\begin{align*} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\ &\vdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n &= b_m \end{align*}

• $\begin{aligned} 2x + 3y &= 1 \\ 3x - y &= 2 \end{aligned}$
• $\begin{aligned} ax + by &= r \\ cx + dy &= s \end{aligned}$

• $ \begin{aligned} x_1 + x_2 + x_3 &= 3 \\ x_1 + x_2 - x_3 &= 1 \\ x_1 - x_2 + x_3 &= 1 \end{aligned}$

	C	H	O
LHS	\(x_1\)	\(2x_2\)	\(2x_1+x_2\)
RHS	\(6x_4\)	\(12x_4\)	\(2x_3+6x_4\)

\[ \begin{aligned} x_1 &= 6x_4 \\ 2x_2 &= 12x_4 \\ 2x_1 + x_2 &= 2x_3 + 6x_4 \end{aligned} \]

$\rightarrow$

\[ \begin{aligned} x_1 - 6x_4 &= 0 \\ 2x_2 - 12x_4 &= 0 \\ 2x_1 + x_2 - 2x_3 - 6x_4 &= 0 \end{aligned} \]

We represent the system

using the augmented matrix

and solve this system using row operations

There are three allowed row operations:

1. Interchanging two rows, i.e.

   $ \left( \begin{array}{rrr|r} 0 & -4 & 8 & 4 \\ 1 & -1 & 1& 0 \\ 2 & 2 & 3 & 4 \end{array} \right) $ $ \begin{array}{rrr|r} R1 \\ R2 \\ R3 \end{array} $ $\quad\rightarrow\quad$ $ \left( \begin{array}{rrr|r} 1 & -1 & 1 & 0 \\ 0 & -4 & 8 & 4 \\ 2 & 2 & 3 & 4 \end{array} \right) $ $ \begin{array}{rrr|r} \color{red}{R2} \\ \color{red}{R1} \\ R3 \end{array} \qquad \qquad \qquad $

2. Rescaling one row, i.e.

   $ \left( \begin{array}{rrr|r} 1 & -1 & 1& 0 \\ 0 & -4 & 8 & 4 \\ 2 & 2 & 3 & 4 \end{array} \right) $ $ \begin{array}{rrr|r} R1a \\ R2a \\ R3 \end{array} $ $\quad\rightarrow\quad$ $ \left( \begin{array}{rrr|r} 1 & -1 & 1 & 0 \\ 0 & 1 & -2 & -1 \\ 2 & 2 & 3 & 4 \end{array} \right) $ $ \begin{array}{r} R1a \\ \color{red}{-\frac{1}{4}R2a} \\ R3 \end{array}\qquad \qquad $

3. Adding a multiple of one row to another row, i.e.

   $ \left( \begin{array}{rrr|r} 1 & -1 & 1 & 0 \\ 0 & 1 & -2 & -1 \\ 2 & 2 & 3 & 4 \end{array} \right) $ $ \begin{array}{r} R1a \\ R2b \\ R3 \end{array} $ $\quad\rightarrow\quad$ $ \left( \begin{array}{rrr|r} 1 & -1 & 1 & 0 \\ 0 & 1 & -2 & -1 \\ 0 & 4 & 1 & 4 \end{array} \right) $ $ \begin{array}{l} R1a \\ R2b \\ \color{red}{R3 - 2 R1a} \end{array}\qquad \qquad $

These operations are allowed because they produce a new set of equations with the same solution.

1. Interchanging two rows: Hopefully obvious 😃

2. Rescaling one row:

3.Adding a multiple of one row to another row:

A square matrix is in upper triangular form if and only if (iff) $a_{ij} = 0$ whenever $i > j.$

That is, whenever all entries below the main diagonal are zero.

Example:

Our objective using row reduction on an augmented matrix:

With equations in this form, we obtain the solution by back-substitution.

The bottom line gives: $\;x_n = r_n.$

The line above then gives:

and so on...

Hence $x_2=1$, and back-substitution in $x_1 - x_2 = 0\;$ gives $\;x_1 -1 = 0$ $\; \Ra\; x_1 = 1.$

Therefore $x_1 = 1$ and $x_2=1.$

Hence $x_2=1$, and back-substitution in $x_1+ \frac{1}{2}x_2 =\frac{3}{2}\;$ gives $\;x_1 +\frac{1}{2} = \frac{3}{2}$ $\; \Ra\; x_1 = 1.$

Therefore $x_1 = 1$ and $x_2=1.$

We found that

It's easy and wise to check your answer...

We found that

						
>> A = [3 2 -1 -2; 3 1 -1 -5; 3 2 1 2]
  

						
A =
3 2 -1 -2
3 1 -1 -5
3 2  1  2
  

						
>> rref(A)
  

						
ans =
1 0 0 -2
0 1 0  3
0 0 1  2
  

Definition (Row Echelon Form)

A matrix is in row echelon form if

Let's try some simple examples...

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Examples:

From the second equation: $\color{blue}{x_2}=-2\color{red}{x_3}.$

Therefore from the first equation: $\color{blue}{x_1}=1-\color{red}{x_3}- \color{blue}{x_2}$ $= 1 + \color{red}{x_3}.$

So $\color{blue}{x_2}=-2\color{red}{x_3}$ and $\color{blue}{x_1}=1+\color{red}{x_3}.$

That is, once $\color{red}{x_3}$ is given, $\color{blue}{x_1}$ and $\color{blue}{x_2}$ are uniquely determined.

Examples:

From the second and third equations: $\color{blue}{x_3}=-\color{red}{x_4} - 2\color{blue}{x_5}$ $=-\color{red}{x_4} + 6.$

Therefore $\,\color{blue}{x_1}= -2-\color{red}{x_2},\, \color{blue}{x_3} = -\color{red}{x_4}+6, \,$ and $\,\color{blue}{x_5}=-3.$

Definition (Reduced Row Echelon Form)

A matrix is in reduced row echelon form if

						
>> rref([1 1 1 1 1 1; 0 0 1 1 2 0; 0 0 0 0 1 -3])
  

						
ans =
1 1 0 0 0 -2
0 0 1 1 0  6
0 0 0 0 1 -3
  

What could the period of a pendulum depend on?

• Pendulum length $l$, measured in $\text{m}$.
• Pendulum mass $\text{m}$, measured in $\text{kg}$.
• Strength of the earth's gravitational field, measured as the acceleration due to gravity, $g$, in $\text{m}/\text{s}^2$.

We assume that $T \propto l^a\text{m}^bg^c,$ where $T$ is the period of the pendulum in seconds.

Comparing units, we get

Equating the exponents:

Row reduction gives:

So $a = 1/ 2, \, b = 0$ and $c =-1/ 2,$ and the formula for the period becomes

Let's include a fourth quantity, and see what happens.

• Pendulum length $l$, measured in $\text{m}$.
• Pendulum mass $\text{m}$, measured in $\text{kg}$.
• Strength of the earth's gravitational field, measured as the acceleration due to gravity, $g$, in $\text{m}/\text{s}^2$.
• Initial pendulum height $h_0$ (above equilibrium point)

Now we assume that $T \propto l^a\text{m}^bg^c h_0^d.$

Comparing units, we get

Equating the exponents:

Row reduction gives:

So $a = 1/ 2-d, \, b = 0$ and $c =-1/ 2,$ with free variable $d$. Thus

Main points about row reduction

To solve systems of linear equations

1. Reduce the system to row echelon form
2. Interpret the results in terms of row-types
   • Rows of the form $(0\; 0\; \cdots 0 \; | \;1)$ $\Ra$ inconsistent system
   • As many non-zero rows as variables $\Ra$ unique solution
   • Fewer non-zero rows than variables $\Ra$ family of solutions $\qquad\qquad\qquad$