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Question: Let \(\left( x,y,z \right)\) be points with integer coordinates satisfying the system of homogeneous...

Let (x,y,z)\left( x,y,z \right) be points with integer coordinates satisfying the system of homogeneous equations:
3xyz=03x-y-z=0, 3x+z=0-3x+z=0, 3x+2y+z=0-3x+2y+z=0. Then the number of such points which lie inside a sphere of radius 1010 centered at the origin is

Explanation

Solution

For this problem we need to calculate the number of points which satisfies the given coordinate system and the given condition. For this we will first consider the given coordinate system and solve those equations. By observing the equation, we can simply add first and second equations to get the value of yy. Now we will substitute this value in the third equation to get the value of zz in terms of xx. By assuming the value of xx to a constant say kk we can write the standard point which satisfies the given coordinate system. Now we will write the equation of the circle having radius 1010 and center at origin and substitute the calculated point in the equation and check for the number points that satisfies the condition.

Complete step by step solution:
Given coordinate system is 3xyz=03x-y-z=0, 3x+z=0-3x+z=0, 3x+2y+z=0-3x+2y+z=0.
Adding the first and second equation we can have
3xyz+(3x+z)=0+03x-y-z+\left( -3x+z \right)=0+0
Removing the parentheses by using mathematical rule that when we multiply a positive sign with a negative sign, we will get negative sign, then we will have
3xyz3x+z=0 y=0 \begin{aligned} & 3x-y-z-3x+z=0 \\\ & \Rightarrow y=0 \\\ \end{aligned}
Substituting the above calculated yy in the third equation, then we will get
3x+2(0)+z=0 3x+z=0 \begin{aligned} & -3x+2\left( 0 \right)+z=0 \\\ & \Rightarrow -3x+z=0 \\\ \end{aligned}
From the above equation we can write the value of zz as
z=3xz=3x
Let us assume the value of xx as constant say kk, then the value of zz will be
z=3kz=3k
Hence the point which satisfies the given coordinate system is (x,y,z)=(k,0,3k)\left( x,y,z \right)=\left( k,0,3k \right).
Now the equation of the sphere having radius 1010 and center at origin is given by
x2+y2+z2=102{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{10}^{2}}
If a point lies in the sphere then the point should satisfy the following equation.
x2+y2+z2<100{{x}^{2}}+{{y}^{2}}+{{z}^{2}}<100
Substituting the point (k,0,3k)\left( k,0,3k \right) in the above expression, then we will get
k2+02+(3k)2<100 k2+9k2<100 10k2<100 \begin{aligned} & {{k}^{2}}+{{0}^{2}}+{{\left( 3k \right)}^{2}}<100 \\\ & \Rightarrow {{k}^{2}}+9{{k}^{2}}<100 \\\ & \Rightarrow 10{{k}^{2}}<100 \\\ \end{aligned}
Divide the above expression with 1010 on both sides, then we will get
k2<10{{k}^{2}}<10
We can write the values of kk which satisfy the above expression are k=0,±1,±2,±3k=0,\pm 1,\pm 2,\pm 3.
Hence the number of points which satisfies all the given conditions is 77.

Note: In this problem we have the simple equations in the coordinate system, so we have simplified the equations in a simple manner. If there are different kinds of equations which cannot be solved in a simple manner, then we need to use any matrix method like crammers’ rule or gauss Jordan method or matrix inverse method to find the solution.