Lesson 1.1: Parameterized Linear Systems & 3-Variable Elimination

Master Advanced Linear Systems with Parameters

Dive deep into parameterized linear systems! Learn to classify solutions by parameter values, master 3-variable elimination techniques, and interpret geometric meanings. Develop systematic approaches to handle complex constraints and real-world modeling scenarios.

Learning Objectives

Classify solution types by parameter conditions

Apply systematic elimination to 3-variable systems

Interpret geometric meaning of solutions

Evaluate solutions under real constraints

Use equivalence transformations (A-REI.5/6)

Apply to real-world modeling problems

Core Concepts & Theoretical Foundation

Parameterized Systems Classification

For a system $\begin{cases} ax + by = c \\ dx + ey = f \end{cases}$ with parameters:

Unique Solution: When $ae - bd \neq 0$ (determinant ≠ 0)

No Solution: When $ae - bd = 0$ but $af - cd \neq 0$

Infinitely Many Solutions: When $ae - bd = 0$ and $af - cd = 0$

Geometric Interpretation Theorem

Theorem: The solution set of a linear system corresponds to the intersection of geometric objects:

• 2-variable system: Intersection of two lines in the plane

• 3-variable system: Intersection of three planes in 3D space

• Unique solution: Single point of intersection

• No solution: Parallel objects (no intersection)

• Infinitely many: Objects coincide or intersect along a line

Elimination Strategy for 3-Variable Systems

Systematic Approach:

1. Choose elimination variable: Select variable with simplest coefficients

2. Create 2-variable system: Use two equations to eliminate chosen variable

3. Solve reduced system: Find values for remaining two variables

4. Back-substitute: Use found values to determine eliminated variable

5. Verify solution: Check all three original equations

Detailed Worked Examples

Example 1: Parameterized 2-Variable System

Solve the system $\begin{cases} ax + y = 3 \\ x + ay = 1 \end{cases}$ and classify solutions by parameter $a$ .

Step 1: Apply elimination method

Multiply first equation by $a$ : $a^2x + ay = 3a$

Subtract second equation: $(a^2x + ay) - (x + ay) = 3a - 1$

Simplify: $(a^2 - 1)x = 3a - 1$

Step 2: Classify by parameter values

Case 1: $a^2 \neq 1$ (i.e., $a \neq \pm 1$ )

• $x = \frac{3a - 1}{a^2 - 1} = \frac{3a - 1}{(a-1)(a+1)}$

• Substitute into first equation: $y = 3 - a \cdot \frac{3a - 1}{a^2 - 1} = \frac{a - 3}{a^2 - 1}$

• Result: Unique solution $\left(\frac{3a-1}{a^2-1}, \frac{a-3}{a^2-1}\right)$

Case 2: $a = 1$

• System becomes: $\begin{cases} x + y = 3 \\ x + y = 1 \end{cases}$

• Result: No solution (contradiction)

Case 3: $a = -1$

• System becomes: $\begin{cases} -x + y = 3 \\ x - y = 1 \end{cases}$

• Result: No solution (contradiction)

Geometric Interpretation: When $a \neq \pm 1$ , the lines intersect at a unique point. When $a = \pm 1$ , the lines are parallel (no intersection).

Example 2: 3-Variable System with Systematic Elimination

Solve the system: $\begin{cases} x + 2y - z = 4 \\ 2x - y + z = 1 \\ -x + y + 2z = 3 \end{cases}$

Step 1: Choose elimination variable (z)

Add equations 1 and 2: $(x + 2y - z) + (2x - y + z) = 4 + 1$

Result: $3x + y = 5$ ... (Equation A)

Multiply equation 2 by 2: $4x - 2y + 2z = 2$

Add to equation 3: $(4x - 2y + 2z) + (-x + y + 2z) = 2 + 3$

Result: $3x - y + 4z = 5$ ... (Equation B)

Step 2: Solve 2-variable system

Add equations A and B: $(3x + y) + (3x - y + 4z) = 5 + 5$

Simplify: $6x + 4z = 10$ → $3x + 2z = 5$

From equation A: $y = 5 - 3x$

Step 3: Express in terms of one variable

Let $x = t$ , then:

• $z = \frac{5 - 3t}{2}$

• $y = 5 - 3t$

Step 4: Find specific solution

Substitute $t = 1$ : $x = 1, y = 2, z = 1$

Verify: $1 + 2(2) - 1 = 4$ ✓, $2(1) - 2 + 1 = 1$ ✓, $-1 + 2 + 2(1) = 3$ ✓

Solution: $(x, y, z) = (1, 2, 1)$ . This represents the unique intersection point of three planes in 3D space.

Example 3: Real-World Constraint Analysis

A restaurant uses chicken (C), vegetables (V), and rice (R) to make meal packages. Package 1: 2C + 3V + 1R, Package 2: 3C + 5V + 2R, Package 3: 4C + 7V + 3R. Available: 60 units chicken, 95 units vegetables. Find maximum packages and rice needed.

Step 1: Set up system

Let $k_1, k_2, k_3$ = number of packages 1, 2, 3

Chicken constraint: $2k_1 + 3k_2 + 4k_3 \leq 60$

Vegetable constraint: $3k_1 + 5k_2 + 7k_3 \leq 95$

Non-negativity: $k_1, k_2, k_3 \geq 0$

Step 2: Find feasible solutions

If only Package 1: $k_1 \leq \min(\frac{60}{2}, \frac{95}{3}) = \min(30, 31.67) = 30$

If only Package 2: $k_2 \leq \min(\frac{60}{3}, \frac{95}{5}) = \min(20, 19) = 19$

If only Package 3: $k_3 \leq \min(\frac{60}{4}, \frac{95}{7}) = \min(15, 13.57) = 13$

Step 3: Mixed strategy analysis

Try $k_1 = 10, k_2 = 10$ :

Chicken used: $2(10) + 3(10) = 50 \leq 60$ ✓

Vegetables used: $3(10) + 5(10) = 80 \leq 95$ ✓

Remaining: 10 chicken, 15 vegetables

Can add: $k_3 = \min(\frac{10}{4}, \frac{15}{7}) = 2$ more packages

Step 4: Calculate rice requirement

Total rice needed: $1(10) + 2(10) + 3(2) = 36$ units

Optimal Solution: 10 Package 1, 10 Package 2, 2 Package 3. Total: 22 packages, requiring 36 units of rice.

Advanced Techniques & Problem-Solving Strategies

Matrix Approach to Parameter Analysis

For system $\begin{cases} ax + by = c \\ dx + ey = f \end{cases}$ , the coefficient matrix is:

\begin{bmatrix} a & b \\ d & e \end{bmatrix}

The determinant $\det = ae - bd$ determines solution type:

• $\det \neq 0$ → Unique solution
• $\det = 0$ → No solution or infinitely many

Cramer's Rule for Parameterized Systems

When $\det \neq 0$ , solutions are:

$x = \frac{\det_x}{\det} = \frac{ce - bf}{ae - bd}$

$y = \frac{\det_y}{\det} = \frac{af - cd}{ae - bd}$

This provides direct formulas for parameterized solutions.

Geometric Visualization Strategy

For 3-variable systems, visualize plane intersections:

• Three planes intersect at point: Unique solution

• Two planes parallel, third intersects both: No solution

• All three planes parallel: No solution

• Three planes intersect along line: Infinitely many solutions

• All three planes identical: Infinitely many solutions

Common Pitfalls & Error Prevention

Pitfall 1: Division by Zero in Parameters

Error: Dividing by $a^2 - 1$ without checking if $a = \pm 1$ .

Solution: Always check parameter values that make denominators zero before dividing.

Pitfall 2: Inconsistent Elimination Order

Error: Randomly choosing which variable to eliminate first in 3-variable systems.

Solution: Choose the variable with the simplest coefficients to minimize arithmetic errors.

Pitfall 3: Ignoring Real-World Constraints

Error: Accepting negative solutions for quantities that must be positive.

Solution: Always check if solutions make sense in the given context.

Pitfall 4: Forgetting to Verify Solutions

Error: Not checking that found solutions satisfy all original equations.

Solution: Always substitute back into all original equations to verify.

Comprehensive Practice Problems

Problem 1: Parameter Analysis

For what values of $k$ does the system $\begin{cases} (k+1)x + 2y = 3 \\ 2x + (k+1)y = 1 \end{cases}$ have:

a) A unique solution?

b) No solution?

c) Infinitely many solutions?

Show Solution

Step 1: Calculate determinant: $\det = (k+1)^2 - 4 = k^2 + 2k - 3 = (k+3)(k-1)$

Step 2: For unique solution: $\det \neq 0$ → $k \neq -3, 1$

Step 3: For no/infinitely many solutions: $\det = 0$ → $k = -3$ or $k = 1$

Step 4: Check each case to determine which gives no solution vs. infinitely many.

Problem 2: 3-Variable System

Solve: $\begin{cases} x + 2y + 3z = 6 \\ 2x + 5y + 2z = 4 \\ 6x - 3y + z = 2 \end{cases}$

Show Solution

Strategy: Eliminate x first using equations 1 and 2, then 1 and 3.

Step 1: $2(1) - (2): -y + 4z = 8$

Step 2: $6(1) - (3): 15y + 17z = 34$

Step 3: Solve the 2-variable system to find y and z, then back-substitute for x.

Problem 3: Real-World Application

A factory produces three products A, B, C using resources X, Y, Z. Product A uses (2,1,3) units, Product B uses (1,3,2) units, Product C uses (3,2,1) units. Available resources: (100, 120, 90) units. Find the maximum number of each product that can be produced.

Show Solution

Constraints:

• $2a + b + 3c \leq 100$ (Resource X)

• $a + 3b + 2c \leq 120$ (Resource Y)

• $3a + 2b + c \leq 90$ (Resource Z)

Strategy: Use linear programming or find intersection points of constraint boundaries.

Problem 4: Advanced Parameter System

Consider the system: $\begin{cases} ax + by = c \\ bx + ay = d \end{cases}$ where $a \neq b$ . Express the solution in terms of a, b, c, d and interpret geometrically.

Show Solution

Solution:

• $x = \frac{ac - bd}{a^2 - b^2} = \frac{ac - bd}{(a-b)(a+b)}$

• $y = \frac{ad - bc}{a^2 - b^2} = \frac{ad - bc}{(a-b)(a+b)}$

Geometric meaning: The lines have slopes $-\frac{a}{b}$ and $-\frac{b}{a}$ , which are negative reciprocals when $a \neq b$ , ensuring they intersect at a unique point.

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