Linear Algebra 1.1

Linear equations

They are basically equations in the form: a1x1 + a2x2 + ... + an xn = b.

xn is a variable whereas an is a constant (any number).

Example: 5x1 + 15x2 + 29x3 = 100.

Solutions of linear equations

Step 1: Draw equations in a graph
Step 2: Find intersection points of the equations

Step 3: Judge it yourself

Augmented matrix of linear equations

Let's say we got this set of linear equations:
x + y + 2z = 9
2x + 4y - 3z = 1
3x + 6y - 5z = 0

Won't it be cool to put it in a matrix like this?

1
1
2
9
2
4
-3
1
3
6
-5
0

Don't be confused.
We simply put the coefficients and the constants on the right hand side of the equations in the matrix.
Augmented matrix makes it easier to perform operations on an equation set as it filters out the variables and unifies the equations into one.

Solving linear equations

We need to perform 3 different operations on an augmented matrix to make our way towards the solution of any set of linear equations:

  1. Multiply a row through by a nonzero constant.
  2. Interchange two rows.
  3. Add a constant times one row to another.
These are called elementary row operations on a matrix.

Example:
If we were to solve the set of equations from the last section:
x + y + 2z = 9
2x + 4y - 3z = 1
3x + 6y - 5z = 0
We'd have an augmented matrix out of it like this:

1
1
2
9
2
4
-3
1
3
6
-5
0

Based on elementary row operations, our goal is to turn the augmented matrix into something like this:
a
x
x
d
0
b
x
e
0
0
c
f

The augmented matrix in this form is said to be in row echelon form.


Here, a, b and c are called pivot points.
x is any number.
We turned the matrix into this form so we can easily solve the equations.
However, let's assume that we know the values:
1
1
2
9
0
1
-7/2
-17/2
0
0
1
3

Since, these are coefficients, we can say that:
That's nice and all but how did we turn an augmented matrix into reduced row echelon form?
The method for doing so is called Guassian elimination!

Gaussian Elimination