β$ , $a = α - i β$ . If the system of equations $4 i x + (1 + i) y = 0$ and $8 (cos \frac{2 π}{3} + i sin \frac{2 π}{3}) x + \overset{a}{ˉ} y = 0$ has more than one solution then $\frac{α}{β}$ is equal to :

If $α + β + γ = 2 π$ , then the system of equations $x + (cos γ) y + (cos β) z = 0$ , $(cos γ) x + y + (cos α) z = 0$ , $(cos β) x + (cos α) y + z = 0$ has :

Let $λ$ and $α$ be real. Find the set of all values of $λ$ for which the system of linear equations $λ x + (sin α) y + (cos α) z = 0, x + (cos α) y + (sin α) z = 0, - x + (sin α) y - (cos α) z = 0$ has a non-trivial solution. For $λ = 1$ , find all values of $α$ .

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 :

For a real number $a$ , if the system $1 a a^{2} a 1 a a^{2} a 1 x y z = 1 - 1 1$ of linear equations, has infinitely many solutions, then $1 + a + a^{2} =$

Consider the system of linear equations in $x, y, z$ : $(sin 3 θ) x - y + z = 0, (cos 2 θ) x + 4 y + 3 z = 0, 2 x + 7 y + 7 z = 0$ . Find the values of $θ$ for which this system has nontrivial solutions.

Let the system of equations $x + 5 y - z = 1$ , $4 x + 3 y - 3 z = 7$ , $24 x + y + λ z = μ$ , $λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, If $x, y, z$ are integers and satisfy $7 \leq x + y + z \leq 77$ , is

Let $S_{1}$ and $S_{2}$ be respectively the sets of all $a \in R - {0}$ for which the system of linear equations $a x + 2 a y - 3 a z = 1$ , $(2 a + 1) x + (2 a + 3) y + (a + 1) z = 2$ , $(3 a + 5) x + (a + 5) y + (a + 2) z = 3$ has unique solution and infinitely many solutions. Then

The number of $θ \in (0, 4 π)$ for which the system of linear equations $3 (sin 3 θ) x - y + z = 2$ , $3 (cos 2 θ) x + 4 y + 3 z = 3$ , $6 x + 7 y + 7 z = 9$ has no solution is :

The number of values of $k$ , for which the system of equations : $(k + 1) x + 8 y = 4 k; k x + (k + 3) y = 3 k - 1$ has no solution, is