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My intent is to create a thread for people interested in Linear Algebra using Matrices. However, I will explicitly state that I am only a student of this class myself and that many things could end up being wrong or an improper way to teach.

I will merely be going along with the class, mostly using excerpts and questions from the book, "Linear Algebra and its Applications: Third Edition," by David C. Lay. So truthfully, this is more for myself. Looking things up and explaining it to others seems to be the best way to learn.

If people have any questions or comments, feel free to share. Also, I know there are many knowledgeable people on this board, so be sure to correct me or make suggestions. This will be very similar to my Intro. to Differential Equations thread.

Some background;

A linear equations is one that has only base variables standing alone or scaled. In most cases I will be giving examples using the variable x with subscripts of whole numbers (i.e 1, 2, ..., n) A linear equation is there for something in this form;

a

where the a's can be any real or complex number and the x's any variable.

For example: 3x

is linear while: 2x

A system of linear equation is then just a set or combination of linear equations usually written one above the other with the symbol "{" denoting the set.

---

A system of a certain number of constants and equations can have three possibilities for solutions; No solution (inconsistent), one solution, or an infinite amount of solutions.

In general, if there are more variables than equations, then the system has infinite many solutions, and if the number of variables equal to the number of equations, the system will only have one solution. But this is not something to memorize, you will easily see this and the exceptions to this very soon. It can easily be shown using two variables and graphing the equations, the intersection will provide the solution.

Starting with the system;

3x

1x

A system can be solved by solving for one variable first, either by substitution or addition. Let's look at substitution of this example first.

Taking the second equation it can be rewritten as;

1x

this can then be substituted into the first equations;

3(1 - 1x

Now that there is only one variable the equation can be solved;

3 - 3x

x

This can then be substituted into any of the original equations, let's use the second since it is much simpler;

1x

1x

Therefore the answer;

x

x

This can become extremely tedious with many places for error as the variables increase so let us look at a different method.

Any equation can be multiplied by a scalar and still be the same equations, so let us take the original and multiply the second by negative three (-3) to obtain;

3x

-3x

An addition of the two equation can the be simply be preformed to obtain;

-x

and then this solution be plugged back in to obtain the same answers;

x

x

The idea with this is to cancel out variables by addition of appropriate scalar quantities. But using this form can become very confusing and cluttering so let us transform this into a matrix which will allow for a much more organized and clean look.

Let us simply take the coefficients of the last example and write a coefficient matrix;

http://www.angelfire.com/tx6/extravagantdreams/matrix1.jpg [Broken]

This is a 2X2 matrix, where the first number denotes the number of rows and the second the number of columns. You will later see why and how this can be done.

To actually solve this problem we will need the numbers for which each equation is equivilant to. Let us just add these to the end and make what is called an augmented matrix.

http://www.angelfire.com/tx6/extravagantdreams/matrix2.jpg [Broken]

Then just writting the coefficients of the previous system we see that;

3x

-3x

http://www.angelfire.com/tx6/extravagantdreams/matrix3.jpg [Broken]

3x

0x

http://www.angelfire.com/tx6/extravagantdreams/matrix4.jpg [Broken]

The idea of a matrix is to have the first term in the first row equal to one and then adding a scalar quantity of that row to the ones below to cancel that first term. Then turning the second number in the second row to a one and adding a scalar quantity of that row to the rows below it to cancel the second term. This goes on until all rows are done and matrix such as this has been created;

http://www.angelfire.com/tx6/extravagantdreams/matrix5.jpg [Broken]

Where the "k" denotes any real number.

This can be accomplished by using Elementary Row Operations:

1. Replacement - Replace one row by the sum of itself and a multiple of another row.

2. Interchange - Interchange two rows

3. Scaling - Multiply all entries in a row by a nonzero constant.

Let us try an example:

Given the system;

x

x

-3x

Making sure we line up our x variable coefficient terms correctly we can write this in an augmented matrix

http://www.angelfire.com/tx6/extravagantdreams/matrix6.jpg [Broken]

Since the first term in the first row is already one we can skip 3 of the Elementary Row Operations (ERO). Since the first terms in the subsequent rows are already zero we can also skip 1 of the ERO for the first term. Now lets make the second term in the second row equal one by using 3 of the ERO

http://www.angelfire.com/tx6/extravagantdreams/matrix7.jpg [Broken]

Now let us use 1 to create this matrix;

http://www.angelfire.com/tx6/extravagantdreams/matrix8.jpg [Broken]

The result can be writen as;

x

x

x

The solution is then;

x

x

x

Some extra links

http://www.ma.utexas.edu/users/kit/les082903web.pdf

http://www.ma.utexas.edu/users/kit/les090303web.pdf

I will merely be going along with the class, mostly using excerpts and questions from the book, "Linear Algebra and its Applications: Third Edition," by David C. Lay. So truthfully, this is more for myself. Looking things up and explaining it to others seems to be the best way to learn.

If people have any questions or comments, feel free to share. Also, I know there are many knowledgeable people on this board, so be sure to correct me or make suggestions. This will be very similar to my Intro. to Differential Equations thread.

Some background;

A linear equations is one that has only base variables standing alone or scaled. In most cases I will be giving examples using the variable x with subscripts of whole numbers (i.e 1, 2, ..., n) A linear equation is there for something in this form;

a

_{1}x_{1}+ a_{2}x_{2}+ ... +a_{n}x_{n}where the a's can be any real or complex number and the x's any variable.

For example: 3x

_{1}- [squ](6)x_{2}= 10is linear while: 2x

_{1}x_{2}+ 6[squ](x_{3}) = 10 is not. This is because of both the x_{1}x_{2}and [squ](x_{3}).A system of linear equation is then just a set or combination of linear equations usually written one above the other with the symbol "{" denoting the set.

---

A system of a certain number of constants and equations can have three possibilities for solutions; No solution (inconsistent), one solution, or an infinite amount of solutions.

In general, if there are more variables than equations, then the system has infinite many solutions, and if the number of variables equal to the number of equations, the system will only have one solution. But this is not something to memorize, you will easily see this and the exceptions to this very soon. It can easily be shown using two variables and graphing the equations, the intersection will provide the solution.

Starting with the system;

3x

_{1}+ 2x_{2}= 61x

_{1}+ 1x_{2}= 1A system can be solved by solving for one variable first, either by substitution or addition. Let's look at substitution of this example first.

Taking the second equation it can be rewritten as;

1x

_{1}= 1 - 1x_{2}this can then be substituted into the first equations;

3(1 - 1x

_{2}) + 2x_{2}= 6Now that there is only one variable the equation can be solved;

3 - 3x

_{2}+ 2x_{2}= 6x

_{2}= -3This can then be substituted into any of the original equations, let's use the second since it is much simpler;

1x

_{1}+ 1(-3) = 11x

_{1}= 4Therefore the answer;

x

_{1}= 4x

_{2}= -3This can become extremely tedious with many places for error as the variables increase so let us look at a different method.

Any equation can be multiplied by a scalar and still be the same equations, so let us take the original and multiply the second by negative three (-3) to obtain;

3x

_{1}+ 2x_{2}= 6-3x

_{1}- 3x_{2}= -3An addition of the two equation can the be simply be preformed to obtain;

-x

_{2}= 3and then this solution be plugged back in to obtain the same answers;

x

_{1}= 4x

_{2}= -3The idea with this is to cancel out variables by addition of appropriate scalar quantities. But using this form can become very confusing and cluttering so let us transform this into a matrix which will allow for a much more organized and clean look.

Let us simply take the coefficients of the last example and write a coefficient matrix;

http://www.angelfire.com/tx6/extravagantdreams/matrix1.jpg [Broken]

This is a 2X2 matrix, where the first number denotes the number of rows and the second the number of columns. You will later see why and how this can be done.

To actually solve this problem we will need the numbers for which each equation is equivilant to. Let us just add these to the end and make what is called an augmented matrix.

http://www.angelfire.com/tx6/extravagantdreams/matrix2.jpg [Broken]

Then just writting the coefficients of the previous system we see that;

3x

_{1}+ 2x_{2}= 6-3x

_{1}- 3x_{2}= -3http://www.angelfire.com/tx6/extravagantdreams/matrix3.jpg [Broken]

3x

_{1}+ 2x_{2}= 60x

_{1}- 1x_{2}= 3http://www.angelfire.com/tx6/extravagantdreams/matrix4.jpg [Broken]

The idea of a matrix is to have the first term in the first row equal to one and then adding a scalar quantity of that row to the ones below to cancel that first term. Then turning the second number in the second row to a one and adding a scalar quantity of that row to the rows below it to cancel the second term. This goes on until all rows are done and matrix such as this has been created;

http://www.angelfire.com/tx6/extravagantdreams/matrix5.jpg [Broken]

Where the "k" denotes any real number.

This can be accomplished by using Elementary Row Operations:

1. Replacement - Replace one row by the sum of itself and a multiple of another row.

2. Interchange - Interchange two rows

3. Scaling - Multiply all entries in a row by a nonzero constant.

Let us try an example:

Given the system;

x

_{1}- 2x_{2}+ x_{3}= 0x

_{2}- 8_{3}= 9-3x

_{2}+ 13x_{3}= -9Making sure we line up our x variable coefficient terms correctly we can write this in an augmented matrix

http://www.angelfire.com/tx6/extravagantdreams/matrix6.jpg [Broken]

Since the first term in the first row is already one we can skip 3 of the Elementary Row Operations (ERO). Since the first terms in the subsequent rows are already zero we can also skip 1 of the ERO for the first term. Now lets make the second term in the second row equal one by using 3 of the ERO

http://www.angelfire.com/tx6/extravagantdreams/matrix7.jpg [Broken]

Now let us use 1 to create this matrix;

http://www.angelfire.com/tx6/extravagantdreams/matrix8.jpg [Broken]

The result can be writen as;

x

_{1}- 2x_{2}+ x_{3}= 0x

_{2}- 4x_{3}= 4x

_{3}= 3The solution is then;

x

_{1}= 29x

_{2}= 16x

_{3}= 3Some extra links

http://www.ma.utexas.edu/users/kit/les082903web.pdf

http://www.ma.utexas.edu/users/kit/les090303web.pdf

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