3: Linear Systems
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1. 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.
2. 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
1. We can multiply an entire row by a nonzero constant
\[cR_i \rightarrow R_i \]
2. We can interchange two rows.
\[R_i \leftrightarrow R_j\]
3. We can replace one row with that row + a multiple of another row
\[cR_j + R_i \rightarrow R_i\]
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
\[\begin{align} &2x &+y & &&= 4 \\ &x & 2y &&+ z &= 0\\ &3x &4y &&+ 2z &= 1 \end{align}\]
Solution:
1. \[\begin{align} &x &2y &&z &= 0 \\ &2x &+y &&&= 4 \\ &3x &4y &&+2z &= 1 \end{align}\]
2. Already done.
3. We replace row two with row two  twice row 1 and replace row three with row three minus three times row
\[\begin{align} &x & 2y &&+ z &= 0 \\ && 5y &&2z&= 4 \\ && 2y && z &= 1 \end{align}\]
4. Not needed
5. We divide row two by 5:
\[\begin{align} &x &2y &&+z &= 0 \\ &&y &&\dfrac{2}{5}z&= \dfrac{4}{5} \\ &&2y &&z &= 1 \end{align}\]
6. We replace row 1 with row 1 + 2 row 2 and replace row three with row three  twice row two.
\[\begin{align} &x &&&+\dfrac{1}{5}z &= \dfrac{8}{5} \\ &&y &&\dfrac{2}{5}z&= \dfrac{4}{5} \\ && && \dfrac{1}{5}z &= \dfrac{3}{5} \end{align}\]
7. We multiply row three by 5
\[\begin{align} &x & &&+\dfrac{1}{5}z &= \dfrac{8}{5} \\ &&y &&\dfrac{2}{5}z&= \dfrac{4}{5} \\ & & &&z &= 3 \end{align}\]
8. We replace row 1 with row 1  \(\dfrac{1}{5}\) row 1 and replace row two with row three + \(\dfrac{2}{5}\) row three.
\[\begin{align} &x & && &= 1 \\ & &y && &= 2 \\ & & && z &= 3 \end{align}\]
3. Exercises
Exercise
\[\begin{align} &4x &3y &&z & = 0 \\ &x &3y &&+2z & = 7 \\ &3x &9y &&z & = 2 \end{align}\]
Exercise
\[\begin{align} &3x &2y &&+z & = 3 \\ &4x &+y &&& = 1 \\ & &11y &&4z & = 9 \end{align}\]
4. Application
Example
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
\(x = \text{ the number of servings of OJ}\)
\(y = \text{ the number of servings of Cereal}\)
\(z = \text{ the number of servings of eggs}\)
Then
\[\begin{align} &&10y &&+20z & = 30 \\ &20x &+15y && & = 30 \\ &100x &+120y &&+100z & = 300 \end{align}\]
We conclude that the breakfast should consist of 1.5 of a serving of OJ, no cereal, and 1.5 servings of eggs.
1.
\[\begin{align} &20x &+15y && & = 30 \\ &&10y &&+20z & = 30 \\ &100x &+120y &&+100z & = 300 \end{align}\]
2.
\[\begin{align} &x &+\dfrac{3}{4}y && & = \dfrac{3}{2} \\ &&10y &&+20z & = 30 \\ &100x &+120y &&+100z & = 300 \end{align}\]
3.
\[\begin{align} &x &+\dfrac{3}{4}y && & = \dfrac{3}{2} \\ &&10y &&+20z & = 30 \\ &&45y &&+100z & = 150 \end{align}\]
4. Already Done
5.
\[\begin{align} &x &+\dfrac{3}{4}y && & = \dfrac{3}{2} \\ &&y &&+2z & = 3 \\ &&45y &&+100z & = 150 \end{align}\]
6 .
\[\begin{align} &x & && \dfrac{3}{2}z& = \dfrac{3}{4} \\ &&y &&+2z & = 3 \\ && &&10z & = 15 \end{align}\]
7.
\[\begin{align} &x & && \dfrac{3}{2}z& = \dfrac{3}{4} \\ &&y &&+2z & = 3 \\ && &&z & = \dfrac{3}{2} \end{align}\]
8.
\[\begin{align} &x & && & = \dfrac{3}{2} \\ &&y && & = 0 \\ && &&z & = \dfrac{3}{2} \end{align}\]
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.