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# 03 Augmented matrix, pivots, EF, REF, RREF, and row equivalence.
Ok so using the elementary row operations to solve a system of linear equations is pretty neat. But it seems like there are so many routes to do it. This is true, however, we wonder if we can be even more systematic when approaching these.
First let us simplify our notation. Notice in solving a linear system, really only the coefficients are the ones that are changing. So let us keep track of the coefficients instead, using an **augmented matrix** for the system.
For example, consider the linear system $$
\left\{
\begin{array}{}
3 x & + & 6y & + & 9 z & = 1 \\
x & & & - & z & = 10 \\
2x & - & y & + & 3z & =11
\end{array}
\right.
$$we shall represent it as an **augmented matrix**: $$
\begin{array}{}
x \ \ \ \ \ \ y\ \ \ \ \ \ z\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\begin{bmatrix}
3 & 6 & 9 & \vdots & 1 \\
1 & 0 & -1 & \vdots & 10 \\
2 & -1 & 3 & \vdots & 11
\end{bmatrix}
\end{array}
$$It is called an **augmented matrix** because to the left of the dotted lines is the **coefficient matrix** of the system: $$\begin{bmatrix}
3 & 6 & 9 \\
1 & 0 & -1 \\
2 & -1 & 3
\end{bmatrix},
$$while the **augmented column** records the constants that are on the right side of the equal sign across from the variables: $$
\begin{bmatrix}
1 \\
10 \\
11
\end{bmatrix}.
$$**Important Remark.** Notice, before you set this up, you should organize the same variables into the same columns, and **label the columns corresponding to the variable** to not make a mistake. Also, if a variable does not show up, then the coefficient of it is zero.
Now the goal is to produce an **echelon form**, which we formally define it now.
First we denote a **pivot position** to be the leading nonzero position in **each row** (leading meaning if we read from left to right).
A matrix (augmented or not) is said to be in **echelon form** (EF) if
(1) If there is a row of zero, it not above any rows with nonzero numbers in it. And,
(2) if there are any pivots, from top to bottom, the position of the pivots move strictly to the right.
This gives a "staircase pattern" that we call echelon form EF.
Let us make some further definitions.
A matrix is in **row echelon form** (REF) if
(1) It is already in echelon form (EF), and
(2) each pivot position has a value of 1.
A matrix is in **reduced row echelon form** (RREF) if
(1) It is already in row echelon form (REF), and
(2) each column with a pivot is all zero, except the pivot (which is a 1).
So we have the hierarchy $$
\text{RREF} \subset \text{REF} \subset \text{EF}.
$$
Some examples. We have
$$
\begin{bmatrix}
2 & 3 & 2 \\
0 & 1 & 0
\end{bmatrix} \text{ is EF, but not REF;}
$$ $$
\begin{bmatrix}
1 & 3 & 2 \\
0 & 1 & 0
\end{bmatrix} \text{ is REF, but not RREF;}
$$ $$
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0
\end{bmatrix} \text{ is RREF, hence also REF and EF.}
$$
Let us make some more terminology. If two matrices $A$ and $B$ can be transformed into the other via a sequence of elementary row operations, then we say $A$ and $B$ are **row equivalent**, and that they belong in the same **row equivalence class**. If $A$ and $B$ are row equivalent then we write $$
A \stackrel{\text{row}}{\sim} B .
$$
Some examples. Let us look at matrix $$
A=\begin{bmatrix}
1 & 1 & 2 \\
2 & 1 & 1
\end{bmatrix}.
$$
If we apply $(R_{2}) \to (R_{2}) + 3(R_{1})$ to the matrix $A$, we get $$
B = \begin{bmatrix}
1 & 1 & 2 \\
5 & 4 & 7
\end{bmatrix}.
$$ If we further apply $(R_{1}) \to 10 (R_{1})$ to $B$ we get $$
C = \begin{bmatrix}
10 & 10 & 20 \\
5 & 4 & 7
\end{bmatrix}
$$ Notice we get $B$ from $A$ by one elementary row operation, and we get $C$ from $A$ by two elementary row operations. Since they are all within a sequence of elementary row operations of each other, they are all row equivalent to each other. That is $$
A \stackrel{\text{row}}{\sim} B,\ \ \ B \stackrel{\text{row}}{\sim} C, \ \ \text{and}\ \ \ A \stackrel{\text{row}}{\sim} C.
$$ We can write it in short: $A \stackrel{\text{row}}{\sim} B \stackrel{\text{row}}{\sim} C$, and that $A,B,C$ all belong in the same row equivalence class.
Notice, just because $A \stackrel{\text{row}}{\sim} B$, it does not necessarily mean $A$ equals to $B$. In our example above, $A\neq B$, $B\neq C$, and $A\neq C$ !
So an interesting question arises: How do we know if two matrices are in the same row equivalence class? For example, it is not entirely clear whether there is a sequence of elementary row operations that takes the matrix $$
A=\begin{bmatrix}
1 & 1 & 2 \\
2 & 1 & 1
\end{bmatrix}
$$ to the matrix $$
B=\begin{bmatrix}
2 & 2 & 2 \\
1 & 2 & 1
\end{bmatrix}.
$$
To answer this, there is a nice theorem:
> **Theorem.**
> Matrices $A$ and $B$ are row equivalent if and only if $\text{RREF}(A)=\text{RREF}(B)$.
Here $\text{RREF}(M)$ means the matrix that we obtained by turning matrix $M$ into RREF via elementary row operations. This suggest that the RREF we obtained from a given matrix is unique, using elementary row operations. So we can use it as a fingerprint to identify which row equivalence class it belongs to.
In the example above, where $A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix}2&2&2\\1&2&1\end{bmatrix}$, we can show (you should try it!) that $$
\text{RREF}(A)=\begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & 3
\end{bmatrix}
$$and $$
\text{RREF}(B)=\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0
\end{bmatrix},
$$ which shows $\text{RREF}(A)\neq \text{RREF}(B)$. Hence $A$ and $B$ are **not** row equivalent! And that they are in different row equivalence class! In other words, there is no sequence of elementary row operations that you can ever do to $A$ and turn it into $B$!
Isn't this interesting!