I've realised, any problems I encounter that set-up simultaneous equations, for whatever reason, ALWAYS take the form of a square matrix.

2 by 2, 3 by 3, 4 by 4, yeah for sure know problem. I can spot simultaneous equations that have appeared from thin-air faster than a protective relay can operate.

I've noticed, it would save a lot of time in small problems that appear all over the place when three variables take the form of two rows and three columns.

In all honesty, I can hold my hands up and say; I can't solve a pair of 2 by 3 simultaneous equations.

The closest example I can find is from a Linear Algebra book.

Problem:

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

By elimination,

y - 3z = 0

We give z any value ≠ 0, say z=1, and solve for y, namely y=3. We then solve for x from the second equation, namely x= -y-z, and obtain x=4.


The problem for me here is, the book only gives examples when the equations equate to zero and not a negative or positive constant.


If someone could develop this method further when they equate to a negative or positive constant, it would be much appreciated.

Also, If someone knows a method for a 2 by n problem, when n is greater than 3, that would be great . I think a useful application for this that would save time is partial fractions, thanks.

 

Your example of two equations with three unknowns can be written sort of like this in matrix form:

2 3 -1 0
1 1  1 0

(There are your zeros on the right.)

You can manipulate the rows of numbers directly. For example you can double the second row:

2 3 -1 0
2 2  2 0

and then subtract the second row from the first row

0 1 -3 0
2 2  2 0

and there you see y-3z=0 on the first row. (And it's OK if z is zero by the way.)

Now, if the last column isn't zeros, it doesn't matter. The operations work the same.

 

Thanks, but your equations all equate to zero, i'm not interested in zero .

I gave those two equations from the book as an example for evidence of research and is not the problem i'm stuck with.

 

Perhaps if you were to explain in more detail what you are trying to achieve?

It is not possible to solve a system of 3 unknowns and 2 equations.
The best you can do is define a region of solution. The technique for this is known as linear programming.

 

Perhaps if you were to explain in more detail what you are trying to achieve?

It is not possible to solve a system of 3 unknowns and 2 equations.
The best you can do is define a region of solution. The technique for this is known as linear programming.

Thats exactly what i'm trying to achieve, solve a system of 3 unknowns and two equations.

Ok thanks.

 

It is an iron-clad mathematical RULE that THREE Simulataneous equations are required to solve for 3 unknown variables. The number of unknown varibles must not exceed the number of simultaneous equations, for there to be a complete solution for all variables.

I will eat a bug and stand on my head till my ears trun blue if anyone can prove this rule wrong.

 

It is an iron-clad mathematical RULE that THREE Simulataneous equations are required to solve for 3 unknown variables. The number of unknown varibles must not exceed the number of simultaneous equations, for there to be a complete solution for all variables.

I will eat a bug and stand on my head till my ears trun blue if anyone can prove this rule wrong.

Ok thanks.

So, is that example I gave from the book absolute garbage?

 

I'm frying up that bug right now, even as I check my facts.

 

Thanks, but your equations all equate to zero, i'm not interested in zero .

Check the last paragraph. I pointed out that with the matrix method you can change the zeros to any numbers you want and it doesn't matter--they operations all work out the same. Zeros aren't any easier or harder to work with than other numbers (except when they happen to make individual arithmetic steps easier of course).

 

It is an iron-clad mathematical RULE that THREE Simulataneous equations are required to solve for 3 unknown variables. The number of unknown varibles must not exceed the number of simultaneous equations, for there to be a complete solution for all variables.

I will eat a bug and stand on my head till my ears trun blue if anyone can prove this rule wrong.

This is true for there to be one unique solution (what you called "complete"), otherwise, the situation is that one of the variables will always be written in terms of one of the other variables ( like x = 2z +1) and therefore not having a "unique" (only one) answer for the system. (See the initial discussion, example) .

I'd go fishing with that bug and catch me some fish to fry!!

 

Thats exactly what i'm trying to achieve, solve a system of 3 unknowns and two equations.

Ok thanks.

Take a system such as -2x + y+ 4z = 5 and 1x + 2y +3z = 6. Put that in as a 4 x 2 matrix, solve with graphing calculator using matrix=>math=>rref and you will get an answer that describes x and y in terms of z. That is, x = z - .8 and y = -2z + 3.4. This is true for any value you choose for z. This system has an infinite number of solutions all defined by your choice of z. You are most familiar with systems with only one solution and for that to happen the number of unique equations must be > the number of variables.
I think you got hung up on semantics. "Solve" does not necessarily mean only one solution.

 

Check the last paragraph. I pointed out that with the matrix method you can change the zeros to any numbers you want and it doesn't matter--they operations all work out the same. Zeros aren't any easier or harder to work with than other numbers (except when they happen to make individual arithmetic steps easier of course).

Yeah I know, I missed that part of your post somehow , apoligies.

It doesn't work, I've tried. See for yourself.

After looking back at the example I gave, when the book claims we choose a value for z, I can see how Monster Catfish chose not to eat that bug.

 

Take a system such as -2x + y+ 4z = 5 and 1x + 2y +3z = 6. Put that in as a 4 x 2 matrix, solve with graphing calculator using matrix=>math=>rref and you will get an answer that describes x and y in terms of z. That is, x = z - .8 and y = -2z + 3.4. This is true for any value you choose for z. This system has an infinite number of solutions all defined by your choice of z. You are most familiar with systems with only one solution and for that to happen the number of unique equations must be > the number of variables.
I think you got hung up on semantics. "Solve" does not necessarily mean only one solution.

I understand where your coming from, but how I see it for example, if I was sitting an exam and I can see I can avoid the conventional method in approaching that question and take a "short-cut" by setting up a matrix, the question may only be asking for one solution.

 

I understand where your coming from, but how I see it for example, if I was sitting an exam and I can see I can avoid the conventional method in approaching that question and take a "short-cut" by setting up a matrix, the question may only be asking for one solution.

Solve means find all sets of values of (in this case) x,y and z that satisfy the given relations.

For example how would you answer the simpler question

solve

\({x^2} = 4\)

in an exam?

There are two solutions x= +2 and x=-2


*************************************************************
Note I did post an alternative (geometric) method to matrices for your question.

Each expression of the type

ax + by +cz = d

defines a region of 3D xyz space where the relation is true.

Two of these define a region of intersection

Three define a point of intersection (a full solution)

 

Ok thanks.

I would just like to add, what do all the different shapes mean in linear algebra. I've seen a cone with an inverted cone placed on the top so the tips meet, with planes sliced through the cones.

 

Ok thanks.

I would just like to add, what do all the different shapes mean in linear algebra. I've seen a cone with an inverted cone placed on the top so the tips meet, with planes sliced through the cones.

Those are the visual representations for the conic sections--parabola, ellipse, hyperbola, and circle.