parameter

[Hielly]

hey
1. Find all solutions of the following sets of equations
a) x-y+z=4 and -x+y+z=6

My first attempt
x-y+z=4 (1)
-x+y+z=6 (2)
 
(1)-(2)
2x-2y=-2

let y=parameter
so x= -1 + paramter
so sub in (1)  to get to find z,
-1 + paramter - parameter + z =4
so z=5

However the book says that x=-1+ paramter and y= paramter

So i tried again and did (1)+(2) to find z, which is 5.
Then subbed z into (1), which equals x-y=-1
and let y=parameter, so x=-1+paramter.

Which is the same as the answer at the back of the book.

But why does y have to be parameter and not x ?

Thanks~

[QuantumJG]

hey
1. Find all solutions of the following sets of equations
a) x-y+z=4 and -x+y+z=6

My first attempt
x-y+z=4 (1)
-x+y+z=6 (2)
 
(1)-(2)
2x-2y=-2

let y=parameter
so x= -1 + paramter
so sub in (1)  to get to find z,
-1 + paramter - parameter + z =4
so z=5

However the book says that x=-1+ paramter and y= paramter

So i tried again and did (1)+(2) to find z, which is 5.
Then subbed z into (1), which equals x-y=-1
and let y=parameter, so x=-1+paramter.

Which is the same as the answer at the back of the book.

But why does y have to be parameter and not x ?

Thanks~

There is a VERY simple way to do this!

Put them in a matrix like this in your graphics calculator:

Where the first column is x, second is y, the third is z and the forth is what the equation equals. The first row is the first equation and the second is the second equation.

1 -1 1 4
-1 1 1 6

this will become

1 -1 0 -1
0  0 1  5

if you use rref (convert the matrix to reduced row echelon form).

This means that equivalently you can say:

x -y +   = -1 => x = y - 1
         z =  5

let y = t
    x = t - 1
    z = 5

so you can say all the solutions are:

(x,y,z) = (-1,0,5) + t(1,1,0)

Basically rref will solve any simultaneous linear equations you have.

What you could do if you don't have a graphics calculator to use is:

 x - y + z = 4 ...1
-x +y + z = 6 ...2

what I would have done is let eqn 3 = eqn 2 + eqn 1

=> 2z = 10 => z = 5 ...3

sub eqn 3 into eqn 1 to give eqn 4

=> x - y + 5 = 4 => x - y = -1 ...4

which gives:

z = 5 ...3
x - y = -1 ...4

[/0]

Generally it doesn't matter which one you choose to be the free variable,

But the way it's normally done is, if you turn the equation into an augmented matrix:



(equation 2 replaced by eq2 + eq1)



Here the boxed '1's are the pivot columns, since they are the left-most non-zero number in their respective row. The x-column has a pivot, the z-column has a pivot, but the y-colum doesn't. So 'y' is considered a 'free' variable, and is used as the parameter. I think this is mostly just a tradition though.

[Hielly]

A tradition. So whenever i come across these questions i would set y equal the parameter?

Okay i just did the next questions which is  2x-y+x=6 and x-z=3, the answer says that z=parameter

Im confused now.. :S

[/0]

If you're doing methods it's really not something to worry about, they're not going to get uptight about which one becomes the parameter, especially since it's correct regardless of which one is the parameter.

But anyway, by tradition:



(eq2 replaced by 2*eq2 - eq1)

Here the pivots are in x and y, so z is the free variable

[QuantumJG]

A tradition. So whenever i come across these questions i would set y equal the parameter?

Okay i just did the next questions which is  2x-y+x=6 (is it 2x - y + z = 6 ?) and x-z=3, the answer says that z=parameter

Im confused now.. :S

No your understanding isn't exactly correct!

Can I ask you CAS guys if you learn't Gaussian elimanation or row reduction?

Matrices are very powerful for solving simulataneous linear equations, because programmers have written algorithms for programs to follow to solve a linear equation with several variables.

so we have:

 x       - z = 3 ...1
2x - y + z = 6 ...2

I would first off let eqn 3 = eqn 2 - 2*eqn 1

=> -y + 3z = 0 ...3

so:

x       -   z = 3
   - y + 3z = 0

so basically we can say that:

x = z + 3
y = 3z

so let z = t (parameter)

=> x = 3 + t
=> y = 3t

=> (x,y,z) = (3,0,0) + t*(1,3,1) (every solution)

If you guys did row reduction or gaussian elimination on matrices I could easily tell you what to assign as your parameter

If say I have a matrix in this form:

1 1 1 | 1
0 0 0 | a 

where a =/= 0, then my system of linear equations has no solution! such that one of these linear equations says that 0 = a (something other than zero)

If I have this

1 0 0 | a
0 1 0 | b
0 0 1 | c

then I have a unique solution (note: you can only have a unique solution if your system of linear equations has no parameters, this will usually mean that for every variable you will need a linear equation)

Then you have linear equations with parameters!

say I have:

1 0 1|a
0 1 1|b

then the parameter is z

or

1 1 0|a
0 0 1|b

the the parameter is y

or (you probably wont see this in methods)

1 1 1| a
0 0 0| 0

then y and z are parameters.

The last case comes from having two linear equations where one is redundant. If you focus on the other 2 you will see that the parameter is basically a value that is not a leading entry (i.e. there is a number other than 0 before it).

I hope this helps! If you do linear algebra in uni you will look at this in fair detail.

[Hielly]

If you're doing methods it's really not something to worry about, they're not going to get uptight about which one becomes the parameter, especially since it's correct regardless of which one is the parameter.

But anyway, by tradition:



(eq2 replaced by 2*eq2 - eq1)

Here the pivots are in x and y, so z is the free variable

Hey thanks, how do you know which numbers are in the square box?

[/0]

The pivots are the first non-zero entries in their respective rows.
You can't have two pivots in the same column though

[Gloamglozer]

Can I ask you CAS guys if you learn't Gaussian elimanation or row reduction?

Nope.  CAS just teaches how to add, subtract and multiply matrices, touches on parameters and transition matrices.

[Hielly]

how about this

1 1 1 1   |4
1 3 3 0   |2
1 1 2 -1  |6

so 1,3,1 ?

[/0]

In your calculator go into the MATH menu and in 'Matrix' select rref( then type in your matrix

Code: [Select]

rref([1,1,1,1,4;1,3,3,0,2;1,1,2,-1,6])
It will use Gaussian elimination to reduce your matrix to



(where the last column is the RHS of the equations)

So if the 1st column is x, 2nd is y, 3rd is z, and 4th is w,
The pivots are in column x, y an z, so you w becomes the free variable, and you can immediately use the matrix to express x,y,z in terms of w (the parameter).

You don't need to know how to get to rref by hand (unless you do linear algebra in university)


Similarly, if you used the rref( function on your original system:

Code: [Select]

rref([1,-1,1,4;-1,1,1,6])


(once again the last column is the RHS of the equations, since I don't know how to do augmented matrices on the calculator)

So you can see that x and z have pivots, and y is the free variable

[Hielly]

Thanks!

[Hielly]

Also i can't seem to find the inverse of
2   1  -1
3   4  -(a+1)
10 8   a-4

on the cas calc, it says error:invalid list or matrix.

[QuantumJG]

Also i can't seem to find the inverse of
2   1  -1
3   4  -(a+1)
10 8   a-4

on the cas calc, it says error:invalid list or matrix.

Is this a 3 x 3 matrix?

[Hielly]

yeah