DWG NO. 7.4 — Lesson 4 of 5

Solving Systems Using Matrices

Unit 7: Systems of Equations and Matrices · ~30 min

Objective Write a system of equations as an augmented matrix and use row operations to reduce it to row-echelon form, then solve by back-substitution.

Elimination in Lesson 7.1 worked fine for two equations, but it gets unwieldy fast for three or more. Matrices, introduced in Lesson 7.3, fix that: strip away the variable names and keep only the coefficients, and elimination becomes a mechanical process on rows of numbers — Gaussian elimination. This is the same idea every graphing calculator and computer algebra system uses under the hood.

The augmented matrix

Write the coefficients of each equation as a row, and attach the constants as one extra column, separated by a dashed divider. The system

x + 2y = 7 3x − y = 7

becomes the augmented matrix

1	2	7
3	−1	7

Each row is one equation; the dashed line separates coefficients from constants

Elementary row operations

Three operations on the rows of a matrix always produce an equivalent system — the same solution set, just repackaged:

Swap Interchange two rows.
Scale Multiply every entry of a row by a nonzero constant.
Combine Add a multiple of one row to another row.

Row-echelon form and back-substitution

The goal is row-echelon form: each row's leading (leftmost nonzero) entry is a 1, and every entry below that leading 1 is 0, producing a staircase of zeros in the lower-left. Once the matrix is in this form, translate the bottom row back into an equation and solve — it involves only the last variable. Then work upward, substituting each solved variable into the row above it.

Row-echelon form: leading 1s on the diagonal, zeros in the staircase below (* = any value)

Worked Example 1 · A 2×2 system

Problem Solve x + 2y = 7 and 3x − y = 7 using an augmented matrix.

1Write the augmented matrix:

1	2	7
3	−1	7

2R2 → R2 − 3R1 to clear the first entry of row 2: (3−3, −1−6, 7−21) = (0, −7, −14).

3New matrix:

1	2	7
0	−7	−14

4Row 2 says −7y = −14, so y = 2. Back-substitute into row 1: x + 2(2) = 7 → x = 3.

Solution: (3, 2)

Worked Example 2 · A 3×3 system

Problem Solve x + y + z = 6, 2y + 5z = −4, 2x + 5y − z = 27.

1Augmented matrix:

1	1	1	6
0	2	5	−4
2	5	−1	27

2R3 → R3 − 2R1 clears column 1 of row 3: (0, 3, −3, 15).

3Matrix so far:

1	1	1	6
0	2	5	−4
0	3	−3	15

4R3 → R3 − 1.5R2 clears column 2 of row 3: (0, 0, −3−7.5, 15−(−6)) = (0, 0, −10.5, 21).

5Row 3 gives −10.5z = 21 → z = −2.

6Back-substitute into row 2: 2y + 5(−2) = −4 → 2y = 6 → y = 3.

7Back-substitute into row 1: x + 3 + (−2) = 6 → x = 5.

Solution: (5, 3, −2)

Guided practice

1.Write the system 2x − y = 5 and x + 3y = 6 as an augmented matrix.
Show answer

Rows are the coefficients, separated by the constants: [[2, −1 | 5], [1, 3 | 6]].
2.Perform R1 → R1 − 2R2 on the matrix [[4, 1 | 9], [2, −1 | 3]]. What is the new row 1?
Show answer

(4−2·2, 1−2·(−1), 9−2·3) = (0, 3, 3). New row 1: (0, 3 | 3).
3.A row-echelon matrix ends in the row (0, 0, 5 | 20). What does this tell you about the last variable?
Show answer

The row translates to 5z = 20, so z = 4.
4.Solve x + y = 4 and 3x − y = 8 by reducing to row-echelon form.
Show answer

Augmented matrix [[1, 1 | 4], [3, −1 | 8]]. R2 → R2 − 3R1 gives (0, −4 | −4), so −4y = −4 → y = 1. Back-substitute: x + 1 = 4 → x = 3. Solution: (3, 1).
5.Why does swapping two rows of an augmented matrix never change the solution of the system?
Show answer

Each row is just one equation of the system written differently; the order the equations are listed in doesn't change what values of the variables satisfy all of them simultaneously.

← 7.3 Introduction to Matrices Unit Overview 7.5 Determinants and Cramer's Rule →