Solutions for Systems of Linear Equations & elementary operations

Linear equations:

$ax=b$ if a and b are 0, there are infinite solutions if a is 0 and b isn’t, there is no solution. if a and b are not zero, $x=b/a$ (this is called a “unique solution”)

ex. $2x-3y=5$ (same as) $y=2/3x-5/3$

Linear equations can be more complex, like this.

$a_1{x}_1+a_2{x}_2+...+a_n{x}_n=b$

That’s called “a linear equation in n variables” The a’s are “real numbers called the coefficients of $x_1$ etc respectively.

b, in this content is “the constant term”

Example: $2x_1-3x_2+5x_3=7$ coefficients are 2, -3, 5 constant term: 7

System of linear equations

A system of linear equations is a set of 1 or more (m) linear equations. m from the system doesn’t have to be the same number as n above.

The solution to the linear equation $a_1{x}_1+a_2{x}_2+...+a_n{x}_n=b$ is $s_1,s_2,...s_n$ provided $a_1{s}_1+a_2{s}_2+...+a_n{s}_n=b$

Example: x=-2, y=5, z=0 and x= 0, y=4, z=-1

are both solutions to the system

$x+y+z=3$ $2x+y+3z=1$

Systems without a solution are called an “inconsistent system”

$$\{\begin{array}{l}x+y=7\\ \\ x+y=-3\end{array}$$

These lines are parallel and never intersect. Also, you can’t add two numers and get 7 and -3. Therefore, there is no solution and it’s inconsistent.

Example: Show that for arbitray values of s & t:

$x_1=t-s+1$ $x_2=t+s+2$ $x_3=s$ $x_4=t$

is a solution to the system

$$\{\begin{array}{l}{x}_1-2x_2+3x_3+x_4=-3\\ \\ 2x_1-x_2+3x_3-x_4=0\end{array}$$

Solution: Make substitutions

$$\begin{array}{r}{x}_1-2x_2+3x_3+x_4=-3\\ \\ (t-s+1)-2(t+s+2)+3s+t=-3\\ \\ t-s+1-2t-2s-4+3s+t=-3\\ \\ t-2t+t+3s-2s-s+1-4=-3\end{array}$$ $$\begin{array}{r}2(t-s+1)-(t+s+2)+3s-t=0\\ \\ 2t-2s+2-t-s-2+3s-t=0\\ \\ 2t-t-t-2s-s+3s+2-2=0\\ \\ 0=0\end{array}$$

The name for those quantities used for subtitution are called “parameters” of the system. If it’s generified, it’s said to e given in the “parametric form” and is called the “general solution” to the system.

Example:

$3x-y+2z=6$

$$\begin{array}{r}y=3x+2z-6\\ \\ x=s\\ \\ z=t\end{array}$$

If you plot the two lines.. any intersection would be a unique solution. If they’re parallel, there are no solutions. If they’re the same line, there are infinitely many solutions.

Consider:

$$\{\begin{array}{l}2x_1+2x_2-3_x+x_4=-1\\ \\ 2x_1-x_3+2x_4=0\\ \\ 3x_1+x_2+2x_3+5x_4=2\\ \end{array}$$

The array of numbers

$$\left[\begin{array}{cccc}3& 2& -1& 1|1\\ & & & \\ 2& 0& -1& 2|0\\ & & & \\ 3& 1& 2& 5|2\\ & & & \end{array}\right]$$

are called the “augmented matrix of the system”

$$\left(\begin{array}{cccc}3& 2& -1& 1\\ & & & \\ 2& 0& -1& 2\\ & & & \\ 3& 1& 2& 5\\ & & & \end{array}\right)$$

^ This is the coefficient matrix of the system

$$\left(\begin{array}{c}-1\\ \\ 0\\ \\ 2\\ \end{array}\right)$$

^ Called the constant matrix of the system

Two systems are equivalent if they have the same set of solutions. The goal is to have a system that’s easy to solve.

EX: Solve the system

$$\{\begin{array}{l}x+2y=-2\\ \\ 2x+y=7\\ \end{array}$$ $$\left[\begin{array}{cc}1& 2|-2\\ & \\ 2& 1|7\end{array}\right]$$

first, subtract 2x the first equation from the second.

$$\{\begin{array}{l}x+2y=7\\ \\ 2x+4y=-4\end{array}$$

0-3y=11

$$\{\begin{array}{l}x+2y=7\\ \\ -3y=11\end{array}$$ $$\left[\begin{array}{cc}1& 2|7\\ & \\ 0& -3|11\end{array}\right]$$

Which is equlvant to

$$\{\begin{array}{l}x+2y=7\\ \\ y=-\frac{11}{3}\end{array}$$ $$\left[\begin{array}{cc}1& 2|7\\ & \\ 0& 1|-\frac{11}{3}\end{array}\right]$$

Finally, subtract 2x the first again to get another equivalent

$$\{\begin{array}{l}2x+y=7\\ \\ y=-\frac{11}{3}\end{array}$$

so…

we eventually get to:

x+0 = -2+22/3

which simplifies to x=16/3

and we end up with x = 16/3 y = -11/3

The “doubling” isn’t a general rule. It was something useful for this formula to start to remove the variables and simplify.

”elementary row operations”

Elementary row operation rules

The elementary row operations are:

1. Interachange: [interchange two rows]
2. Scaling: [multiply one row by a non-zero number]
3. Replacement: [add a multiple of one row to a different row]

Theorem

Suppose that the sequence of elementary row operations is performed on a system of linear equations, the resulting system must be [equivalent]. This means they have [the same set of solutions] as the original system.

Example: Find all solutions to this:

$$\{\begin{array}{l}3x+4y+z=1\\ \\ 2x+3y=0\\ \\ 4x+3y-z=-2\end{array}$$ $$\left(\begin{array}{ccc}3& 4& 1|1\\ & & \\ 2& 3& 0|0\\ & & \\ 4& 3& -1|-2\\ & & \end{array}\right)$$

The goal is to get a matrix that looks like:

$$\left(\begin{array}{ccc}1& 0& 0\\ & & \\ 0& 1& 0\\ & & \\ 0& 0& 1\\ & & \end{array}\right)$$

So to get that first row of 1,0,0, we can subtract row 2 from row 1.

$$\left[\begin{array}{ccc}3& 4& 1|1\\ & & \\ 2& 3& 0|0\\ & & \\ 4& 3& -1|-2\\ & & \end{array}\right]$$

r1=r1-r2

$$\left[\begin{array}{ccc}1& 1& 1|1\\ & & \\ 2& 3& 0|0\\ & & \\ 4& 3& -1|-2\\ & & \end{array}\right]$$

r2 = r2 - 2*r1

$$\left[\begin{array}{ccc}1& 1& 1|1\\ & & \\ 0& 1& -2|-2\\ & & \\ 4& 3& -1|-2\\ & & \end{array}\right]$$

r3: r3 - 4*r1

$$\left[\begin{array}{ccc}1& 1& 1|1\\ & & \\ 0& 1& -2|-2\\ & & \\ 0& -1& -5|-6\\ & & \end{array}\right]$$

r1: r1-r2

$$\left[\begin{array}{ccc}1& 0& 3|3\\ & & \\ 0& 1& -2|-2\\ & & \\ 0& -1& -5|-6\\ & & \end{array}\right]$$

r3: r3+r2

$$\left[\begin{array}{ccc}1& 0& 3|3\\ & & \\ 0& 1& -2|-2\\ & & \\ 0& 0& -7|-8\\ & & \end{array}\right]$$

r3 = -1/7 r3

$$\left[\begin{array}{ccc}1& 0& 3|3\\ & & \\ 0& 1& -2|-2\\ & & \\ 0& 0& 1|8/7\\ & & \end{array}\right]$$

r1 = r1 - 3r3

$$\left[\begin{array}{ccc}1& 0& 0|-3/7\\ & & \\ 0& 1& -2|-2\\ & & \\ 0& 0& 1|8/7\\ & & \end{array}\right]$$

r2 = r2 + r3*2

$$\left[\begin{array}{ccc}1& 0& 0|-3/7\\ & & \\ 0& 1& 0|2/7\\ & & \\ 0& 0& 1|8/7\\ & & \end{array}\right]$$

The matrix with the 1’s on the diagonal is called an “identity matrix”.

Definitions

Parts of a linear equation

$ax=b$

Given $ax=b$,

In this example, $b$ is called the [constant term]. In this example, $a$ is called the [coefficient].

Solutions

No solution

Systems without a solution are called [an “inconsistent system”]

Ex:

$$\{\begin{array}{l}x+y=7\\ \\ x+y=-3\end{array}$$

Substitutions

The name for the quantities used for subtitution are called [“parameters”] of the system.

If the quantities used for subtitution are generified, it’s said to be given in the [“parametric form”] and is called the [“general solution”] to the system.

Graphing linear equations

If you plot the two lines.. any intersection would be a [unique solution]. If they’re parallel, there are [no solutions]. If they’re the same line, there are [infinitely many solutions].

Matrices

Consider:

$$\{\begin{array}{l}2x_1+2x_2-3_x+x_4=-1\\ \\ 2x_1-x_3+2x_4=0\\ \\ 3x_1+x_2+2x_3+5x_4=2\\ \end{array}$$

The array of numbers

$$\left[\begin{array}{cccc}3& 2& -1& 1|1\\ & & & \\ 2& 0& -1& 2|0\\ & & & \\ 3& 1& 2& 5|2\\ & & & \end{array}\right]$$

are called [the “augmented matrix of the system”]

Matrices

Consider:

$$\{\begin{array}{l}2x_1+2x_2-3_x+x_4=-1\\ \\ 2x_1-x_3+2x_4=0\\ \\ 3x_1+x_2+2x_3+5x_4=2\\ \end{array}$$

The array of numbers

$$\left[\begin{array}{cccc}3& 2& -1& 1\\ & & & \\ 2& 0& -1& 2\\ & & & \\ 3& 1& 2& 5\\ & & & \end{array}\right]$$

are called [the “coefficient matrix of the system”]

Matrices

Consider:

$$\{\begin{array}{l}2x_1+2x_2-3_x+x_4=-1\\ \\ 2x_1-x_3+2x_4=0\\ \\ 3x_1+x_2+2x_3+5x_4=2\\ \end{array}$$

The array of numbers

$$\left[\begin{array}{c}-1\\ \\ 0\\ \\ 2\\ \end{array}\right]$$

are called [the “constant matrix of the system”]

Back Substitution

[Back substitution] is the method of solving for one variable and plugging it in to another equation to determine the other variable.

Homework