Echelon Form

Echelon form is a matrix form where the leading coefficient of each row is always to the right of the leading coefficient of the row above it. The leading coefficient of a row is the first non-zero element in that row.

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Lets start with an example:

$$\begin{array}{rl}{x}_1-4x_2+3x_3+x_4& =6\\ 4x_3-x_5& =-3\\ x_4+x_5& =-1\end{array}$$

Definition

The lead variable of a linear equation is the first variable with a nonzero coefficient. Equations of the form $0=b$ have no lead variable.

Definition

A system of linear equations is in Echelon Form if the lead variable of each equation is to the right of the lead variable of all previous equations and all equations $0=b$ are at the end of the last.

Applying it to this example, we note that the lead variable would be $x_1$

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Back Substitution

Example

$$\begin{array}{rl}-2x_1-2x_2& =0\\ -x_2+x_3-3x_4& =1\\ 3x_4-x_5& =1\end{array}$$

Let’s start with $x_5$ :

• is it the lead variable of the last equation? no $\to$ therefore it’s a “free” variable.
  • Instead of choosing a value for $x_5$, we are going to pick a Parameter $P$

Now let’s do $x_4$:

• $x_4$ is the lead variable of its line, therefore:

$$x_4:\frac{1}{2}+\frac{1}{3}PLEAD!!!$$

Now $x_3$:

• $x_3$ is NOT the lead variable, therefore it is “free” and we assign another Parameter $q$

Now $x_2$ :

• $x_2$ IS a lead variable, therefore:

$$x_2:-2-p+qLEAD!!!!$$

And now finally $x_1$:

• $x_1$ IS a lead variable, we need to solve for this variable using the information we have gathered thus far.

$$x_1:2+p-qLEAD!!!!$$

We are left with a set of values for the x variants:

$$\begin{array}{rlrlrlr}{x}_5& :& & P& & & FREE\\ x_4& :& & \frac{1}{2}+\frac{1}{3}p& & & LEAD\\ x_3& :& & Q& & & FREE\\ x_2& :& & -2-p+q& & & LEAD\\ x_1& :& & 2+p-q& & & LEAD\end{array}$$

And thus, we get the following general solution:

Solution

$$\left(2+p-q,-2-p+q,q,\frac{1}{3}+\frac{1}{3}p,p\right)$$

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More Examples

Example

$$\begin{array}{rl}x+y+z& =0\\ x+2y+3z& =8\\ 2x+3y+4z& =13\end{array}$$

$$-R_1+R_2,-2R_1+R_3$$ $$\begin{array}{rl}x+y+z& =6\\ y+2z& =2\\ y+2z& =1\end{array}$$

Solving:

$$-R_2+R_3$$ $$\begin{array}{rl}x+y+z& =6\\ y+2z& =2\\ 0& =-1\end{array}$$

Now you might notice that this is an impossible solution, and therefore this system of equations cannot be solved.

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General Strategy

We want to be able to easily determine if a system of equations is impossible long before we go through all the effort of solving it just to realize we can’t. How can we do this?

$$SYSTEM\to EchelonSystem$$

	SOME	NONE
SOME	No solution	$\infty$ solutions
NONE	No solution	$1$ solution