Matrix Additon

Matrix addition is fairly simple, and is done entry-wise.

• Add the following matrices:
I need to add the pairs of entries, and then simplify for the final answer:

So the answer is:

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Up until now, you've been able to add any two things you felt like: numbers, variables, equations, and so forth. But addition doesn't always work with matrices.

• Perform the indicated operation, or explain why it is not possible.
Since matrices are added entry-wise, I have to add the 1 and the 4, the 2 and 5, the 0 and the7, and the 3 and the 8. But what do I add to the 6 and to the 9? There are no corresponding entries in the first matrix that can be added to these entries in the second matrix. So the answer is:

I can't add these matrices, because they're not the same size.

This is always the case: To be able to add two matrices, they must be of the same size. If they are not the same size (if they do not have the same "dimensions"), then the addition is "not defined" (doesn't make mathematical sense).

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Subtraction works entry-wise, too.   Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved

• Given the following matrices, find A – Band A – C, or explain why you can not.
A and B are the same size, each being 2 × 3 matrices, so I can subtract, working entry-wise:

However, A and C are not the same size, since A is 2 × 3 and C is 2 × 2. So this subtraction is not defined.


A – C is not defined, because A and C are not the same size.

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Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense), can be turned into homework problems.

• Find the values of xand y given the following equation:
First, I'll simplify the left-hand side a bit by adding entry-wise:

Since matrix equality works entry-wise, I can compare the entries to create simple equations that I can solve. In this case, the 1,2-entries tell me that x + 6 = 7, and the 2,1-entries tell me that 2y – 3 = –5. Solving, I get:

x + 6 = 7 
x = 1
2y – 3 = –5 
2y = –2 
y = –1
Bibliography: http://www.purplemath.com/modules/mtrxadd.htm

Operations with Matrices


In the previous posts, we used matrices to solve linear equations. Matrices, however, can be used to do much more than that. There is a rich mathematical theory of matrices, and it's applications are numerous.

Equality of Matrices

For two matrices to be equal, they must be of the same size and have all the same entries in the same places. For instance, suppose you have the following two matrices:

These matrices cannot be the same, since they are not the same size. Even if A and B are the following two matrices:

...they are still not the same. Yes, A and B each have six entries, and the entries are even the same numbers, but that is not enough for matrices. A is a 3 × 2 matrix and B is a 2 × 3matrix, and, for matrices, 3 × 2 does not equal2 × 3! It doesn't matter if A and B have the same number of entries or even the same numbers as entries. Unless A and B are the same size and the same shape and have the same values in exactly the same places, they are not equal.
This property of matrix equality can be turned into homework questions. You will be given two matrices, and you will be told that they are equal. You will need to use this equality to solve for the values of variables.

• Given that the following matrices are equal, find the values of x and y.
For A and B to be equal, they must have the same size and shape (which they do; they're each2 × 2 matrices) and they must have the same values in the same spots. Then a_1,1 must equalb_1,1, a_1,2 must equal b_1,2, and so forth. The entries a_1,2 and a_2,1 are clearly equal, respectively, to entries b_1,2 and b_2,1 "by inspection" (that is, "just by looking at them"). But a_1,1 = 1 is not obviously equal to b_1,1 = x. For A to equal B, I must have a_1,1 = b_1,1, so it must be that 1 = x. Similarly, I must have a_2,2 = b_2,2, so then 4 must equal y. Then the solution is:


x = 1,    y = 4   Co








pyright © Elizabeth Stapel 2003-2011 All Rights Reserved

Network Analysis

Network analysis is another application for systems of linear equations. Networks are composed of junctions and branches and are used as models in many diverse fields such as traffic analysis and electrical engineering.

Determine the amount of traffic between each of the four intersections.


x1 =?, x2 =?, x3 =?, x4 =?

At each intersection, the incoming traffic has to match the outgoing traffic.

Intersection A: x4 + 610 = x1 + 450
Intersection B: x1 + 400 = x2 + 640
Intersection C: x2 + 600 = x3
Intersection D: x3 = x4 + 520



x4 + 610 = x1 + 450
x1 + 400 = x2 + 640
x2 + 600 = x3
x3 = x4 + 520


x1 + x4 = −160
x1 − x2 = 240
x2 − x3 = −600
x3 − x4 = 520

APPLICATIONS OF SYSTEMS OF LINEAR EQUATIONS


Systems of Linear equations arise in a wide variety of applications and are one of the central themes in linear algebra.  There are mainly two applications of systems of linear equation.


Polynomial Curve Fitting

Suppose a collection of data is represented by n points in the xy-plane,

         (x1, y1), (x2, y2),.....,(xn, yn)

and you are asked to find a polynomial function of degree n - 1

          

whose procedure is called polynomial curve fitting. If all x-coordinates of the points are distinct, then there is precisely one polynomial function of degree n - 1 (or less) that fits the n points.

Solving Systems of Linear Equations Using Matrices

The matrix method of solving systems of linear equations is just the elimination method in disguise. By using matrices, the notation becomes a little easier.

Suppose you have a system of linear equations such as:

The first step is to convert this into a matrix. Make sure all equations are in standard form (Ax + By = C), and use the coefficients of each equation to form each row of the matrix. It may help you to separate the right column with a dotted line.

Next, we use the matrix row operations to change the 2 × 2 matrix on the left side to the identity matrix. First, we want to get a zero in Row 1, Column 2. So, add 4 times Row 2 to Row 1.

Next we want a 1 in the top left corner.

Now we want a zero in the bottom left corner.

Finally, we want a 1 in Row 2, Column 2.

Now that we have the 2 × 2 identity matrix on the left, we can read off the solutions from the right column:

x = 3

y = –1

The same method can be used for n linear equations in n unknowns; in this case you would create an n × (n + 1) matrix, and use the matrix row operations to get the identity n × n matrix on the left side.

Important Note: If the equations represented by your original matrix represent parallel lines, you will not be able to get the identity matrix using the row operations. In this case, the solution either does not exist or there are infinitely many solutions to the system.

Matrices

In mathematics, a matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. The individual items in a matrix are called its elements or entries.

Matrix Row Operations

There are 3 basic operations used on the rows of a matrix when you are using the matrix to solve a system of linear equations. The goal is usually to get the left part of the matrix to look like the identity matrix.

The three operations are:

• Switching Rows
• Multiplying a Row by a Number
• Adding Rows

Switching Rows

You can switch the rows of a matrix to get a new matrix.

In the example shown above, we move Row 1 to Row 2, Row 2 to Row 3, and Row 3 to Row 1. (The reason for doing this is to get a 1 in the top left corner.)

Multiplying a Row by a Number

You can multiply any row by a number. (This means multiplying every entry in the row by the same number.)

In this example, we have multiplied Row 3 of the matrix by 1/3. (This gives us the 1 we need in Row 3, Column 3.)

Adding Rows

You can also add two rows together, and replace a row with the result.

For example, in the matrix that resulted in the last example, we can add rows 2 and 3 together, entry by entry:

Then, we replace Row 2 with the result.

Adding Multiples of Rows

We said there were only three operations, and there are. But by using the last two operations in combination, we can add whole multiples of rows to other rows, to make things go faster.

Back up a step, so we have the matrix:

Now instead of just adding Row 2 + Row 3, add Row 2 + (2 × Row 3):

Then replace Row 2 with the result.

This way, we get a 0 in Row 2, Column 3.

We can do this again to get a 0 in Row 2, Column 1. Here, we multiply Row 1 by –2, add it Row 2, and replace Row 2 with the result.

We'll show a few more steps, to get the 3 × 3 identity matrix on the left (and thus solve the system).

The next step is to add Row 2 + (4 × Row 3) to get a 0 in Row 2, Column 3.

Next, we need a zero in Row 1, Column 3.

The last step is just an application of the second operation, multiplying a row by a number.

We now have the solution as ordered triple (1, 0, –2).

Important Note: If the equations represented by your original matrix represent identical or parallel lines, you will not be able to get the identity matrix using these row operations. In this case, the solution either does not exist or there are infinitely many solutions to the system.

Solutions of a Linear equation (Part 2)

The addition method of solving systems of equations is also called the method of elimination. This method is similar to the method you probably learned for solving simple equations.
If you had the equation "x + 6 = 11", you would write "–6" under either side of the equation, and then you'd "add down" to get "x = 5" as the solution.

x + 6 = 11
    –6    –6
x       =   5

You'll do something similar with the addition method.

• Solve the following system using addition.
2x + y = 9
3x – y = 16
Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down:

2x + y = 9
3x – y = 165x      = 25
Now I can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so I'll back-solve in that one:

2(5) + y = 9
  10 + y = 9
          y = –1
Then the solution is (x, y) = (5, –1).

It doesn't matter which equation you use for the backsolving; you'll get the same answer either way. If I'd used the second equation, I'd have gotten:

3(5) – y = 16
  15 – y = 16
        –y = 1
          y = –1

...which is the same result as before.

• Solve the following system using addition.
x – 2y = –9x + 3y = 16
Note that the x-terms would cancel out if only they'd had opposite signs. I can create this cancellation by multiplying either one of the equations by –1, and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the –1 through the entire equation. (That means both sides of the "equals" sign!)
I'll multiply the second equation.

The "–1R₂" notation over the arrow indicates that I multiplied row 2 by –1. Now I can solve the equation "–5y = –25" to get y = 5. Back-solving in the first equation, I get:

x – 2(5) = –9
x – 10 = –9
x = 1
Then the solution is (x, y) = (1, 5).

A very common temptation is to write the solution in the form "(first number I found, second number I found)". Sometimes, though, as in this case, you find the y-value first and then the x-value second, and of course in points the x-value comes first. So just be careful to write the coordinates for your solutions correctly. Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved

Bibliography: http://www.purplemath.com/modules/systlin5.htm

This post is about the solution of a Linear System (Part 1).

The solution of a Linear system is he ordered pair ( for 2 variables), ordered triple ( for 3 variables) and so on, that satisfies or makes each equation of a system true. 

We can classify Linear systems based on it's solution. 

1. Consistent - we can say that the system is a consistent one if it has 1 or more solution. There are two types of consistent system, the consistent independent system and the consistent dependent. The system is said to be consistent independent if it has exactly one solution. On the other hand, the system is said to be 

consistent dependent if it has an infinite number of solutions.

2. Inconsistent - we can say that the system is an inconsistent one if it has no solution.

To solve a Linear system, we can use many techniques like:

-Graphing method

-Substitution method

-Elimination method

- Matrix method

But the commonly used technique in Linear algebra is the Elimination method and the matrix method.

In Linear Algebra, we deal with systems of Linear Equation. But what is a system of Linear equation??

One definition is: 

"A "system" of equations is a set or collection of equations that you deal with all together at once. Linear equations (ones that graph as straight lines) are simpler than non-linear equations, and the simplest linear system is one with two equations and two variables." (read more at purplemath.com) 

For example:

y = 3x – 2
y = –x – 6

The 2 equations above is an example of a system of Linear equation. We deal with the two equations together at once. We can solve the system by many ways. But the easiest one is transforming the system into the "Row Echelon Form." I will post it later, so don't worry.

Now we know what a system of Linear equation is. My next post will be about how to find the solution of a system. Stay Tuned!!

Hi!! This is my humble blog. My aim is to post some topics and lessons about Linear Algebra and show some tips on how to solve some problems. Enjoy learning ^U^