Solving a system of linear equations

[Specter]

Homework Statement

Determine whether the following system of equations has a single point of intersection. If so, find the point of intersection.

4x+y-9z=0
x+2y+3z=0
2x-3y-5=0

Homework Equations

n₁⋅(n₂×n₃)

The Attempt at a Solution

I have to pick a variable, use a pair of equations to eliminate the variable. Then I have to eliminate the same variable but with a different pair of equations. I tried doing this but I am not sure how correct it is.

n₁=(1,1,-9)
n₂=(1,2,3)
n₃=(2,-3,0)

n₁⋅(n₂×n₃)

=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.4x+y-9z=0 [1]
x+2y+3z=0 [2]
2x-3y-5=0 [3]

4x+y-9z=0 [1]
4x+8y+12z=0 [4] Eqn [2] x 4 to eliminate x.
Subtract and the new equation is 7y+3z=0 [5].

Use a different pair of equations to eliminate x. I used equations 1 and 3.

4x+y-9z=0 [1]
4x-6z-10=0 [6] (Eqn 3x2)
Subtract and the new equation is -5y-9z-10=0 [7]

The new system with 2 eqns and 2 variables:

7y+3z=0
-5y-9z-10=0

If this is correct, I sort of know where to go from here. I would solve for y and z.

 

[SammyS]

Homework Statement

Determine whether the following system of equations has a single point of intersection. If so, find the point of intersection.

4x+y-9z=0
x+2y+3z=0
2x-3y-5=0

Homework Equations

n₁⋅(n₂×n₃)

The Attempt at a Solution

I have to pick a variable, use a pair of equations to eliminate the variable. Then I have to eliminate the same variable but with a different pair of equations. I tried doing this but I am not sure how correct it is.

n₁=(1,1,-9)
n₂=(1,2,3)
n₃=(2,-3,0)

n₁⋅(n₂×n₃)

=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.4x+y-9z=0 [1]
x+2y+3z=0 [2]
2x-3y-5=0 [3]

4x+y-9z=0 [1]
4x+8y+12z=0 [4] Eqn [2] x 4 to eliminate x.
Subtract and the new equation is 7y+3z=0 [5].

Use a different pair of equations to eliminate x. I used equations 1 and 3.

4x+y-9z=0 [1]
4x-6z-10=0 [6] (Eqn 3x2)
Subtract and the new equation is -5y-9z-10=0 [7]

The new system with 2 eqns and 2 variables:

7y+3z=0
-5y-9z-10=0

If this is correct, I sort of know where to go from here. I would solve for y and z.

Your equation [5] is incorrect.

12z − ( −9z) ≠ 3z

 

[Ray Vickson]

Homework Statement

Determine whether the following system of equations has a single point of intersection. If so, find the point of intersection.

4x+y-9z=0
x+2y+3z=0
2x-3y-5=0

Homework Equations

n₁⋅(n₂×n₃)

The Attempt at a Solution

I have to pick a variable, use a pair of equations to eliminate the variable. Then I have to eliminate the same variable but with a different pair of equations. I tried doing this but I am not sure how correct it is.

n₁=(1,1,-9)
n₂=(1,2,3)
n₃=(2,-3,0)

n₁⋅(n₂×n₃)

=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.4x+y-9z=0 [1]
x+2y+3z=0 [2]
2x-3y-5=0 [3]

4x+y-9z=0 [1]
4x+8y+12z=0 [4] Eqn [2] x 4 to eliminate x.
Subtract and the new equation is 7y+3z=0 [5].

Use a different pair of equations to eliminate x. I used equations 1 and 3.

4x+y-9z=0 [1]
4x-6z-10=0 [6] (Eqn 3x2)
Subtract and the new equation is -5y-9z-10=0 [7]

The new system with 2 eqns and 2 variables:

7y+3z=0
-5y-9z-10=0

If this is correct, I sort of know where to go from here. I would solve for y and z.

Use Gaussian elimination; that is the standard method we almost always apply in such problems. It does not matter if you have 2 variables, 3 variables or 100 variables---the method does not change; only the amount of work is different in such cases.

Anyway, if your third equation really is $2x-3y=5-0$ (exactly as written), you get one solution, but if it was supposed to be $2x-3y-5z=0$ you get a different solution. Both problems can be tackled successfully via Gaussian elimination.

 

[YoungPhysicist]

You can get the relation $x = \frac{5+3y}{2}$,
from the last equation ,then you can substitute x away in the two previous equations and get two equations each with two variables.

Then you can do the process again,and get the final equation with only one variable.then use the relations to find the other variables.

 

[Ray Vickson]

Another way to solve this is by substitution:
You can get the relation $x = \frac{5+3y}{2}$,
from the last equation ,then you can substitute x away in the two previous equations and get two equations each with two variables.

Then you can do the process again,and get the final equation with only one variable.then use the relations to find the other variables.

Yes, indeed, and that is what is meant by "Gaussian elimination".

 

[YoungPhysicist]

Yes, indeed, and that is what is meant by "Gaussian elimination".

Oops! My previous post edited.

 

[Specter]

You can get the relation $x = \frac{5+3y}{2}$,
from the last equation ,then you can substitute x away in the two previous equations and get two equations each with two variables.

Then you can do the process again,and get the final equation with only one variable.then use the relations to find the other variables.

Thanks! I'll try it.

 

[Specter]

Use Gaussian elimination; that is the standard method we almost always apply in such problems. It does not matter if you have 2 variables, 3 variables or 100 variables---the method does not change; only the amount of work is different in such cases.

Anyway, if your third equation really is $2x-3y=5-0$ (exactly as written), you get one solution, but if it was supposed to be $2x-3y-5z=0$ you get a different solution. Both problems can be tackled successfully via Gaussian elimination.

Thank you. And yes the third equation is 2x-3y=5-0. I'll try to do it the way you have told me but I haven't learned it like that so I might run into some trouble.

 

[Specter]

You can get the relation $x = \frac{5+3y}{2}$,
from the last equation ,then you can substitute x away in the two previous equations and get two equations each with two variables.

Then you can do the process again,and get the final equation with only one variable.then use the relations to find the other variables.

Whats the reason in choosing eqn [3] and not [1] or [2] to solve for x? How do I know which of the equations to choose when solving the question using this method?

 

[YoungPhysicist]

Whats the reason in choosing eqn [3] and not [1] or [2] to solve for x? How do I know which of the equations to choose when solving the question using this method?

Eqn 3 is the only equation in the system that has exact two variables,so it’s the only equation that can give the relation between two and only two variables.

 

[Specter]

Eqn 3 is the only equation in the system that has exact two variables,so it’s the only equation that can give the relation between two and only two variables.

Thanks! So I solved for x the way I was told to, getting x=5+3y/2

Next I substitute the x value into equation [2] so solve for z, giving me z=- 5+7y/6

Then I substitute the x and z values into equation [1] to solve for y, giving me y=-1.

I'm not sure what to do next, I see people getting different answers than me. Is there more to the question?

Edit: Never mind I just have to substitute the values into x and z to get my coordinates.

 

[Mark44]

Thanks! So I solved for x the way I was told to, getting x=5+3y/2

This is incorrect.
From your third equation, 2x - 3y = 5, solving for x gives 2x = 3y + 5, so $x = \frac{3y + 5} 2$. Maybe that's what you meant, but what you wrote above says $x = 5 + \frac{3y} 2$, and this is different from what I wrote.

Next I substitute the x value into equation [2] so solve for z, giving me z=- 5+7y/6

Then I substitute the x and z values into equation [1] to solve for y, giving me y=-1.

I'm not sure what to do next, I see people getting different answers than me. Is there more to the question?

I get a different value for y, not -1.
Edit: Your value for y is correct. I missed a minus sign when I wrote down the original system of equations.
Once you get a solution, you can (and should) substitute the three values in your original system of equations, from which you should get three true statements.

 

[Specter]

This is incorrect.
From your third equation, 2x - 3y = 5, solving for x gives 2x = 3y + 5, so $x = \frac{3y + 5} 2$. Maybe that's what you meant, but what you wrote above says $x = 5 + \frac{3y} 2$, and this is different from what I wrote.I get a different value for y, not -1.

Once you get a solution, you can (and should) substitute the three values in your original system of equations, from which you should get three true statements.

I keep getting y=-1, using $x = \frac{5 + 3y} 2$ and $z= - \frac{5+7y} 6$

I can show my work if that would help! I haven't used LaTeX before... is there an extension for chrome or is it used by typing out the code?

Thanks

 

[Mark44]

I keep getting y=-1, using $x = \frac{5 + 3y} 2$ and $z= - \frac{5+7y} 6$

Your value for x is correct. I missed a sign when I wrote down your original system.

What values do you get for x and z?

I can show my work if that would help! I haven't used LaTeX before... is there an extension for chrome or is it used by typing out the code?

There's no extension -- you write down the LaTeX script. See our tutorial here -- https://www.physicsforums.com/help/latexhelp/

 

[Specter]

Your value for x is correct. I missed a sign when I wrote down your original system.

What values do you get for x and z?
There's no extension -- you write down the LaTeX script. See our tutorial here -- https://www.physicsforums.com/help/latexhelp/

Thanks. I'll take a look.

I got x=1 and $z= \frac{1} 3$

So the point of intersection would be $(1,-1,\frac{1} 3$)

 

[Mark44]

Thanks. I'll take a look.

I got x=1 and $z= \frac{1} 3$

So the point of intersection would be $(1,-1,\frac{1} 3$)

Yes, that's the solution, which you can check.
Is 4(1) + 1(-1) -9(1/3) = 0?
Is 1 + 2(-1) + 3(1/3) = 0?
Is 2(1) - 3(-1) = 5?
If all three are true, you've found the solution.

 

[Mark44]

Determine whether the following system of equations has a single point of intersection.

Did you do this part? Possibly it entails finding the determinant of the matrix of coefficients, but I don't know if you have covered that yet.

 

[Specter]

Did you do this part? Possibly it entails finding the determinant of the matrix of coefficients, but I don't know if you have covered that yet.

Oh I'm pretty sure what my teacher means by determine if there is a single point of intersection was this:

n1=(1,1,-9)
n2=(1,2,3)
n3=(2,-3,0)

n1⋅(n2×n3)

=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.

Thats all I can think of because I haven't learned what you just mentioned.

 

[SammyS]

Oh I'm pretty sure what my teacher means by determine if there is a single point of intersection was this:

n1=(1,1,−9)
n2=(1,2,3)
n3=(2,−3,0)

n1⋅(n2×n3)

=(4,1,−9)⋅[(1,2,3)×(2,−3,0)]
=(4,1,−9)⋅(9,6,−7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.

That's all I can think of because I haven't learned what you just mentioned.

The product you calculated, known as the scalar triple product, is equivalent to the determinant of the coefficient matrix mentioned by @Mark44 .

So, yes, you did determine that there is a single point of intersection.

By the way: The method you were initially using in your Original Post, was correct, but your algebra contained errors.

I can follow up on that if you're interested.

 

[Specter]

The product you calculated, known as the scalar triple product, is equivalent to the determinant of the coefficient matrix mentioned by @Mark44 .

So, yes, you did determine that there is a single point of intersection.

By the way: The method you were initially using in your Original Post, was correct, but your algebra contained errors.

I can follow up on that if you're interested.

Sure that would be great! Thanks!

 

[SammyS]

The Attempt at a Solution

I have to pick a variable, use a pair of equations to eliminate the variable. Then I have to eliminate the same variable but with a different pair of equations. I tried doing this but I am not sure how correct it is.

=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.

The above is fine.

In the following, you choose to eliminate $\ x \$ in two equations, which is a fairly common approach. However, as you point out above, you can pick any variable and in the case of these three equations, equation [3] already has $\ z \$ eliminated. I'll say more on that later^†.

4x+y−9z=0 [1]
x+2y+3z=0 [2]
2x−3y = 5 [3] . . . (Note: I added 5 to get the constant on the right side. Less confusing that way. : Sammys ) .

4x+y−9z=0 [1]
4x+8y+12z=0 [4] : ( 4 × Eqn [2] to eliminate x. )
Subtract and the new equation is 7y+3z=0 [5].

Apparently, you intended to eliminate $\ x\$ by subtracting Eqn [1] from Eqn [4]. (As I stated in Post #2, the result you give as Eqn [5] is incorrect.)
Specifically, 12z − (−9z) → 21z, not 3z .

I find that I make fewer arithmetic errors, if I do the elimination by adding equations rather than subtracting them. So for your system of equations, get Eqn [4] by multiplying Eqn [2] by −4 .
Then add equations [1] and [4].

That gives: $\ 0x - 7y - 21z = 0 \$ for Eqn [5].

( Maybe divide this by −7 & use this instead. $\ \ \ y + 3z = 0 \$ , Eqn [5*]

Furthermore, this could also be written: $\ 3z = -y \,.$ Going even further, plug this back into Eqn[2] to get

$x + 2y + (-y) =0$
$x + y =0 \$ ( Eqn [5**] )​

Again, more on this later^† .​

)​

Use a different pair of equations to eliminate x. I used equations [1] and [3].

4x+y−9z=0 [1]
4x−6z−10=0 [6] : ( 2 × Eqn [3] )
Subtract and the new equation is −5y−9z−10=0 [7]

Again, an arithmetic (or algebra) mistake due to incorrect subtracting.
That should have given: −7y + 9z − 10 = 0 for Eqn [7]

The new system with 2 eqns and 2 variables:

7y+3z=0
-5y-9z-10=0

If this is correct, I sort of know where to go from here. I would solve for y and z.

In writing the "system", it's customary to write a complete set of three equations. So, any of the original equations, Eqn [2] is recommended b/c the coefficient of x is 1 .

With the corrected equations, the system can be written as follows.
$\left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\ \quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\ \ \ \ -7y + 9z &= 10 & \text{ [7*]} \nonumber \end{align} \right.$

Use elimination with Eqn's [5*] and [7*] to obtain an equation in which $y$ is eliminated. Easily done by taking 7 times Eqn [5*] and adding Eqn [7*] to that.
You get $\ 30z = 10 \$ which simplifies to $\ 3z = 1 \,.$

The "system becomes:
$\left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\ \quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\ \ \ \ 3z &= 1 & \text{ [8]} \nonumber \end{align} \right.$

The above system may be said to be in upper triangular form and the system can be solved by "back substitution".

^† This post is way long enough. I'll do the "later" stuff in another post. Basically, it involves using the observation that eliminating $z$ from an additional equation ( using elimination on Eqn's [1] &[2] ) gets you a result with much less computation.

 

[Specter]

The above is fine.

In the following, you choose to eliminate $\ x \$ in two equations, which is a fairly common approach. However, as you point out above, you can pick any variable and in the case of these three equations, equation [3] already has $\ z \$ eliminated. I'll say more on that later^†. Apparently, you intended to eliminate $\ x\$ by subtracting Eqn [1] from Eqn [4]. (As I stated in Post #2, the result you give as Eqn [5] is incorrect.)
Specifically, 12z − (−9z) → 21z, not 3z .

I find that I make fewer arithmetic errors, if I do the elimination by adding equations rather than subtracting them. So for your system of equations, get Eqn [4] by multiplying Eqn [2] by −4 .
Then add equations [1] and [4].

That gives: $\ 0x - 7y - 21z = 0 \$ for Eqn [5].

( Maybe divide this by −7 & use this instead. $\ \ \ y + 3z = 0 \$ , Eqn [5*]

Furthermore, this could also be written: $\ 3z = -y \,.$ Going even further, plug this back into Eqn[2] to get

$x + 2y + (-y) =0$
$x + y =0 \$ ( Eqn [5**] )​

Again, more on this later^† .​

)​

Again, an arithmetic (or algebra) mistake due to incorrect subtracting.
That should have given: −7y + 9z − 10 = 0 for Eqn [7]In writing the "system", it's customary to write a complete set of three equations. So, any of the original equations, Eqn [2] is recommended b/c the coefficient of x is 1 .

With the corrected equations, the system can be written as follows.
$\left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\ \quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\ \ \ \ -7y + 9z &= 10 & \text{ [7*]} \nonumber \end{align} \right.$

Use elimination with Eqn's [5*] and [7*] to obtain an equation in which $y$ is eliminated. Easily done by taking 7 times Eqn [5*] and adding Eqn [7*] to that.
You get $\ 30z = 10 \$ which simplifies to $\ 3z = 1 \,.$

The "system becomes:
$\left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\ \quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\ \ \ \ 3z &= 1 & \text{ [8]} \nonumber \end{align} \right.$

The above system may be said to be in upper triangular form and the system can be solved by "back substitution".

^† This post is way long enough. I'll do the "later" stuff in another post. Basically, it involves using the observation that eliminating $z$ from an additional equation ( using elimination on Eqn's [1] &[2] ) gets you a result with much less computation.

Wow! ! I need to make it a habit to check my arithmetic over and over. Sometimes I feel overwhelmed and can't find the mistakes if that makes any sense. I appreciate you taking the time to type all of this out and explain it to me. Thank you, it helps a bunch!

 

[Mark44]

I need to make it a habit to check my arithmetic over and over.

Well, yes, but if you get to a solution, you should always check that it is a solution to the original problem. We all make mistakes, especially in these kinds of problems where there are many arithmetic operations. If your solution checks, then that's a good sign that all of your work was correct.