Another Method of Solving Systems of Linear Equations

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Cramer's Method

Timothy Foster 09 Oct 2008 17:35

One way to solve linear systems is Cramer's method.
In Cramer's method, you can easily solve for a variable using matrices. Let's say you had a system
x+y=2
2x+3y=6
First, set up a matrix for the variables. This is done my putting the corresponding coefficients in their appropriate element.

(1)

\begin{align} \left[ \begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array} \right] \end{align}

Then, make a matrix with the answers, but instead, put those answers in the variable matrix in place of the variable you want to solve for. In his case, we want x, so we will put the answers in x's spot in the matrix.

(2)

\begin{align} \left[ \begin{array}{cc} 2 & 1 \\ 6 & 3 \end{array} \right] \end{align}

Now that you have these two matrices, the value of x is determined by the quotient of the second matrix and the first.

(3)

\begin{align} {\left| \begin{array}{cc} 2 & 1 \\ 6 & 3 \end{array} \right| \over \left| \begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array} \right| } \end{align}

The determinant of the first matrix is 0, and the determinant of the second matrix is 1. When this is solved, we get x=0. We can then substitute this value into an equation and find y=2. So, our solution is (0,2).

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~Timothy

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Timothy Foster - @tfAuroratide
Auroratide.com - Go here if you're nerdy like me

Re: Cramer's Method

Wisp 09 Oct 2008 18:00

By the way, this method can be used for systems of equations with more than two variables and with more than two equations. Just make sure that when using Cramer's method, you begin by writing out the matrix with all the coeffecients (the first matrix seen up there in Timothy's example), then writing out the matrix with the answers (2 and 6 in the example) out to the side somewhere else. This way you can just go back and look at them for reference.

Re: Cramer's Method

Timothy Foster 09 Oct 2008 19:32

Right. This method is very good for systems with more equations and variables. May I caution, though, that it will not work if you don't know how to find the determinant of matrices with larger dimensions. It gets rather complicated. For longer systems of equations, I would use the rref, but that can't be done without a calculator, unlike in Cramer's method.

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~Timothy

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Timothy Foster - @tfAuroratide
Auroratide.com - Go here if you're nerdy like me

Re: Cramer's Method

lphillips 10 Oct 2008 03:04

rref can be done by hand… It's called row reduction. Maybe someone will research the hand method and post it.

Re: Cramer's Method

Wisp 10 Oct 2008 01:37

Hey, this is an assignment, remember? Leave some things for other people to say until they've all posted! XD

Row reduction/hand method/Gaussian elimination.

Superkidthezooligist 12 Oct 2008 19:02

Row reduction, also known as Gaussian elimination, is an algorithm(I watch numb3rs(-:) for solving a system of linear equations
In row reduction a sequence of elementary row operations are applied to a matrix to transform it to row "echelon" form. These operations are:

row scaling: the multiplication a row by a nonzero scalar

row exchange: the exchanges of two rows

and row replacement: the addition of a multiple of one row to another row

( A pivot is the leading non-zero entry of some row.)
If there is a pivot in the last column, the system is inconsistent ; there will be no solutions.
If that is not the case, then the general solution will have degrees of freedom, where is the number of columns from to that have no pivot. To be more precise, the general solution will have the form of one particular solution plus an arbitrary linear combination of linearly independent -vectors.

A spin-off of the Gaussian elimination method is the Gauss-Jordan elimination method. In this variation you reduce to "echelon" form, and then if the system proves to be consistent, you continue to apply the elementary row operations until the augmented matrix is in reduced "echelon" form. This means that not only does each pivot have all zeroes below it, but that each pivot also has all zeroes above it.

In essence, Gauss-Jordan elimination performs the back substitution; the values of the unknowns can be read off directly from the terminal augmented matrix. This method is slower than the gaussian method but is good for solving systems on paper.

Source:http://planetmath.org/encyclopedia/GaussianElimination.html

Q-what is the "echelon" form? do you know mr. P?
SAM B

Re: Row reduction/hand method/Gaussian elimination.

lphillips 14 Oct 2008 01:27

Echelon is another big word that simply means formation. In this situation, it refers to a very specific matrix with 1s in the diagonal form the top-left and 0s below the diagonal.

Like so:

(1)

\begin{align} \left[ \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 0 & 1 & 5 & 6 \\ 0 & 0 & 1 & 7 \end{array} \right] \end{align}

Dig my LaTex skeelz…

Re: Row reduction/hand method/Gaussian elimination.

Wisp 14 Oct 2008 17:30

"with 1s in the diagonal form the top-left…"

Form as in the form of the matrix, or did you mean 'from'? And what do 1s and 0s mean?

Last edited on 14 Oct 2008 18:51 by Wisp Show more

Re: Row reduction/hand method/Gaussian elimination.

Timothy Foster 14 Oct 2008 18:03

He meant from.
Having something in this form prepares you to reduce it into something more friendly.

(1)

\begin{align} \left[ \begin{array}{cccc} 1 & 0 & 0 & 41 \\ 0 & 1 & 0 & -29 \\ 0 & 0 & 1 & 7 \end{array} \right] \end{align}

This is Mr. Phillips matrix reduced. This translates to a system solution of (41,-29,7), so this form is very useful.

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~Timothy

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Timothy Foster - @tfAuroratide
Auroratide.com - Go here if you're nerdy like me

Re: Row reduction/hand method/Gaussian elimination.

Superkidthezooligist 15 Oct 2008 19:14

nice…
thank you mr P! 8-)

Re: Cramer's Method

agoodwyn 14 Oct 2008 22:32

The Gaussian Method

My source was purple math

math is hard.

Solving three-variable, three-equation linear systems is more difficult, at least initially, than solving the two-variable systems, because the computations involved are more messy. You will need to be very neat in your working, and you should plan to use lots of scratch paper. The method for solving these systems is an extension of the two-variable solving-by-addition method, so make sure you know this method well and can use it consistently correctly.

Though the method of solution is based on addition/elimination, trying to do actual addition tends to get very messy, so there is a systematized method for solving the three-or-more-variables systems. This method is called "Gaussian elimination" (with the equations ending up in what is called "row-echelon form").

Let's start simple, and work our way up to messier examples.

Solve the following system of equations.
5x + 4y – z = 0
10y – 3z = 11
z = 3

It's fairly easy to see how to proceed in this case. I'll just back-substitute the z-value from the third equation into the second equation, solve the result for y, and then plug z and y into the first equation and solve the result for x.

10y – 3(3) = 11
10y – 9 = 11
10y = 20
y = 2

5x + 4(2) – (3) = 0
5x + 8 – 3 = 0
5x + 5 = 0
5x = –5
x = –1

Then the solution is (x, y, z) = (–1, 2, 3).

The reason this system was easy to solve is that the system was "triangular"; this refers to the equations having the form of a triangle, because of the lower equations containing only the later variables.

The point is that, in this format, the system is simple to solve. And Gaussian elimination is the method we'll use to convert systems to this upper triangular form, using the row operations we learned when we did the addition method.

Re: Cramer's Method

Timothy Foster 14 Oct 2008 23:57

Good method, but such a situation is probably rare. If you happen upon such a system, take advantage of this!
Also, the method works in other ways, like in this system:

(1)

\begin{align} 2x+4y = 13 \\ 2y+3 = 12 \\ 5y-3x+2z = 23 \end{align}

As you can see, this is also a triangular system, since an equation contains only y, another contains y and some other variable, and the last contains all of them. So this similarly can be solved the same way.

Also, it works for even more complex systems, such as

(2)

\begin{align} 2a = 2 \\ 3a+4b = 11 \\ -2a-3b+c = -5 \\ 4a-2b+c-2d = -5 \\ 5a+2b-3c-2d+3e = 7 \end{align}

It works for this monster as well.

Step One

Solve the first equation, and substitute that into the next one.
2a=2
a=1
3(1)+4b=11

Step Two

Solve second equation, and sub both values into third equation.
3+4b=11
4b=8
b=2
-2(1)-3(2)+c=-5

Step Three

Solve third equation, and sub the three values into fourth equation.
-2-6+c=-5
c=3
4(1)-2(2)+3-2d=-5

Step Four

Solve fourth equation, and sub all the values into final equation.
4-4+3-2d=-5
-2d=-8
d=4
5(1)+2(2)-3(3)-2(4)+3e=7

Step Five

Solve final equation.
5+4-9-8+3e=7
3e=15
e=5

Solution

Arrange all of the values into a five-dimensional coordinate for your final answer.
Solution: (1,2,3,4,5)

So, if you see such a triangular pattern, take advantage!
Wow, I took too long on this…

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~Timothy

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Timothy Foster - @tfAuroratide
Auroratide.com - Go here if you're nerdy like me

Re: Cramer's Method

severed_heart 16 Oct 2008 14:23

WHOA :O
srry for breaking into the Alg 2 disscusion or whatever this is, but THATS AWSOME

Re: Cramer's Method

Superkidthezooligist 15 Oct 2008 19:17

Timothy, How in the world of the internets did you get that effect????
(the whole step 1, step 2 thing)
That is Wickedly awesometastical!!!

Re: Cramer's Method

Timothy Foster 20 Oct 2008 21:03

So far, there are four participants in this homework assignment. Is anyone else going to do something? Just a reminder…

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~Timothy

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Timothy Foster - @tfAuroratide
Auroratide.com - Go here if you're nerdy like me