The Three Row Operations on Augmented Matrices

An augmented matrix combines the coefficient matrix of a linear system with its constant column. The three elementary row operations on the matrix correspond exactly to the three elementary operations on the equations — so we can manipulate the matrix directly instead of rewriting the variables each time.

What is an Augmented Matrix?

The augmented matrix of a system of linear equations is made up of the matrix of the coefficients of the unknowns, augmented by the matrix of the constants.

Consider the following \( 2 \times 2 \) system in the unknowns \( x \) and \( y \):

\[ \begin{cases} a x + b y = c \\ d x + e y = f \end{cases} \]

In matrix form:

\[ \begin{bmatrix} a & b \\ d & e \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c \\ f \end{bmatrix}. \]

The augmented matrix combines the coefficients and constants into a single matrix, separated by a vertical line:

\[ \left[ \begin{array}{cc|c} a & b & c \\ d & e & f \end{array} \right]. \]

Key idea. The vertical bar separates the coefficient matrix \( \begin{bmatrix} a & b \\ d & e \end{bmatrix} \) from the constant column \( \begin{bmatrix} c \\ f \end{bmatrix} \). Row operations applied to the augmented matrix preserve the solution set of the original system — which is why we can use them to reduce the system to a simpler form.

Examples of Augmented Matrices

Example 1 — Write Augmented Matrices

Write the augmented matrix for each system.

System a)

\[ \begin{cases} x - y + 2z = 1 \\ 2x + y - z = 0 \\ -x + 2y + 3z = 4 \end{cases} \]

Augmented matrix a)

\[ \left[ \begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 2 & 1 & -1 & 0 \\ -1 & 2 & 3 & 4 \end{array} \right] \]

System b)

\[ \begin{cases} 3x + 4y = -2 \\ -2x + 5y = 7 \end{cases} \]

Augmented matrix b)

\[ \left[ \begin{array}{cc|c} 3 & 4 & -2 \\ -2 & 5 & 7 \end{array} \right] \]

Note: the augmented matrix contains all the information in the system of equations, so we can manipulate the matrix directly instead of rewriting the variables.

Example 2 — Extract Systems from Augmented Matrices

Write the system of equations whose augmented matrix is given.

a)

\[ \left[ \begin{array}{cc|c} 0 & 1 & 4 \\ -2 & 0 & 0 \end{array} \right] \quad\Longrightarrow\quad \begin{cases} y = 4 \\ -2x = 0 \end{cases} \]

b)

\[ \left[ \begin{array}{ccc|c} -9 & 0 & 0 & 4 \\ 0 & 3 & 0 & -1 \\ 1 & -3 & 9 & -7 \end{array} \right] \quad\Longrightarrow\quad \begin{cases} -9x = 4 \\ 3y = -1 \\ x - 3y + 9z = -7 \end{cases} \]

c)

\[ \left[ \begin{array}{cccc|c} -4 & 0 & -1 & 0 & -4 \\ 7 & 0 & -2 & 7 & -2 \\ 4 & 0 & 0 & -8 & 0 \\ -1 & -5 & 7 & 0 & -1 \end{array} \right] \quad\Longrightarrow\quad \begin{cases} -4x - z = -4 \\ 7x - 2z + 7w = -2 \\ 4x - 8w = 0 \\ -x - 5y + 7z = -1 \end{cases} \]

The Three Row Operations on Augmented Matrices

The three elementary operations on the equations of a system have three completely equivalent operations on the rows of the augmented matrix.

Operations on Equations Operations on Rows of Augmented Matrix
  1. Interchange two equations.
  2. Add a multiple of one equation to another.
  3. Multiply an equation by a nonzero constant.
  1. Interchange two rows.
  2. Add a multiple of one row to another.
  3. Multiply a row by a nonzero constant.
Why the three row operations preserve solutions. Each operation corresponds to an invertible transformation of the equations: swapping two equations, adding a multiple of one equation to another, or scaling by a nonzero constant. None of these changes the set of \( (x, y, z, \dots) \) that satisfies the system.

Step-by-Step Reduction Example

Example 3. Apply operations to the equations, and the same operations to the rows of the augmented matrix, to show that the two systems obtained correspond perfectly.

System of Equations Corresponding Augmented Matrix
Given: \[ \begin{cases} x + y - z = -3 \\ -x + 3y + 2z = 5 \\ x + 5z = 9 \end{cases} \] \[ \left[ \begin{array}{ccc|c} 1 & 1 & -1 & -3 \\ -1 & 3 & 2 & 5 \\ 1 & 0 & 5 & 9 \end{array} \right] \]
Step 1 — Add equation (1) to equation (2) / Add row (1) to row (2)
\[ \begin{cases} x + y - z = -3 \\ 4y + z = 2 \\ x + 5z = 9 \end{cases} \] \[ \begin{matrix} \\ R_2 + R_1 \\ \\ \end{matrix} \left[ \begin{array}{ccc|c} 1 & 1 & -1 & -3 \\ 0 & 4 & 1 & 2 \\ 1 & 0 & 5 & 9 \end{array} \right] \]
Step 2 — Add \(-1\) times equation (1) to equation (3) / Add \(-1\) times row (1) to row (3)
\[ \begin{cases} x + y - z = -3 \\ 4y + z = 2 \\ -y + 6z = 12 \end{cases} \] \[ \begin{matrix} \\ \\ R_3 - R_1 \end{matrix} \left[ \begin{array}{ccc|c} 1 & 1 & -1 & -3 \\ 0 & 4 & 1 & 2 \\ 0 & -1 & 6 & 12 \end{array} \right] \]
Step 3 — Interchange equations (2) and (3) / Interchange rows (2) and (3)
\[ \begin{cases} x + y - z = -3 \\ -y + 6z = 12 \\ 4y + z = 2 \end{cases} \] \[ \begin{matrix} \\ R_3 \\ R_2 \end{matrix} \left[ \begin{array}{ccc|c} 1 & 1 & -1 & -3 \\ 0 & -1 & 6 & 12 \\ 0 & 4 & 1 & 2 \end{array} \right] \]
Step 4 — Add \(4\) times equation (2) to equation (3) / Add \(4\) times row (2) to row (3)
\[ \begin{cases} x + y - z = -3 \\ -y + 6z = 12 \\ 25z = 50 \end{cases} \] \[ \begin{matrix} \\ \\ R_3 + 4R_2 \end{matrix} \left[ \begin{array}{ccc|c} 1 & 1 & -1 & -3 \\ 0 & -1 & 6 & 12 \\ 0 & 0 & 25 & 50 \end{array} \right] \]
Step 5 — Multiply equation (2) by \(-1\) / Multiply row (2) by \(-1\)
\[ \begin{cases} x + y - z = -3 \\ y - 6z = -12 \\ 25z = 50 \end{cases} \] \[ \begin{matrix} \\ -R_2 \\ \\ \end{matrix} \left[ \begin{array}{ccc|c} 1 & 1 & -1 & -3 \\ 0 & 1 & -6 & -12 \\ 0 & 0 & 25 & 50 \end{array} \right] \]
Step 6 — Multiply equation (3) by \(\tfrac{1}{25}\) / Multiply row (3) by \(\tfrac{1}{25}\)
\[ \begin{cases} x + y - z = -3 \\ y - 6z = -12 \\ z = 2 \end{cases} \] \[ \begin{matrix} \\ \\ \tfrac{1}{25}R_3 \end{matrix} \left[ \begin{array}{ccc|c} 1 & 1 & -1 & -3 \\ 0 & 1 & -6 & -12 \\ 0 & 0 & 1 & 2 \end{array} \right] \]
Back-substitution. From the final system: \( z = 2 \), then \( y = -12 + 6z = -12 + 12 = 0 \), then \( x = -3 - y + z = -3 - 0 + 2 = -1 \). The solution is \( (x, y, z) = (-1, 0, 2) \).

Row operations reduce augmented matrices to row echelon form (REF) — or, with additional steps, to reduced row echelon form (RREF) — which is used to solve systems of equations, find matrix inverses, and determine vector spaces.

Practice Questions & Solutions

Question Set 1 — Find the Augmented Matrix

Find the augmented matrix for each of the following systems.

  1. \[ \begin{cases} x + 2y = 2 \\ -2x + 4y = -1 \end{cases} \]
  2. \[ \begin{cases} -x + 2y + z = -5 \\ -x - \tfrac{1}{3}y = -7 \\ -\tfrac{2}{5}y - 9z = 0 \end{cases} \]
  3. \[ \begin{cases} 2x + z - w = 0 \\ -\tfrac{1}{2}y - 4z = -4 \\ x - 7y = 7 \\ -4y - 5z - \tfrac{1}{5}w = 2 \end{cases} \]

Solutions to Set 1

  1. \[ \left[ \begin{array}{cc|c} 1 & 2 & 2 \\ -2 & 4 & -1 \end{array} \right] \]
  2. \[ \left[ \begin{array}{ccc|c} -1 & 2 & 1 & -5 \\ -1 & -\tfrac{1}{3} & 0 & -7 \\ 0 & -\tfrac{2}{5} & -9 & 0 \end{array} \right] \]
  3. \[ \left[ \begin{array}{cccc|c} 2 & 0 & 1 & -1 & 0 \\ 0 & -\tfrac{1}{2} & -4 & 0 & -4 \\ 1 & -7 & 0 & 0 & 7 \\ 0 & -4 & -5 & -\tfrac{1}{5} & 2 \end{array} \right] \]
Question Set 2 — Find the System from the Matrix

Write the system of linear equations given each augmented matrix.

  1. \[ \left[ \begin{array}{cc|c} 1 & -2 & -8 \\ 9 & -1 & 0 \end{array} \right] \]
  2. \[ \left[ \begin{array}{ccc|c} -1 & 1 & -1 & -6 \\ 1 & 0 & -2 & 4 \\ 0 & -1 & 7 & -4 \end{array} \right] \]
  3. \[ \left[ \begin{array}{cccc|c} -\tfrac{2}{3} & -1 & 1 & -8 & 2 \\ 0 & -\tfrac{1}{2} & -4 & \tfrac{1}{3} & 7 \\ -1 & 3 & 0 & 0 & 7 \\ 0 & -5 & 8 & \tfrac{4}{5} & 2 \end{array} \right] \]

Solutions to Set 2

  1. \[ \begin{cases} x - 2y = -8 \\ 9x - y = 0 \end{cases} \]
  2. \[ \begin{cases} -x + y - z = -6 \\ x - 2z = 4 \\ -y + 7z = -4 \end{cases} \]
  3. \[ \begin{cases} -\tfrac{2}{3}x - y + z - 8w = 2 \\ -\tfrac{1}{2}y - 4z + \tfrac{1}{3}w = 7 \\ -x + 3y = 7 \\ -5y + 8z + \tfrac{4}{5}w = 2 \end{cases} \]

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