Introduction

When solving a system of linear equations using an augmented matrix, we rely on three legal moves called elementary row operations. These moves let us simplify the matrix without changing the solution set of the system it represents.

Because you already know what an augmented matrix is, we will focus on:

• What each row operation does
• How it affects the augmented matrix
• Why it preserves the solution set
• Simple examples using small systems

These operations form the backbone of Gaussian elimination and reduced row‑echelon form.

Why Row Operations Matter

Elementary row operations preserve the meaning of the system:

• The left block (coefficients) and right block (constants) change in a controlled way.
• The system of equations represented by the augmented matrix remains equivalent.
• Row operations help us:
  • eliminate variables
  • create pivots
  • detect inconsistent or dependent systems

What Pivots Are

Before delving into the row operations, it might be useful to have a brief detour to define what we mean by pivot.

A pivot is a leading nonzero entry in a row of a matrix after you have used row operations to simplify it.
Pivots help you understand the structure of the system and guide the elimination process.

How to recognize a pivot

A pivot is:

• The first nonzero entry in a row (from left to right)
• Usually turned into 1 using a scaling operation
• Used to eliminate entries below (and sometimes above) it

In an augmented matrix, pivots always appear in the coefficient block, not the augmented column.

Example

Consider the matrix: $$\left( \begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 5 & 1 \end{array} \right)$$ The pivots are:

• Row 1 pivot: the $1$ in column 1
• Row 2 pivot: the $1$ in column 2
• Row 3 pivot: the $5$ in column 3

Each pivot is the first nonzero entry in its row.

The Three Legal Moves

1. Row Swapping ($R_i \leftrightarrow R_j$)

You may exchange two rows of the augmented matrix.

• This corresponds to swapping two equations.
• The solution set does not change.
• Often used to bring a nonzero pivot into position.

Example:
Swap rows 1 and 2: $$\left( \begin{array}{cc|c} 0 & 1 & 3 \\ 4 & 2 & 6 \end{array} \right) \;\longrightarrow\; \left( \begin{array}{cc|c} 4 & 2 & 6 \\ 0 & 1 & 3 \end{array} \right)$$

2. Row Scaling ($R_i \to kR_i$, $k \neq 0$)

You may multiply an entire row by a nonzero constant.

• This corresponds to multiplying an equation by a constant.
• The augmented part must be multiplied as well.
• Useful for creating leading 1s.

Example:
Multiply row 1 by $\tfrac12$: $$\left( \begin{array}{cc|c} 2 & 4 & 10 \\ 1 & -3 & 2 \end{array} \right) \;\longrightarrow\; \left( \begin{array}{cc|c} 1 & 2 & 5 \\ 1 & -3 & 2 \end{array} \right)$$

3. Row Replacement ($R_i \to R_i + kR_j$)

You may add a multiple of one row to another row.

• This corresponds to replacing an equation with a combination of equations.
• The augmented part must be included in the operation.
• This is the main tool for eliminating variables.

Example:
Use $R_2 \to R_2 - 3R_1$: $$\left( \begin{array}{cc|c} 1 & 1 & 4 \\ 3 & 2 & 9 \end{array} \right) \;\longrightarrow\; \left( \begin{array}{cc|c} 1 & 1 & 4 \\ 0 & -1 & -3 \end{array} \right)$$

Common Mistakes to Avoid

• Multiplying a row by 0 (not allowed — destroys information)
• Forgetting to apply the operation to the augmented column
• Replacing the wrong row (only the target row changes)
• Trying to use column operations (these do not preserve the solution set)

Exercises

1. Perform the row swap $R_1 \leftrightarrow R_2$ on $$\left( \begin{array}{cc|c} 5 & 1 & 7 \\ -3 & 4 & 2 \end{array} \right)$$

   Solution

   Row swap

   $$\left( \begin{array}{cc|c} 5 & 1 & 7 \\ -3 & 4 & 2 \end{array} \right) \to \left( \begin{array}{cc|c} -3 & 4 & 2 \\ 5 & 1 & 7 \end{array} \right)$$

2. Multiply row 2 by $-2$ in $$\left( \begin{array}{cc|c} 1 & 0 & 3 \\ 2 & -5 & 1 \end{array} \right)$$

   Solution

   Scale row 2 by $-2$

   $$\left( \begin{array}{cc|c} 1 & 0 & 3 \\ 2 & -5 & 1 \end{array} \right) \to \left( \begin{array}{cc|c} 1 & 0 & 3 \\ -4 & 10 & -2 \end{array} \right)$$

3. Apply $R_2 \to R_2 + 4R_1$ to $$\left( \begin{array}{cc|c} 1 & -1 & 2 \\ 3 & 2 & 5 \end{array} \right)$$

   Solution

   Row replacement

   $$R_2 \to R_2 + 4R_1$$ $$\left( \begin{array}{cc|c} 1 & -1 & 2 \\ 3 & 2 & 5 \end{array} \right) \to \left( \begin{array}{cc|c} 1 & -1 & 2 \\ 7 & -2 & 13 \end{array} \right)$$

4. Eliminate the 6 below the pivot in $$\left( \begin{array}{cc|c} 2 & 1 & 3 \\ 6 & 5 & 9 \end{array} \right)$$ using legal row operations.

   Solution

   Eliminate the 6

   Use $R_2 \to R_2 - 3R_1$: $$\left( \begin{array}{cc|c} 2 & 1 & 3 \\ 6 & 5 & 9 \end{array} \right) \to \left( \begin{array}{cc|c} 2 & 1 & 3 \\ 0 & 2 & 0 \end{array} \right)$$

5. Convert the first row of $$\left( \begin{array}{cc|c} 5 & 10 & 15 \\ 1 & -3 & 4 \end{array} \right)$$ into a row with leading 1.

   Solution

   Make a leading 1

   Scale row 1 by $\tfrac15$: $$\left( \begin{array}{cc|c} 5 & 10 & 15 \\ 1 & -3 & 4 \end{array} \right) \to \left( \begin{array}{cc|c} 1 & 2 & 3 \\ 1 & -3 & 4 \end{array} \right)$$

6. True or false: The operation $R_1 \to R_1 + 0R_2$ is legal.

   Solution

   True or false

   True — it is legal, but it changes nothing.

7. Perform the sequence:
   • Swap rows 1 and 2
   • Then scale the new row 1 by $\tfrac12$
   on $$\left( \begin{array}{cc|c} 0 & 4 & 8 \\ 6 & 2 & 10 \end{array} \right)$$

   Solution

   Sequence of operations

   Start: $$\left( \begin{array}{cc|c} 0 & 4 & 8 \\ 6 & 2 & 10 \end{array} \right)$$ Swap rows: $$\left( \begin{array}{cc|c} 6 & 2 & 10 \\ 0 & 4 & 8 \end{array} \right)$$ Scale row 1 by $\tfrac12$: $$\left( \begin{array}{cc|c} 3 & 1 & 5 \\ 0 & 4 & 8 \end{array} \right)$$