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Introduction To Augmented Matrices

What is a matrix?

A matrix, according to the dictionary, is:

matrix - a rectangular array of mathematical elements(as the coefficients of simultaneous linear equations) that can be combined to form sums and products with similar arrays having a appropriate number of rows and columns

More simply put, it's just a array of numbers mainly used for manipulating linear equations through different functions.

What are some terms related to matrix?

This is a example matrix:

3 5 | 9

2 3 | 5

Describing this matrix, we can say it has two rows and three columns. Be aware that the first row is represented by:

R₁

While the second row is represented by:

R₂

In addition, matrices can contain any number of rows and columns. Just know that any line of numbers joined with another becomes a augmented matrix. Augmented matrices that have two rows and three columns have a simplified form, called the identity matrix:

1 0 | x

0 1 | y

Also known as reduced row-echelon form, this will come in handy when we discuss solving system of equations later.

How do we manipulate the values in matrices?

Elementary row operations can be used to manipulate values in a matrix. There are actually three simple ones we'll be using:

We can swap the values of any two rows. For example, say we have the following matrix:

3 5 | 9

2 3 | 5

Swapping the two rows returns:

2 3 | 5

3 5 | 9
We can multiply all columns in a row by a number that isn't zero. Using the same original example as above:

R₁ * 5

15 25 | 45

2 3 | 5
Add the values in one row to the product of the values in another row. This one is possibly the most complex of the the three, but here's an example of how it can be represented:

R₁ + (5 * R₂)

Let's use this with our original example. Multiplying our second row by two gives us:

3 5 | 9

10 15 | 25

However, this doesn't actually change the value of the second row. All the columns in the second row stay the same, in this operation. We'll just calculating what to add to the values in the first row:

13 20 | 34

2 3 | 5

Notice how we added the values in the second row to the first(3 + 10 = 13, 5 + 15 = 20, 9 + 25 = 34), but didn't change the second row? This is how this operation works.

An example of how this can be used?

We can use matrices to solve system of equations. For example, take a look at these two linear equations:

3x + 5y = 9

2x + 3y = 5

What values for x and y work for both of these? There's actually a very systematic way of solving this. But first, let's convert these to a matrix:

3 5 | 9

2 3 | 5

Looks similar, right? It's actually just the example I used in the previous sections. To solve for x and y, we need to get this into reduced row-echelon form, which we discussed earlier. There's a very systematic way of doing this, and that is Guass-Jordan elimination. The steps are as follows:

Convert the value in the first row, first column to a one. We can simply multiply the entire row by the first row, first column's inverse:

inverse = 1/3

1 5/3 | 3

2 3 | 5
Next, we need to convert the value in the second row, first column to a zero. We can achieve this by multiplying the opposite value(E.G 2 = -2, -2 = 2) in the second row, second column by the first row and then adding that to the second row.

opposite = -2

1 5/3 | 3

0 -1/3 | -1
Then, we need to convert the value in the second row, second column to a one. We can do this by multiplying the second row by the inverse of the second row, second column, similar to step one.

inverse = -3

1 5/3 | 3

0 1 | 3
Finally, to convert the value in the first row, second column to a zero, we can multiply the opposite value of the value in the first row, second column by the second row and add it to the first row, similar to step two:

opposite = -5/3

1 0 | -2

0 1 | 3
Now we can check our answer, (-2, 3). Plugging the values into our first equation shows us we are correct:

3x + 5y = 9

3(-2) + 5(3) = 9

-6 + 15 = 9

9 = 9

And this series of steps can be applied to every system of equations. If you aren't sure if there are infinite solutions, try putting in random values for x and y. If you aren't sure if there are no solutions, try plugging in the values for x and y.

Here's a program that demonstrates this: Solving System Of Equations[Challenge]