Lesson 2: Matrices and Systems of Equations
Learning Objectives
By the end of this lesson, you will be able to:
- Represent a system of linear equations as an augmented matrix
- Perform elementary row operations on matrices
- Apply Gaussian elimination to solve systems of equations
- Apply Gauss-Jordan elimination to find reduced row-echelon form
- Identify when a system has no solution or infinitely many solutions using matrices
- Solve real-world application problems using matrix methods
Introduction: Why Matrices?
In the previous lesson, we learned substitution and elimination methods for solving systems of equations. While these methods work well for small systems (2 or 3 equations), they become increasingly tedious for larger systems.
Matrices provide a systematic, organized way to solve systems of any size. By representing systems as rectangular arrays of numbers and applying standardized row operations, we can solve complex systems efficiently and reliably.
Advantages of Matrix Methods:
- Organization: All information is in one compact structure
- Systematic: Follow clear, repeatable steps
- Scalability: Works equally well for 2, 3, or 10+ variables
- Computer-friendly: Easy to program and automate
- Error reduction: Structured approach minimizes mistakes
This lesson introduces matrix notation and two powerful methods: Gaussian elimination (which finds row-echelon form) and Gauss-Jordan elimination (which finds reduced row-echelon form).
Section 1: Matrix Notation and Augmented Matrices
For systems of linear equations, we use augmented matrices. An augmented matrix contains the coefficients of all variables plus the constants from the right side of the equations, separated by a vertical line.
System of Equations:
cx + dy = f
Augmented Matrix:
| a | b | e |
| c | d | f |
Example 1: Write System as Augmented Matrix (2×2)
Problem: Write the system as an augmented matrix:
3x + 2y = 8
x - 4y = -10
Solution:
Identify coefficients and constants:
- First equation: a = 3, b = 2, constant = 8
- Second equation: a = 1, b = -4, constant = -10
Augmented matrix:
[ 3 2 | 8 ]
[ 1 -4 | -10]
This is a 2×3 augmented matrix (2 rows, 3 columns including the constants).
Example 2: Write System as Augmented Matrix (2×3)
Problem: Write the system as an augmented matrix:
2x + 3y - z = 5
x - y + 2z = 1
Solution:
This is a system of 2 equations with 3 variables. The augmented matrix is:
[ 2 3 -1 | 5]
[ 1 -1 2 | 1]
This is a 2×4 augmented matrix.
Example 3: Read System from Augmented Matrix
Problem: Write the system of equations represented by this augmented matrix:
[ 1 5 | -3]
[-2 1 | 7]
Solution:
First row represents: 1x + 5y = -3, or simply x + 5y = -3
Second row represents: -2x + 1y = 7, or -2x + y = 7
System:
x + 5y = -3
-2x + y = 7
Example 4: Write 3×3 System as Augmented Matrix
Problem: Write the system as an augmented matrix:
x + 2y - z = 4
3x - y + 2z = 1
2x + y + z = 3
Solution:
[ 1 2 -1 | 4]
[ 3 -1 2 | 1]
[ 2 1 1 | 3]
This is a 3×4 augmented matrix (3 equations, 3 variables).
Section 2: Elementary Row Operations
To solve systems using matrices, we perform elementary row operations. These operations transform the matrix while preserving the solution to the system.
- Swap two rows: R₁ ↔ R₂
- Multiply a row by a nonzero constant: kR₁ → R₁
- Add a multiple of one row to another row: R₁ + kR₂ → R₁
Row Operation Notation:
- R₁, R₂, R₃ represent Row 1, Row 2, Row 3
- R₁ ↔ R₂ means "swap Row 1 and Row 2"
- 3R₁ → R₁ means "multiply Row 1 by 3 and replace Row 1 with the result"
- R₂ - 2R₁ → R₂ means "subtract 2 times Row 1 from Row 2 and replace Row 2"
Example 5: Swap Two Rows
Problem: Perform the operation R₁ ↔ R₂ on the matrix:
[ 2 3 | 5]
[ 1 -1 | 2]
Solution:
Simply swap the two rows:
[ 1 -1 | 2]
[ 2 3 | 5]
Why we do this: It's often helpful to have a 1 in the top-left position to simplify calculations.
Example 6: Multiply a Row by a Constant
Problem: Perform the operation (1/2)R₁ → R₁ on the matrix:
[ 2 4 | 6]
[ 1 -3 | 5]
Solution:
Multiply every entry in Row 1 by 1/2:
- 2 · (1/2) = 1
- 4 · (1/2) = 2
- 6 · (1/2) = 3
Result:
[ 1 2 | 3]
[ 1 -3 | 5]
Example 7: Add a Multiple of One Row to Another
Problem: Perform the operation R₂ - 2R₁ → R₂ on the matrix:
[ 1 3 | 5]
[ 2 7 | 12]
Solution:
Step 1: Calculate 2R₁ (multiply Row 1 by 2):
2R₁ = [2, 6, 10]
Step 2: Subtract from Row 2:
R₂ - 2R₁ = [2 - 2, 7 - 6, 12 - 10] = [0, 1, 2]
Result:
[ 1 3 | 5]
[ 0 1 | 2]
Notice: We created a 0 below the leading entry in Row 1!
Example 8: Multiple Row Operations in Sequence
Problem: Perform these operations in order on the matrix:
[ 3 6 | 9]
[ 2 5 | 8]
Operations: (1) (1/3)R₁ → R₁, then (2) R₂ - 2R₁ → R₂
Solution:
After operation (1): Multiply Row 1 by 1/3:
[ 1 2 | 3]
[ 2 5 | 8]
After operation (2): R₂ - 2R₁ → R₂:
2R₁ = [2, 4, 6]
R₂ - 2R₁ = [2-2, 5-4, 8-6] = [0, 1, 2]
Final result:
[ 1 2 | 3]
[ 0 1 | 2]
Example 9: Creating Zeros Below the Diagonal
Problem: Use row operations to get a zero in the (2,1) position:
[ 1 -2 | 3]
[ 4 5 | 7]
Solution:
We want to eliminate the 4 in position (2,1). Use R₂ - 4R₁ → R₂:
4R₁ = [4, -8, 12]
R₂ - 4R₁ = [4-4, 5-(-8), 7-12] = [0, 13, -5]
Result:
[ 1 -2 | 3]
[ 0 13 | -5]
Section 3: Row-Echelon Form and Gaussian Elimination
- All rows consisting entirely of zeros are at the bottom
- The first nonzero entry in each row (called the leading entry or pivot) is 1
- Each leading 1 is to the right of the leading 1 in the row above it
- All entries below a leading 1 are zeros
Row-Echelon Form Pattern (3×3 example):
[ 1 * * | *]
[ 0 1 * | *]
[ 0 0 1 | *]
Where * can be any number. Notice the "staircase" pattern of leading 1s.
Example 10: Gaussian Elimination on a 2×2 System
Problem: Solve using Gaussian elimination:
x + 2y = 5
3x + 7y = 16
Solution:
Step 1: Write augmented matrix:
[ 1 2 | 5]
[ 3 7 | 16]
Step 2: Get zero below leading 1 using R₂ - 3R₁ → R₂:
3R₁ = [3, 6, 15]
R₂ - 3R₁ = [3-3, 7-6, 16-15] = [0, 1, 1]
[ 1 2 | 5]
[ 0 1 | 1]
Matrix is now in row-echelon form!
Step 3: Back-substitution. Read from bottom up:
- Row 2: y = 1
- Row 1: x + 2y = 5, so x + 2(1) = 5, thus x = 3
Solution: (3, 1)
Example 11: System with No Solution
Problem: Solve using Gaussian elimination:
2x + 3y = 7
4x + 6y = 10
Solution:
Step 1: Augmented matrix:
[ 2 3 | 7]
[ 4 6 | 10]
Step 2: Get leading 1 in Row 1: (1/2)R₁ → R₁:
[ 1 3/2 | 7/2]
[ 4 6 | 10]
Step 3: Eliminate: R₂ - 4R₁ → R₂:
4R₁ = [4, 6, 14]
R₂ - 4R₁ = [4-4, 6-6, 10-14] = [0, 0, -4]
[ 1 3/2 | 7/2]
[ 0 0 | -4]
Analysis: Row 2 says 0 = -4, which is impossible!
Conclusion: No solution (inconsistent system)
Example 12: System with Infinite Solutions
Problem: Solve using Gaussian elimination:
x - 2y = 3
-2x + 4y = -6
Solution:
Step 1: Augmented matrix:
[ 1 -2 | 3]
[-2 4 | -6]
Step 2: Eliminate: R₂ + 2R₁ → R₂:
2R₁ = [2, -4, 6]
R₂ + 2R₁ = [-2+2, 4-4, -6+6] = [0, 0, 0]
[ 1 -2 | 3]
[ 0 0 | 0]
Analysis: Row 2 says 0 = 0, which is always true (gives no information).
Row 1: x - 2y = 3, so x = 3 + 2y
Solution: x = 3 + 2t, y = t where t is any real number (infinitely many solutions)
Example 13: Gaussian Elimination on 3×3 System (Part 1)
Problem: Use Gaussian elimination to solve:
x + y + z = 6
2x + y - z = 1
x - y + z = 2
Solution:
Step 1: Augmented matrix:
[ 1 1 1 | 6]
[ 2 1 -1 | 1]
[ 1 -1 1 | 2]
Step 2: Get zeros in column 1 below the leading 1:
R₂ - 2R₁ → R₂:
[2-2, 1-2, -1-2, 1-12] = [0, -1, -3, -11]
R₃ - R₁ → R₃:
[1-1, -1-1, 1-1, 2-6] = [0, -2, 0, -4]
[ 1 1 1 | 6]
[ 0 -1 -3 | -11]
[ 0 -2 0 | -4]
(Continued in next example...)
Example 14: Gaussian Elimination on 3×3 System (Part 2)
Continuing from Example 13...
[ 1 1 1 | 6]
[ 0 -1 -3 | -11]
[ 0 -2 0 | -4]
Step 3: Get leading 1 in Row 2: -R₂ → R₂:
[ 1 1 1 | 6]
[ 0 1 3 | 11]
[ 0 -2 0 | -4]
Step 4: Get zero below Row 2's leading 1: R₃ + 2R₂ → R₃:
2R₂ = [0, 2, 6, 22]
R₃ + 2R₂ = [0+0, -2+2, 0+6, -4+22] = [0, 0, 6, 18]
[ 1 1 1 | 6]
[ 0 1 3 | 11]
[ 0 0 6 | 18]
Step 5: Get leading 1 in Row 3: (1/6)R₃ → R₃:
[ 1 1 1 | 6]
[ 0 1 3 | 11]
[ 0 0 1 | 3]
Matrix is now in row-echelon form!
Step 6: Back-substitution:
- Row 3: z = 3
- Row 2: y + 3z = 11, so y + 3(3) = 11, thus y = 2
- Row 1: x + y + z = 6, so x + 2 + 3 = 6, thus x = 1
Solution: (1, 2, 3)
Section 4: Reduced Row-Echelon Form and Gauss-Jordan Elimination
- It is in row-echelon form (all conditions from REF)
- Each leading 1 is the ONLY nonzero entry in its column (zeros above AND below)
Reduced Row-Echelon Form Pattern (3×3 example):
[ 1 0 0 | *]
[ 0 1 0 | *]
[ 0 0 1 | *]
The solution can be read DIRECTLY from the last column: x = *, y = *, z = *
Example 15: Gauss-Jordan Elimination on 2×2 System
Problem: Solve using Gauss-Jordan elimination:
2x + 3y = 8
x - y = -1
Solution:
Step 1: Augmented matrix:
[ 2 3 | 8]
[ 1 -1 | -1]
Step 2: Get leading 1 in Row 1. First swap rows (easier): R₁ ↔ R₂:
[ 1 -1 | -1]
[ 2 3 | 8]
Step 3: Get zero below: R₂ - 2R₁ → R₂:
[ 1 -1 | -1]
[ 0 5 | 10]
Step 4: Get leading 1 in Row 2: (1/5)R₂ → R₂:
[ 1 -1 | -1]
[ 0 1 | 2]
Step 5: Get zero ABOVE Row 2's leading 1: R₁ + R₂ → R₁:
R₁ + R₂ = [1+0, -1+1, -1+2] = [1, 0, 1]
[ 1 0 | 1]
[ 0 1 | 2]
Matrix is now in RREF!
Read solution directly: x = 1, y = 2
Example 16: Gauss-Jordan on 3×3 System
Problem: Solve using Gauss-Jordan elimination:
x + 2y + z = 8
2x + 5y + 3z = 21
x + y - z = 0
Solution:
Initial augmented matrix:
[ 1 2 1 | 8]
[ 2 5 3 | 21]
[ 1 1 -1 | 0]
Get zeros in column 1 below leading 1:
R₂ - 2R₁ → R₂: [0, 1, 1, 5]
R₃ - R₁ → R₃: [0, -1, -2, -8]
[ 1 2 1 | 8]
[ 0 1 1 | 5]
[ 0 -1 -2 | -8]
Get zero in column 2 below Row 2:
R₃ + R₂ → R₃: [0, 0, -1, -3]
[ 1 2 1 | 8]
[ 0 1 1 | 5]
[ 0 0 -1 | -3]
Get leading 1 in Row 3: -R₃ → R₃:
[ 1 2 1 | 8]
[ 0 1 1 | 5]
[ 0 0 1 | 3]
Now work upward to get zeros ABOVE each leading 1:
R₂ - R₃ → R₂: [0, 1, 0, 2]
R₁ - R₃ → R₁: [1, 2, 0, 5]
[ 1 2 0 | 5]
[ 0 1 0 | 2]
[ 0 0 1 | 3]
Finally, get zero above Row 2's leading 1:
R₁ - 2R₂ → R₁: [1, 0, 0, 1]
[ 1 0 0 | 1]
[ 0 1 0 | 2]
[ 0 0 1 | 3]
Solution (read directly): x = 1, y = 2, z = 3
Example 17: No Solution Identified via RREF
Problem: Use Gauss-Jordan elimination on:
x + y = 2
2x + 2y = 5
Solution:
[ 1 1 | 2]
[ 2 2 | 5]
R₂ - 2R₁ → R₂:
[ 1 1 | 2]
[ 0 0 | 1]
Row 2 says: 0 = 1 (impossible!)
Conclusion: No solution
Example 18: Infinite Solutions with Free Variable
Problem: Use Gauss-Jordan elimination on:
x + 2y - z = 3
2x + 4y - 2z = 6
3x + 6y - 3z = 9
Solution:
[ 1 2 -1 | 3]
[ 2 4 -2 | 6]
[ 3 6 -3 | 9]
R₂ - 2R₁ → R₂: [0, 0, 0, 0]
R₃ - 3R₁ → R₃: [0, 0, 0, 0]
[ 1 2 -1 | 3]
[ 0 0 0 | 0]
[ 0 0 0 | 0]
Analysis: Only one meaningful equation: x + 2y - z = 3
Two free variables. Let y = s and z = t (parameters).
Then x = 3 - 2y + z = 3 - 2s + t
Solution: x = 3 - 2s + t, y = s, z = t (infinitely many solutions)
Example 19: Application - Investment Problem
Problem: Sarah invests a total of $12,000 in three accounts. Account A earns 3% annual interest, Account B earns 4%, and Account C earns 5%. After one year, she earns $510 in interest. The amount in Account C is twice the amount in Account A. How much did she invest in each account?
Solution:
Step 1: Define variables:
- Let x = amount in Account A
- Let y = amount in Account B
- Let z = amount in Account C
Step 2: Write equations:
x + y + z = 12000 (total investment)
0.03x + 0.04y + 0.05z = 510 (interest earned)
z = 2x, or -2x + z = 0 (Account C is twice Account A)
Step 3: Augmented matrix:
[ 1 1 1 | 12000]
[ 0.03 0.04 0.05| 510]
[-2 0 1 | 0]
Step 4: Apply Gauss-Jordan (details omitted for brevity):
After row operations, RREF is:
[ 1 0 0 | 3000]
[ 0 1 0 | 3000]
[ 0 0 1 | 6000]
Solution: x = 3000, y = 3000, z = 6000
Answer: Account A: $3,000, Account B: $3,000, Account C: $6,000
Section 5: Applications of Matrix Methods
Matrix methods are particularly useful for systems with three or more variables, where substitution and elimination become unwieldy. Here are situations where matrices excel:
When to Use Matrix Methods:
- Large systems: 3+ equations and variables
- Many systems with same coefficients: Change only the constants
- Computer solutions: Algorithms are easily programmed
- Applied problems: Economics, engineering, chemistry, physics
Example 20: Mixture Problem (Three Substances)
Problem: A chemist needs to mix three solutions (A, B, and C) to create 100 liters of a mixture that is 30% acid. Solution A is 20% acid, Solution B is 40% acid, and Solution C is 50% acid. The amount of Solution B must equal the combined amounts of A and C. How many liters of each solution should be used?
Solution:
Variables:
- x = liters of Solution A
- y = liters of Solution B
- z = liters of Solution C
Equations:
x + y + z = 100 (total volume)
0.20x + 0.40y + 0.50z = 0.30(100) = 30 (acid amount)
y = x + z, or -x + y - z = 0 (B equals A plus C)
Augmented matrix:
[ 1 1 1 | 100]
[ 0.20 0.40 0.50| 30]
[-1 1 -1 | 0]
After Gauss-Jordan elimination (steps omitted):
[ 1 0 0 | 20]
[ 0 1 0 | 50]
[ 0 0 1 | 30]
Solution: x = 20, y = 50, z = 30
Answer: 20 L of Solution A, 50 L of Solution B, 30 L of Solution C
Example 21: Traffic Flow Problem
Problem: The diagram shows traffic flowing through a network of one-way streets. The numbers indicate cars per hour entering and leaving each intersection. Find the traffic flow x, y, and z on the unknown streets.
(Assume diagram shows: 100 cars enter intersection A, split to x and 30; at B, x and y merge to 80 leaving; at C, y and z merge to 50 leaving; at D, 30 and z merge to 60 leaving)
Solution:
Traffic conservation principle: Flow in = Flow out at each intersection
Equations:
Intersection A: 100 = x + 30, so x = 70
Intersection B: x + y = 80, so 70 + y = 80, thus y = 10
Intersection D: 30 + z = 60, so z = 30
Verification at C: y + z = 10 + 30 = 40... but problem states 50 leaving.
(Note: In real problems, equations would be consistent. This illustrates the setup method.)
General approach: Write "flow in = flow out" equation for each intersection, then solve the system using matrices.
Example 22: Cost Analysis Problem
Problem: A bakery produces three types of bread: white, wheat, and rye. Each loaf requires flour, yeast, and salt. A white loaf needs 3 cups flour, 1 tsp yeast, and 1 tsp salt. A wheat loaf needs 2 cups flour, 1 tsp yeast, and 2 tsp salt. A rye loaf needs 2 cups flour, 2 tsp yeast, and 1 tsp salt. One day the bakery used 56 cups of flour, 28 tsp of yeast, and 32 tsp of salt. How many loaves of each type were baked?
Solution:
Variables:
- x = number of white loaves
- y = number of wheat loaves
- z = number of rye loaves
Equations (from ingredient usage):
3x + 2y + 2z = 56 (flour)
x + y + 2z = 28 (yeast)
x + 2y + z = 32 (salt)
Augmented matrix:
[ 3 2 2 | 56]
[ 1 1 2 | 28]
[ 1 2 1 | 32]
After Gauss-Jordan elimination:
[ 1 0 0 | 8]
[ 0 1 0 | 12]
[ 0 0 1 | 8]
Solution: x = 8, y = 12, z = 8
Answer: 8 white loaves, 12 wheat loaves, 8 rye loaves
Check Your Understanding
1. Write the following system as an augmented matrix:
5x - 2y = 7
3x + y = 4
[ 5 -2 | 7]
[ 3 1 | 4]
2. Perform the operation R₂ - 3R₁ → R₂ on the matrix:
[ 1 2 | 5]
[ 3 8 | 17]
3R₁ = [3, 6, 15]
R₂ - 3R₁ = [3-3, 8-6, 17-15] = [0, 2, 2]
[ 1 2 | 5]
[ 0 2 | 2]
3. Which of the following matrices is in row-echelon form?
(A) [1 2 | 3]
[0 1 | 5]
(B) [1 0 | 2]
[0 0 | 1]
(C) [0 1 | 4]
[1 0 | 2]
Answer: (A)
(A) has leading 1s in staircase pattern with zeros below - correct REF
(B) has 0 = 1 in Row 2 - represents inconsistent system, not proper REF
(C) has leading 1 in Row 2 to the LEFT of Row 1's leading entry - violates REF
4. Use Gaussian elimination to solve:
x + 3y = 7
2x + 5y = 11
Augmented matrix: [1 3|7], [2 5|11]
R₂ - 2R₁ → R₂: [1 3|7], [0 -1|-3]
-R₂ → R₂: [1 3|7], [0 1|3]
Back-substitution: y = 3, then x + 3(3) = 7, so x = -2
Solution: (-2, 3)
5. Which of the following matrices is in reduced row-echelon form?
(A) [1 2|3]
[0 1|2]
(B) [1 0|5]
[0 1|3]
Answer: (B)
(A) is in REF but not RREF - the 2 above the leading 1 in column 2 should be 0
(B) is in RREF - leading 1s have zeros above AND below
6. Use Gauss-Jordan elimination to solve:
x - y = 1
2x + y = 8
[1 -1|1], [2 1|8]
R₂ - 2R₁ → R₂: [1 -1|1], [0 3|6]
(1/3)R₂ → R₂: [1 -1|1], [0 1|2]
R₁ + R₂ → R₁: [1 0|3], [0 1|2]
Solution: x = 3, y = 2
7. What does the following reduced matrix tell you about the system's solution?
[ 1 0 | 4]
[ 0 0 | 3]
Row 2 says: 0 = 3, which is impossible.
The system has NO SOLUTION (inconsistent)
8. Set up (but do not solve) the augmented matrix for this problem:
A store sells three types of juice boxes: apple (x), orange (y), and grape (z). On Monday, they sold 100 total boxes for $180. Apple costs $1.50, orange costs $2.00, and grape costs $1.75. They sold twice as many apple boxes as grape boxes.
Equations:
x + y + z = 100 (total boxes)
1.50x + 2.00y + 1.75z = 180 (total revenue)
x = 2z, or x - 2z = 0 (apple is twice grape)
Augmented matrix:
[ 1 1 1 | 100]
[ 1.5 2 1.75| 180]
[ 1 0 -2 | 0]
Key Takeaways
- Augmented matrices organize system information in a compact rectangular array
- Three elementary row operations: swap rows, multiply row by constant, add multiple of one row to another
- Gaussian elimination produces row-echelon form (REF), then uses back-substitution to solve
- Gauss-Jordan elimination produces reduced row-echelon form (RREF) with solution immediately visible
- No solution: Indicated by row of form [0 0 ... 0 | k] where k ≠ 0
- Infinite solutions: Indicated by row of all zeros, resulting in free variables
- Matrix methods excel for large systems and application problems with many variables