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# 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:

$cR_j + R_i \rightarrow R_i$

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

1. We interchange rows so that the top left corner is nonzero (preferably a 1).
2. We multiply row 1 by the appropriate number to make the top left corner a 1.
3. 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.
4. We interchange the two bottom rows so that the middle coefficient nonzero.
5. We multiply row two appropriately so that the middle number is a 1.
6. We use row two to make the middle top and bottom zero.
7. Make the bottom right 0.
8. Use row three to make the top and middle rights 0.

\begin{align} &x & && &= 1 \\ & &y && &= 2 \\ & & && z &= 3 \end{align}

### 3. Exercises

\begin{align} &4x &-3y &&-z & = 0 \\ &x &-3y &&+2z & = 7 \\ &3x &9y &&-z & = -2 \end{align}

\begin{align} &3x &-2y &&+z & = 3 \\ &4x &+y &&& = 1 \\ & &11y &&-4z & = -9 \end{align}

## 4. 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

$$x = \text{ the number of servings of OJ}$$
$$y = \text{ the number of servings of Cereal}$$
$$z = \text{ the number of servings of eggs}$$

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}

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.