Linear Systems

Geometry of Systems of Equations
We know that for two by two linear systems of equation, the geometry is that
of two lines that either intersect, are parallel, or are the same line. If
they intersect then there is exactly one solution, if they are parallel then
there are no solutions, and if they are the same line, then there are infinitely
many solutions. For three by three systems, the situation is different. The solution
set is either the empty set, a point, a line, or a whole plane. For four by four systems, the geometry becomes four dimensional and is rough
to comprehend, but is still useful.

The Algebra of Linear Systems.
In this class we will perform algebra on linear systems in a new way. For
example, for a three by three system, we line the equations up to form three
rows. We will manipulate the rows to simplify the equations. There
are three operations, called row operations that we can perform:
Row Operations

We can multiply an entire row by a nonzero constant
cR_{i} > R_{i}

We can interchange two rows.
R_{i} <> R_{j
}

We can replace one row with that row + a multiple of another row
cR_{j } + R_{i} > R_{i}

We follow the following steps that use row operations to solve a system of
equations.

We interchange rows so that the top left corner is nonzero (preferably
a 1).

We multiply row 1 by the appropriate number to make the top left corner
a 1.

We replace rows two and three with the appropriate multiples of
row 1 + rows two and three to make the two bottom rows begin with 0.

We interchange the two bottom rows so that the middle coefficient
nonzero.

We multiply row two appropriately so that the middle number is a
1.

We use row two to make the middle top and bottom zero.

Make the bottom right 0.

Use row three to make the top and middle rights 0.
Example:
2x + y = 4
x  2y + z = 0
3x  4y + 2z = 1

x  2y + z = 0
2x + y = 4
3x  4y + 2z = 1

Already done.

We replace row two with row two  twice row 1 and replace row three with row three minus
three times row 1.
x  2y + z = 0
5y  2z = 4
2y  z = 1

Not needed

We divide row two by 5:
x  2y + z = 0
y  2/5 z =
4/5
2y 
z = 1

We replace row 1 with row 1 + 2 row 2 and replace row three with
row three  twice row two.
x + 1/5 z =
8/5
y  2/5 z
= 4/5
1/5 z = 3/5

We multiply row three by 5
x + 1/5 z = 8/5
y  2/5 z = 4/5
z = 3

We replace row 1 with row 1  1/5 row 1 and replace row two with
row three + 2/5 row three.
x
= 1
y = 2
z = 3
Exercises:

4x  3y  z = 0
x  3y + 2z = 7
3x + 9y  z = 2

3x  2y + z = 3
4x + y = 1
11y  4z = 9

Application
Your breakfast consists of orange juice, cereal, and eggs with the following
nutritional information:

OJ 
Cereal 
Eggs 
Protein 
0% 
10% 
20% 
Vitamin C 
20% 
15% 
0% 
Calories 
100 
120 
100 
If you must have 30% protein, 30% Vitamin C and 300 calories for your breakfast,
How many servings of OJ, Cereal, and Eggs should you have?
Solution
Let
x = the number of servings of OJ
y = the number of servings of Cerial
z = the number of servings of eggs
Then
10y + 20z = 30
20x +
15y
= 30
100x + 120y + 100z = 300

20x +
15y
= 30
10y + 20z = 30
100x + 120y + 100z = 300

x +
3/4y
= 3/2
10y + 20z = 30
100x + 120y + 100z = 300

x +
3/4y
= 3/2
10y + 20z = 30
45y +
100z = 150

Already Done

x +
3/4y =
3/2
y + 2z = 3
45y +
100z = 150

x  3/2z = 3/4
y
+ 2z = 3
10z = 15

x  3/2z = 3/4
y
+ 2z = 3
z = 3/2

x
= 3/2
y
= 0
z = 3/2
We conclude that the breakfast should consist of 1.5 of a serving of OJ, no
cereal, and 1.5 servings of eggs.
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