Elementary operations

An elementary operation is a single, reversible row (or column) modification that we can apply to a matrix. Three types are allowed, and chaining them is the workhorse of linear algebra: they reduce matrices to simpler forms, solve linear systems, and compute inverses. This subtopic introduces the three types of operations, gives examples of their use, and prepares the way for the inverse-by-row-operations method in the next subtopic.

The three elementary row operations

For a matrix $A$:

1. Swap two rows. $R_i \leftrightarrow R_j$ swaps rows $i$ and $j$.
2. Scale a row. $R_i \to k R_i$ multiplies row $i$ by a nonzero scalar $k$.
3. Add a multiple of one row to another. $R_i \to R_i + k R_j$ where $i \neq j$.

Each operation is reversible by another operation of the same type. The three corresponding elementary matrices are obtained by applying the operation to the identity matrix.

Column operations

The three types of column operations $C_i \leftrightarrow C_j$, $C_i \to k C_i$, $C_i \to C_i + k C_j$ are defined identically. Applying a row operation to $A$ is the same as multiplying $A$ on the left by the corresponding elementary matrix; column operations correspond to multiplication on the right.

Row-equivalent matrices

Two matrices $A$ and $B$ are row-equivalent if one can be transformed into the other by a finite sequence of elementary row operations. We write $A \sim B$.

If $A \sim B$, the systems $AX = O$ and $BX = O$ have the same solutions. This is what makes row operations useful for solving systems.

Reduced row echelon form (RREF)

A matrix is in row echelon form if

• All zero rows are at the bottom.
• The leading entry (first nonzero entry) of each nonzero row is strictly to the right of the leading entry above.

It is in reduced row echelon form if additionally

• Each leading entry is $1$.
• The leading entry is the only nonzero entry in its column.

Every matrix is row-equivalent to a unique reduced row echelon form. This is the goal of Gaussian elimination.

Worked example: bringing a matrix to RREF

Let $A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 0 \end{pmatrix}$.

Step 1. $R_2 \to R_2 - 2 R_1$, $R_3 \to R_3 - R_1$:

$\begin{pmatrix} 1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -1 & -1 \end{pmatrix}$.

Step 2. $R_2 \to -R_2$:

$\begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & -1 & -1 \end{pmatrix}$.

Step 3. $R_3 \to R_3 + R_2$:

$\begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$.

Step 4. $R_1 \to R_1 - 2 R_2$:

$\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$.

This is the reduced row echelon form.

Application: solving linear systems

The system $\begin{cases} x + 2y + z = 4 \\ 2x + 3y + z = 5 \\ x + y = 1 \end{cases}$ has augmented matrix $\begin{pmatrix} 1 & 2 & 1 & | & 4 \\ 2 & 3 & 1 & | & 5 \\ 1 & 1 & 0 & | & 1 \end{pmatrix}$.

Row-reducing (similar to the example above), we reach $\begin{pmatrix} 1 & 0 & -1 & | & a \\ 0 & 1 & 1 & | & b \\ 0 & 0 & 0 & | & c \end{pmatrix}$ for some $a, b, c$. If $c = 0$, the system has infinitely many solutions parametrised by $z$. If $c \neq 0$, no solution.

Elementary matrices

Each elementary operation corresponds to multiplication by an elementary matrix. For $n = 2$:

• $E_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ swaps rows when multiplied on the left.
• $E_2 = \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$ scales row $1$.
• $E_3 = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ adds $k$ times row $2$ to row $1$.

Elementary matrices are always invertible.

Worked examples

Example 1. Apply $R_2 \to R_2 - 3 R_1$ to $A = \begin{pmatrix} 1 & 2 \\ 4 & 5 \end{pmatrix}$.

New matrix: $\begin{pmatrix} 1 & 2 \\ 1 & -1 \end{pmatrix}$.

Example 2. Reduce $\begin{pmatrix} 2 & 4 \\ 3 & 6 \end{pmatrix}$ to row echelon form.

$R_1 \to \tfrac{1}{2}R_1$: $\begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix}$. $R_2 \to R_2 - 3 R_1$: $\begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}$. The matrix has rank $1$.

Example 3. Find the rank of $\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix}$.

All rows are multiples of $(1, 2, 3)$. Row-reducing gives only one nonzero row. Rank $= 1$.

Example 4. Determine $k$ such that $\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & k \end{pmatrix}$ has rank $2$.

$R_2 \to R_2 - 2R_1$ gives a zero row. $R_3 \to R_3 - R_1$ gives $\begin{pmatrix} 0 & -1 & k - 3 \end{pmatrix}$. Rank $= 2$ iff this row is nonzero , always. (Rank is $2$ regardless of $k$.)

Example 5. Solve by row-reduction: $\begin{cases} x + y = 3 \\ 2x - y = 0 \end{cases}$.

Augmented: $\begin{pmatrix} 1 & 1 & | & 3 \\ 2 & -1 & | & 0 \end{pmatrix}$. $R_2 \to R_2 - 2R_1$: $\begin{pmatrix} 1 & 1 & | & 3 \\ 0 & -3 & | & -6 \end{pmatrix}$. Back-substitute: $y = 2, x = 1$.

Example 6. Show that the elementary matrix $E = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ when applied to $A$ on the left performs $R_1 \to R_1 + 2 R_2$.

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $EA = \begin{pmatrix} a + 2c & b + 2d \\ c & d \end{pmatrix}$. $\checkmark$

Try it yourself

1. Apply $R_2 \to R_2 + R_1$ to $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.
2. Reduce $\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$ to RREF.
3. Reduce $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ to RREF.
4. Find the rank of $\begin{pmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 3 & 6 & 3 \end{pmatrix}$.
5. Solve by elimination: $\begin{cases} x + 2y = 3 \\ 3x + y = 4 \end{cases}$.
6. Write down the elementary matrix that swaps rows $1$ and $3$ in a $3 \times 3$ matrix.
7. Write down the elementary matrix that scales row $2$ by $5$ in a $3 \times 3$ matrix.
8. Apply $R_3 \to R_3 - R_2 - R_1$ to $\begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}$.
9. Show that two elementary matrices in succession need not commute.
10. Reduce $\begin{pmatrix} 1 & -2 & 3 \\ 2 & -3 & 4 \\ 3 & -4 & 5 \end{pmatrix}$ to RREF and state its rank.
11. Find a row-echelon form of $\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 3 \end{pmatrix}$.
12. Show that two row-equivalent matrices have the same rank.
13. Solve $\begin{cases} x + y + z = 1 \\ x + 2y + 3z = 2 \\ 2x + 3y + 4z = 3 \end{cases}$ by row reduction.
14. Determine if $\begin{cases} x + y = 1 \\ 2x + 2y = 3 \end{cases}$ is consistent.

Pitfalls / Tricks

• Row scaling requires a nonzero scalar , multiplying by $0$ would destroy information.
• The order of operations matters; chain them carefully.
• When solving, keep the augmented matrix's bar consistent , entries to the left of $|$ are coefficients, to the right are constants.
• A consistent system has at least one solution; an inconsistent system has none (it reduces to a row $0 \cdots 0 | c$ with $c \neq 0$).

The next subtopic builds the inverse of a matrix by precisely these operations.