Let (x, y, z) be points with integer coordinates satisfying the system of homogeneous equations : 3x

Q: Wait, why do we reject the negative root here? Or do we?

Great question! We don't reject the negative roots. The inequality $x^2 \le 10$ allows $x$ to be any integer whose square is less than or equal to 10. This includes both positive and negative integers: $\{-3, -2, -1, 0, 1, 2, 3\}$. Always remember to check both sides of the number line!

Q: Why can't we just use Cramer's Rule to solve this?

You absolutely could! Cramer's Rule is a valid tool for any system of linear equations. However, for this specific system, the determinant of the coefficient matrix is zero, which means the system has infinitely many solutions (a line of solutions). Cramer's Rule is best for unique solutions. Here, algebraic substitution is much faster and less prone to calculation errors.

Q: How do I know if the third equation is redundant?

In a homogeneous system, if you find a solution set from two equations, you must always verify it against the third. If it satisfies the third equation (like $0=0$), it confirms that the third equation is linearly dependent on the others, meaning it doesn't add new constraints but confirms the line of intersection.

Question

Let $(x, y, z)$ be points with integer coordinates satisfying the system of homogeneous equations : $3x - y - z = 0; -3x + z = 0; -3x + 2y + z = 0$. Then the number of such points for which $x^2 + y^2 + z^2 \le 100$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In the context of JEE Advanced, a homogeneous system of linear equations is defined by the property that the constant term in each equation is zero. Given the system:

$$3x - y - z = 0$$
$$-3x + z = 0$$
$$-3x + 2y + z = 0$$

Because the constant terms are zero, the origin $(0, 0, 0)$ is guaranteed to be a solution. We must determine if these planes intersect at a single point or along a common line.

### The Algebraic Reduction

We begin by simplifying the system. The second equation, $-3x + z = 0$, provides an immediate relationship:

$$z = 3x$$

Substituting this into the first equation, $3x - y - z = 0$, we obtain:

$$3x - y - (3x) = 0 \Rightarrow -y = 0 \Rightarrow y = 0$$

We have determined that for any point on the intersection of these planes, the $y$-coordinate must be zero. Our solution space is confined to the $x-z$ plane.

To verify consistency, we substitute $y = 0$ and $z = 3x$ into the third equation:

$$-3x + 2(0) + (3x) = 0 \Rightarrow 0 = 0$$

This confirms that the equations are linearly dependent. The solution set is a line defined by the parametric coordinates $(x, 0, 3x)$.

### The Sphere Constraint

We now introduce the boundary condition $x^2 + y^2 + z^2 \le 100$, which represents a solid sphere of radius $10$ centered at the origin. We seek integer points on our line that satisfy this inequality.

Substituting the parametric point $(x, 0, 3x)$ into the inequality:

$$x^2 + 0^2 + (3x)^2 \le 100$$
$$x^2 + 9x^2 \le 100$$
$$10x^2 \le 100$$

Dividing by $10$, we arrive at the simplified constraint:

$$x^2 \le 10$$

### The Final Count

We must identify all integer values of $x$ that satisfy $x^2 \le 10$. Testing the integers:

For $x = 0$, $x^2 = 0 \le 10$.
For $x = \pm 1$, $x^2 = 1 \le 10$.
For $x = \pm 2$, $x^2 = 4 \le 10$.
For $x = \pm 3$, $x^2 = 9 \le 10$.

If we test $x = \pm 4$, we obtain $x^2 = 16$, which exceeds the limit of $10$. Thus, the set of possible integer values for $x$ is $\{-3, -2, -1, 0, 1, 2, 3\}$.

Counting these values, we find there are exactly **7** such points.

Enter Numerical Value:

Visualized Solution

Orienting the System

Isolating $z$ from Equation 2

Substituting $z$ into Equation 1

Verifying Consistency

Applying the Sphere Constraint

Solving the Inequality

Counting Integer Solutions

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

Analyzing the Setup

The Algebraic Reduction

The Sphere Constraint

The Final Count

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