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For extra information the crit points are (0,0,0) (-1,1,1) (1,-1,1) (1,1,-1) (-1,-1,-1)

Thanks!

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- Thread starter Typhon4ever
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In summary: You know you are going to get 5 different solutions. Take those 'or' cases separately. In solving the first equation you had the z=0 OR x=(-y/z). First supposing z=0 what can you conclude about x and y? Then after solving the second equation consider the cases x=0, z=1 and z=(-1) separately. Substitute anywhere you like.After solving the first equation you can conclude that x=+-y and y=+-x. After solving the second equation you can conclude that x=+-z and z=+-x. After solving the third equation you can conclude that x=0 and z=+-1.

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For extra information the crit points are (0,0,0) (-1,1,1) (1,-1,1) (1,1,-1) (-1,-1,-1)

Thanks!

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Dick said:

What's the "algebraic" way of doing this? Multiple rounds of substitution?

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Typhon4ever said:What's the "algebraic" way of doing this? Multiple rounds of substitution?

Yes, basically. If e.g. x+yz=0 then either y=(-x/z) or z=0, subsitute that into the other equations. Explain how you got your solutions.

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Dick said:Yes, basically. If e.g. x+yz=0 then either y=(-x/z) or z=0, subsitute that into the other equations. Explain how you got your solutions.

I just looked at this and thought of good numbers that would satisfy the equations. That's why I wanted a more systematic way of doing it. The substitution looks like a real headache to do.

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Typhon4ever said:I just looked at this and thought of good numbers that would satisfy the equations. That's why I wanted a more systematic way of doing it. The substitution looks like a real headache to do.

OK, you just solved by inspection. Doing it systematically isn't that much of headache. Just do it. If I substitute y=(-x/z) into the second relation I get x(1-z^2)=0 after a little algebra. What does that tell you?

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Well I get that x=0 and z=+-1

If I plug into the third one I get x=+-z and z=+-x. Where should I start plugging things in next?

If I plug into the third one I get x=+-z and z=+-x. Where should I start plugging things in next?

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Dick said:OK, you just solved by inspection. Doing it systematically isn't that much of headache. Just do it. If I substitute y=(-x/z) into the second relation I get x(1-z^2)=0 after a little algebra. What does that tell you?

Well I get that x=0 and z=+-1

If I plug into the third one I get x=+-z and z=+-x. Where should I start plugging things in next?

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Typhon4ever said:Well I get that x=0 and z=+-1

That's actually an or, not an and that you want

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Office_Shredder said:That's actually an or, not an and that you want

Oh ok. Thank you. So now that I know x=0 or z=+-1 what should my next step be? Should I plug 1 and -1 into the first equation?

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Typhon4ever said:Oh ok. Thank you. So now that I know x=0 or z=+-1 what should my next step be? Should I plug 1 and -1 into the first equation?

You know you are going to get 5 different solutions. Take those 'or' cases separately. In solving the first equation you had the z=0 OR x=(-y/z). First supposing z=0 what can you conclude about x and y? Then after solving the second equation consider the cases x=0, z=1 and z=(-1) separately. Substitute anywhere you like.

The matrix method is a systematic approach to solving a system of linear equations using matrices. It involves representing the coefficients and constants of the equations as elements in a matrix and performing certain operations to find the solution.

The matrix method is most commonly used when there are multiple equations with the same variables and when the number of equations is greater than the number of variables. It is also useful when the equations are in standard form with variables on one side and constants on the other.

The first step is to write the equations in standard form and identify the coefficients and constants. Then, create a matrix with the coefficients and a separate column matrix with the constants. Next, use row operations to simplify the matrix until it is in reduced row-echelon form. The solution can then be read directly from the matrix.

The matrix method can only be used for systems of linear equations. It also requires the number of equations to be equal to or greater than the number of variables. If these conditions are not met, another method of solving the equations should be used.

Yes, the matrix method can be used to solve systems of equations with any number of variables. However, as the number of variables increases, the matrix becomes larger and more complex, making it more difficult to solve by hand. In these cases, it may be more practical to use a calculator or computer program to perform the matrix operations.

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