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1.1 Introduction to Systems of Linear Equations: Class Notes

Definition

#14

In each part, use parametric equations to describe the solution set of the linear equation.

a. $x+10y=2$
b. $x_1+3x_2-12x_3=3$
c. $4x_1+2x_2+3x_3+x_4=20$
d. $v+w+x-5y+7z=0$

Solution

a. Solve for $x$:

$$x=-10y+2$$

The variable $y$ can be anything, so make it the parameter: let $t=y$. Then

b. Solve for $x_1$:

$$x_1=-3x_2+12x_3+3$$

Let $s=x_2$ and $t=x_3$. Then

c. Solve for $x_1$ (divide every term by 4):

$${\displaystyle x_1=-\frac{1}{2}{x}_2-\frac{3}{4}{x}_3-\frac{1}{4}{x}_4+5}$$

Let $r=x_2$, $s=x_3$, and $t=x_4$. Then

Definition

Definitions

Definition

Each column before the bar holds the coefficients of one unknown: the first column goes with $x_1$, the second with $x_2$, and so on up to $x_n$. The last column holds the constants.

#16

Each linear system has infinitely many solutions. Use parametric equations to describe its solution set.

a.

$$\begin{array}{cc}\hfill {\displaystyle 6x_1+2x_2}& {\displaystyle =-8}\hfill \\ \hfill {\displaystyle 3x_1+x_2}& {\displaystyle =-4}\hfill \end{array}$$

b.

$$\begin{array}{cc}\hfill {\displaystyle 2x-y+2z}& {\displaystyle =-4}\hfill \\ \hfill {\displaystyle 6x-3y+6z}& {\displaystyle =-12}\hfill \\ \hfill {\displaystyle -4x+2y-4z}& {\displaystyle =8}\hfill \end{array}$$

Solution

a. With elimination, subtracting twice the second equation from the first, $E_1-2E_2$, wipes out everything: adding $6x_1+2x_2=-8$ and $-6x_1-2x_2=8$ gives $0=0$.

The same step on the augmented matrix is $R_2\to R_1-2R_2$:

$$\left[\begin{array}{ccc}\hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -8}\hfill \\ \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -4}\hfill \end{array}\right]⟶\left[\begin{array}{ccc}\hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -8}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

Row 1 is the equation $6x_1+2x_2=-8$, so

$${\displaystyle x_1=-\frac{1}{3}{x}_2-\frac{4}{3}}$$

Let $t=x_2$. Then

b. The augmented matrix is

$$\left[\begin{array}{cccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -4}\hfill \\ \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle -12}\hfill \\ \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle 8}\hfill \end{array}\right]$$

Use $R_2\to R_2-3R_1$ and $R_3\to R_3+2R_1$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -4}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

Only row 1 is left: $2x-y+2z=-4$. Solve for $x$:

$${\displaystyle x=\frac{1}{2}y-z-2}$$

Let $s=y$ and $t=z$. Then

Definitions

Definitions

#19

Find all values of $k$ for which the given augmented matrix corresponds to a consistent linear system.

a. $\left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle -4}\hfill \\ \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 8}\hfill & \hfill {\displaystyle 2}\hfill \end{array}\right]$  b. $\left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 8}\hfill & \hfill {\displaystyle -4}\hfill \end{array}\right]$

Solution

a. Use $R_2\to R_2-4R_1$, adding $-4$ times row 1 to row 2:

$$\left[\begin{array}{ccc}4& 8& 2\end{array}\right]+\left[\begin{array}{ccc}-4& -4k& 16\end{array}\right]=\left[\begin{array}{ccc}0& 8-4k& 18\end{array}\right]$$

$$\left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle -4}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 8-4k}\hfill & \hfill {\displaystyle 18}\hfill \end{array}\right]$$

Row 2 is the equation $(8-4k)y=18$. If $8-4k=0$, that is, $k=2$, it says $0=18$, which has no solution. For any other $k$ we can divide by $8-4k$, so the system is consistent.

b. Use $R_2\to R_2-4R_1$ again:

$$\left[\begin{array}{ccc}4& 8& -4\end{array}\right]+\left[\begin{array}{ccc}-4& -4k& 4\end{array}\right]=\left[\begin{array}{ccc}0& 8-4k& 0\end{array}\right]$$

Then ${\displaystyle R_2\to \frac{1}{4}{R}_2}$:

$$\left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2-k}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

Row 2 is the equation $(2-k)y=0$, which $y=0$ always satisfies. So this is consistent for every $k$.

Example

A 3-7-9 diet calls for 3 units of fat, 7 units of protein, and 9 units of carbs in each meal. Suppose an individual has three possible foods to choose from to meet these requirements. Each ounce of the food contains

Food 1: 3 units of fat, 4 units of protein, 1 unit of carbs
Food 2: 2 units of fat, 5 units of protein, 3 units of carbs
Food 3: 4 units of fat, 1 unit of protein, 2 units of carbs

Let $x$, $y$, and $z$ denote the number of ounces of the first, second, and third foods that a person will consume at the main meal. Find a linear system in $x$, $y$, and $z$ whose solution tells how many ounces of each food must be consumed to meet the diet requirements.

Solution

So $x$ goes with Food 1, $y$ with Food 2, and $z$ with Food 3. Each nutrient gives one equation: the units from each food, times the ounces of that food, must add up to the requirement.

Fat: $3x+2y+4z=3$
Protein: $4x+5y+z=7$
Carbs: $x+3y+2z=9$