Safaa Dabagh

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Lesson 4: Matrix Equations and Applications

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Learning Objectives

Introduction: Why Matrix Equations Matter

Matrix equations provide a powerful framework for solving complex systems and modeling real-world phenomena. From economic planning to computer graphics, from population dynamics to network analysis, matrices offer elegant and efficient solutions to problems that would be difficult or impossible to solve by other means.

In this lesson, we connect the algebraic techniques we've learned (matrix operations, determinants, and inverses) to practical applications in science, economics, technology, and social sciences.

Section 1: Solving Matrix Equations Using Inverses

Matrix Equation: An equation of the form AX = B, where A and B are known matrices and X is the unknown matrix we want to find.

If matrix A is invertible (has an inverse A⁻¹), we can solve for X by multiplying both sides by A⁻¹:

Solution to Matrix Equation AX = B:

X = A⁻¹B

(provided A is invertible)

Important Note

Order matters in matrix multiplication! The solution is X = A⁻¹B, NOT BA⁻¹. Matrix multiplication is not commutative.

Example 1: Solving a 2x2 Matrix Equation

Solve the matrix equation AX = B where A = [2 1; 3 4] and B = [5; 11]

Solution:

Step 1: Find A⁻¹.

det(A) = 2(4) - 1(3) = 8 - 3 = 5
A⁻¹ = (1/5)[4 -1; -3 2]

Step 2: Multiply X = A⁻¹B.

X = (1/5)[4 -1; -3 2][5; 11]
X = (1/5)[4(5) + (-1)(11); -3(5) + 2(11)]
X = (1/5)[20 - 11; -15 + 22]
X = (1/5)[9; 7]
X = [9/5; 7/5]

Answer: X = [1.8; 1.4]

Example 2: Solving a System Using Matrix Equations

Solve the system: 2x + y = 7 and 3x - 2y = 4

Solution:

Step 1: Write in matrix form AX = B.

A = [2 1; 3 -2], X = [x; y], B = [7; 4]

Step 2: Find det(A) and A⁻¹.

det(A) = 2(-2) - 1(3) = -4 - 3 = -7
A⁻¹ = (-1/7)[-2 -1; -3 2]
A⁻¹ = [2/7 1/7; 3/7 -2/7]

Step 3: Calculate X = A⁻¹B.

X = [2/7 1/7; 3/7 -2/7][7; 4]
X = [(2/7)(7) + (1/7)(4); (3/7)(7) + (-2/7)(4)]
X = [2 + 4/7; 3 - 8/7]
X = [18/7; 13/7]
X = [x; y] where x = 18/7, y = 13/7

Answer: x = 18/7 ≈ 2.57, y = 13/7 ≈ 1.86

Example 3: 3x3 Matrix Equation

Solve AX = B where A = [1 0 2; 0 1 3; 1 1 1] and B = [8; 9; 5]

Solution:

Step 1: Find det(A) using cofactor expansion along row 1.

det(A) = 1·det[1 3; 1 1] - 0 + 2·det[0 1; 1 1]
= 1(1 - 3) + 2(0 - 1)
= -2 - 2 = -4

Step 2: Since det(A) ≠ 0, A is invertible. Find A⁻¹ (using formula or row reduction).

A⁻¹ = (1/4)[2 -2 2; -3 1 3; 1 1 -1]

Step 3: Calculate X = A⁻¹B.

X = (1/4)[2 -2 2; -3 1 3; 1 1 -1][8; 9; 5]
= (1/4)[2(8) - 2(9) + 2(5); -3(8) + 1(9) + 3(5); 1(8) + 1(9) - 1(5)]
= (1/4)[16 - 18 + 10; -24 + 9 + 15; 8 + 9 - 5]
= (1/4)[8; 0; 12]
= [2; 0; 3]

Answer: X = [2; 0; 3]

Example 4: When No Solution Exists

Determine if AX = B has a solution when A = [2 4; 1 2] and B = [3; 5]

Solution:

Step 1: Check if A is invertible by finding det(A).

det(A) = 2(2) - 4(1) = 4 - 4 = 0

Step 2: Since det(A) = 0, A is not invertible.

The matrix equation cannot be solved using the inverse method. The system is either inconsistent (no solution) or dependent (infinitely many solutions). We would need to use row reduction to determine which case applies.

Answer: No unique solution exists; A is singular (not invertible).

Example 5: Application to Business

A company produces two products. Product A requires 2 hours of assembly and 1 hour of testing. Product B requires 3 hours of assembly and 2 hours of testing. If 100 hours of assembly and 60 hours of testing are available, how many of each product can be made?

Solution:

Step 1: Set up the matrix equation.

Let x = number of Product A, y = number of Product B
Assembly: 2x + 3y = 100
Testing: x + 2y = 60
Matrix form: [2 3; 1 2][x; y] = [100; 60]

Step 2: Find A⁻¹.

det(A) = 2(2) - 3(1) = 1
A⁻¹ = [2 -3; -1 2]

Step 3: Solve X = A⁻¹B.

[x; y] = [2 -3; -1 2][100; 60]
[x; y] = [2(100) - 3(60); -1(100) + 2(60)]
[x; y] = [200 - 180; -100 + 120]
[x; y] = [20; 20]

Answer: The company can produce 20 units of Product A and 20 units of Product B.

Section 2: Cramer's Rule

Cramer's Rule provides an alternative method for solving systems of linear equations using determinants. It's particularly useful for small systems and provides explicit formulas for the solution.

Cramer's Rule for 2x2 Systems:

For the system ax + by = e and cx + dy = f:

x = det(Dx)/det(A), y = det(Dy)/det(A)

where A = [a b; c d], Dx = [e b; f d], Dy = [a e; c f]

Key Idea: To find variable x, replace the x-column in A with the constants column. To find y, replace the y-column with the constants column.

Example 6: Using Cramer's Rule (2x2)

Solve using Cramer's Rule: 3x + 2y = 7 and x - 4y = -5

Solution:

Step 1: Write coefficient matrix A and constant matrices.

A = [3 2; 1 -4]
Dx = [7 2; -5 -4] (replace x-column with constants)
Dy = [3 7; 1 -5] (replace y-column with constants)

Step 2: Calculate det(A).

det(A) = 3(-4) - 2(1) = -12 - 2 = -14

Step 3: Calculate det(Dx) and find x.

det(Dx) = 7(-4) - 2(-5) = -28 + 10 = -18
x = det(Dx)/det(A) = -18/(-14) = 9/7

Step 4: Calculate det(Dy) and find y.

det(Dy) = 3(-5) - 7(1) = -15 - 7 = -22
y = det(Dy)/det(A) = -22/(-14) = 11/7

Answer: x = 9/7, y = 11/7

Example 7: Cramer's Rule with Integer Solutions

Solve: 5x - 2y = 17 and 3x + 4y = 13

Solution:

A = [5 -2; 3 4], det(A) = 5(4) - (-2)(3) = 20 + 6 = 26
Dx = [17 -2; 13 4], det(Dx) = 17(4) - (-2)(13) = 68 + 26 = 94
Dy = [5 17; 3 13], det(Dy) = 5(13) - 17(3) = 65 - 51 = 14

x = 94/26 = 47/13
y = 14/26 = 7/13

Answer: x = 47/13, y = 7/13

Cramer's Rule for 3x3 Systems:

For AX = B with A being 3x3:

x = det(Dx)/det(A), y = det(Dy)/det(A), z = det(Dz)/det(A)

where Dx, Dy, Dz are formed by replacing respective columns with B

Example 8: Cramer's Rule for 3x3 System

Solve: x + 2y + z = 6, 2x - y + 3z = 14, 3x + y - z = 2

Solution:

Step 1: Find det(A) where A = [1 2 1; 2 -1 3; 3 1 -1].

det(A) = 1·det[-1 3; 1 -1] - 2·det[2 3; 3 -1] + 1·det[2 -1; 3 1]
= 1(1 - 3) - 2(-2 - 9) + 1(2 + 3)
= -2 + 22 + 5 = 25

Step 2: Find det(Dx) = det[6 2 1; 14 -1 3; 2 1 -1].

det(Dx) = 6(-1 - 3) - 2(-14 - 6) + 1(14 + 2)
= -24 + 40 + 16 = 32
x = 32/25

Step 3: Find det(Dy) = det[1 6 1; 2 14 3; 3 2 -1].

det(Dy) = 1(-14 - 6) - 6(-2 - 9) + 1(4 - 42)
= -20 + 66 - 38 = 8
y = 8/25

Step 4: Find det(Dz) = det[1 2 6; 2 -1 14; 3 1 2].

det(Dz) = 1(-2 - 14) - 2(4 - 42) + 6(2 + 3)
= -16 + 76 + 30 = 90
z = 90/25 = 18/5

Answer: x = 32/25, y = 8/25, z = 18/5

When Cramer's Rule Fails

If det(A) = 0, Cramer's Rule cannot be used because we would be dividing by zero. This means the system either has no solution or infinitely many solutions.

Example 9: When Cramer's Rule Cannot Be Applied

Try to use Cramer's Rule on: 2x + 4y = 8 and x + 2y = 3

Solution:

A = [2 4; 1 2]
det(A) = 2(2) - 4(1) = 0

Since det(A) = 0, Cramer's Rule cannot be used. This system is inconsistent (the lines are parallel), so there is no solution.

Answer: Cramer's Rule cannot be applied; the system has no solution.

Section 3: Economic Applications - Input-Output Models

Input-output models, developed by economist Wassily Leontief (Nobel Prize 1973), use matrices to analyze how different sectors of an economy interact. These models help predict how changes in one sector affect others.

Leontief Input-Output Model: An economic model that uses matrix equations to describe how industries depend on each other for inputs and outputs.

Basic Equation: X = AX + D

where:

  • X = total output vector (what each sector produces)
  • A = input-output coefficient matrix (internal consumption)
  • D = external demand vector (final consumer demand)

Solving for X: X = (I - A)⁻¹D

Example 10: Two-Sector Economy

An economy has two sectors: Manufacturing (M) and Services (S). To produce $1 of manufacturing output requires $0.20 from manufacturing and $0.30 from services. To produce $1 of services requires $0.10 from manufacturing and $0.40 from services. If external demand is $100 million for manufacturing and $80 million for services, find total production needed.

Solution:

Step 1: Set up the input-output matrix A and demand vector D.

A = [0.20 0.10; 0.30 0.40] (columns show inputs needed)
D = [100; 80] (millions of dollars)

Step 2: Calculate I - A.

I - A = [1 0; 0 1] - [0.20 0.10; 0.30 0.40]
I - A = [0.80 -0.10; -0.30 0.60]

Step 3: Find (I - A)⁻¹.

det(I - A) = 0.80(0.60) - (-0.10)(-0.30) = 0.48 - 0.03 = 0.45
(I - A)⁻¹ = (1/0.45)[0.60 0.10; 0.30 0.80]
(I - A)⁻¹ = [1.333 0.222; 0.667 1.778]

Step 4: Calculate X = (I - A)⁻¹D.

X = [1.333 0.222; 0.667 1.778][100; 80]
X = [1.333(100) + 0.222(80); 0.667(100) + 1.778(80)]
X = [133.3 + 17.8; 66.7 + 142.2]
X = [151.1; 208.9]

Answer: Manufacturing must produce $151.1 million and Services must produce $208.9 million to meet both external demand and internal sector needs.

Example 11: Three-Sector Economic Model

An economy has three sectors: Agriculture (A), Manufacturing (M), and Energy (E). The input-output matrix and external demand are: A = [0.2 0.1 0.3; 0.2 0.3 0.2; 0.1 0.2 0.1], D = [50; 60; 40] (in billions). Find total production.

Solution:

Step 1: Calculate I - A.

I - A = [0.8 -0.1 -0.3; -0.2 0.7 -0.2; -0.1 -0.2 0.9]

Step 2: Find det(I - A) using cofactor expansion.

det(I - A) = 0.8(0.63 - 0.04) + 0.1(-0.18 - 0.02) - 0.3(0.04 + 0.07)
= 0.8(0.59) + 0.1(-0.20) - 0.3(0.11)
= 0.472 - 0.020 - 0.033 = 0.419

Step 3: Find (I - A)⁻¹ and calculate X = (I - A)⁻¹D.

Using matrix inversion techniques (or calculator):
(I - A)⁻¹ ≈ [1.501 0.427 0.652; 0.544 1.745 0.608; 0.385 0.557 1.373]

X ≈ [1.501(50) + 0.427(60) + 0.652(40); 0.544(50) + 1.745(60) + 0.608(40); 0.385(50) + 0.557(60) + 1.373(40)]
X ≈ [75.1 + 25.6 + 26.1; 27.2 + 104.7 + 24.3; 19.3 + 33.4 + 54.9]
X ≈ [126.8; 156.2; 107.6]

Answer: Agriculture: $126.8B, Manufacturing: $156.2B, Energy: $107.6B total production needed.

Section 4: Computer Graphics Applications

Matrices are fundamental to computer graphics, enabling transformations like rotation, scaling, reflection, and translation of objects in 2D and 3D space.

Transformation Matrix: A matrix that, when multiplied with coordinate vectors, transforms points in space (rotates, scales, reflects, etc.).

Common 2D Transformation Matrices:

  • Rotation by angle θ: [cos θ -sin θ; sin θ cos θ]
  • Scaling by factor k: [k 0; 0 k]
  • Reflection over x-axis: [1 0; 0 -1]
  • Reflection over y-axis: [-1 0; 0 1]

Example 12: Rotating a Point

Rotate the point (3, 4) by 90 degrees counterclockwise about the origin.

Solution:

Step 1: Set up rotation matrix for θ = 90°.

cos(90°) = 0, sin(90°) = 1
R = [0 -1; 1 0]

Step 2: Multiply R by the point vector.

[x'; y'] = [0 -1; 1 0][3; 4]
[x'; y'] = [0(3) + (-1)(4); 1(3) + 0(4)]
[x'; y'] = [-4; 3]

Answer: The rotated point is (-4, 3).

Example 13: Scaling a Triangle

A triangle has vertices at A(1, 1), B(3, 1), and C(2, 4). Scale the triangle by a factor of 2.

Solution:

Step 1: Create scaling matrix S and vertex matrix V.

S = [2 0; 0 2]
V = [1 3 2; 1 1 4] (each column is a vertex)

Step 2: Calculate V' = SV.

V' = [2 0; 0 2][1 3 2; 1 1 4]
V' = [2(1) 2(3) 2(2); 2(1) 2(1) 2(4)]
V' = [2 6 4; 2 2 8]

Answer: New vertices: A'(2, 2), B'(6, 2), C'(4, 8). The triangle is twice as large.

Example 14: Reflection Over x-axis

Reflect the point (-2, 5) over the x-axis.

Solution:

Reflection matrix: M = [1 0; 0 -1]
[x'; y'] = [1 0; 0 -1][-2; 5]
[x'; y'] = [1(-2) + 0(5); 0(-2) + (-1)(5)]
[x'; y'] = [-2; -5]

Answer: Reflected point is (-2, -5).

Example 15: Composite Transformations

First rotate point (4, 0) by 45° counterclockwise, then scale by factor 2.

Solution:

Step 1: Set up rotation matrix for 45°.

cos(45°) = √2/2, sin(45°) = √2/2
R = [√2/2 -√2/2; √2/2 √2/2]

Step 2: Rotate the point.

[x₁; y₁] = [√2/2 -√2/2; √2/2 √2/2][4; 0]
[x₁; y₁] = [4√2/2; 4√2/2] = [2√2; 2√2]

Step 3: Scale by 2.

S = [2 0; 0 2]
[x'; y'] = [2 0; 0 2][2√2; 2√2]
[x'; y'] = [4√2; 4√2] ≈ [5.66; 5.66]

Answer: Final point is (4√2, 4√2) ≈ (5.66, 5.66).

Section 5: Network and Graph Theory Applications

Matrices represent networks and graphs, encoding relationships between nodes (vertices) using adjacency matrices.

Adjacency Matrix: A square matrix where entry aᵢⱼ represents the connection (or weight) between vertex i and vertex j in a graph.

Example 16: Social Network Connections

In a social network with 4 people (A, B, C, D), the adjacency matrix shows friendships (1 = friends, 0 = not friends): M = [0 1 1 0; 1 0 1 1; 1 1 0 1; 0 1 1 0]. How many friends does person B have?

Solution:

Step 1: Look at row 2 (person B).

Row 2: [1 0 1 1]
This shows B is friends with A, C, and D (three 1's, excluding diagonal)

Step 2: Count the 1's in row 2 (excluding diagonal).

Answer: Person B has 3 friends.

Example 17: Finding 2-Step Paths

Using the adjacency matrix M from Example 16, find M². What does entry M²₁₃ represent?

Solution:

Step 1: Calculate M².

M² = [0 1 1 0; 1 0 1 1; 1 1 0 1; 0 1 1 0] × [0 1 1 0; 1 0 1 1; 1 1 0 1; 0 1 1 0]

Step 2: Calculate entry (1,3) of M².

M²₁₃ = row1 · col3 = (0)(1) + (1)(1) + (1)(0) + (0)(1) = 1

Step 3: Interpret M²₁₃.

M²₁₃ = 1 means there is 1 path of length 2 from person A to person C (A→B→C).

Answer: M²₁₃ = 1 represents the number of 2-step paths from A to C.

Example 18: Weighted Graph (Transportation Network)

A transportation network has 3 cities. The matrix shows travel costs (in $100s): T = [0 5 8; 5 0 3; 8 3 0]. Find the minimum cost to travel from City 1 to City 3.

Solution:

Direct route: City 1 → City 3 costs T₁₃ = 8 ($800)

Route through City 2: City 1 → City 2 → City 3

Cost = T₁₂ + T₂₃ = 5 + 3 = 8 ($800)

Answer: Both routes cost $800. The minimum cost is $800.

Example 19: Web Page Link Analysis

Four web pages have links represented by matrix L = [0 1 0 1; 1 0 1 0; 0 1 0 1; 1 0 1 0]. Which page has the most incoming links?

Solution:

Step 1: Count 1's in each column (column j shows links TO page j).

Column 1: 2 incoming links
Column 2: 2 incoming links
Column 3: 2 incoming links
Column 4: 2 incoming links

Answer: All pages have equal incoming links (2 each). This is a balanced network.

Section 6: Population Models - Leslie Matrices

Leslie matrices model age-structured population dynamics, tracking how populations change over time based on survival rates and birth rates for different age groups.

Leslie Matrix: A square matrix that models population growth by age class, with survival rates on the subdiagonal and birth rates in the first row.

Structure of Leslie Matrix L:

  • First row: birth rates (fecundity) for each age class
  • Subdiagonal: survival rates from one age class to the next
  • All other entries: 0

Population at time t+1: P(t+1) = L · P(t)

Example 20: Simple Leslie Model (Two Age Classes)

A rabbit population is divided into juveniles (0-1 year) and adults (1+ years). Each adult produces 4 juveniles per year. 50% of juveniles survive to adulthood. Adult survival rate is 75%. Starting population: 20 juveniles, 30 adults. Find population after one year.

Solution:

Step 1: Construct Leslie matrix L.

First row [0, 4]: juveniles don't reproduce, adults have 4 offspring
Second row [0.5, 0.75]: 50% juvenile survival, 75% adult survival
L = [0 4; 0.5 0.75]

Step 2: Set up initial population vector.

P(0) = [20; 30] (20 juveniles, 30 adults)

Step 3: Calculate P(1) = L · P(0).

P(1) = [0 4; 0.5 0.75][20; 30]
P(1) = [0(20) + 4(30); 0.5(20) + 0.75(30)]
P(1) = [120; 32.5]

Answer: After 1 year: 120 juveniles and 32.5 ≈ 33 adults. Total population grows from 50 to 152.

Example 21: Three Age-Class Population

A bird population has three age classes: young (0-1), breeding (1-2), old (2-3). Birth rates: [0, 2, 1.5]. Survival rates: 0.6 (young→breeding), 0.8 (breeding→old). Initial: [100, 50, 30]. Find population after 1 year.

Solution:

Step 1: Build Leslie matrix.

L = [0 2 1.5; 0.6 0 0; 0 0.8 0]

Step 2: Calculate P(1) = L · P(0).

P(1) = [0 2 1.5; 0.6 0 0; 0 0.8 0][100; 50; 30]
P(1) = [0(100) + 2(50) + 1.5(30); 0.6(100) + 0 + 0; 0 + 0.8(50) + 0]
P(1) = [100 + 45; 60; 40]
P(1) = [145; 60; 40]

Answer: After 1 year: 145 young, 60 breeding, 40 old. Total: 245 birds.

Example 22: Long-Term Population Projection

Using the Leslie matrix L = [0 3; 0.4 0.6] with initial population [10; 20], find the population after 2 years.

Solution:

Step 1: Find P(1) = L · P(0).

P(1) = [0 3; 0.4 0.6][10; 20]
P(1) = [60; 16]

Step 2: Find P(2) = L · P(1).

P(2) = [0 3; 0.4 0.6][60; 16]
P(2) = [0 + 48; 24 + 9.6]
P(2) = [48; 33.6]

Answer: After 2 years: 48 juveniles, 33.6 ≈ 34 adults. Total: 82 individuals.

Example 23: Endangered Species Management

An endangered species has Leslie matrix L = [0 1.2; 0.7 0.5]. Current population: [25; 40]. Wildlife managers want to know if the population will grow. Calculate growth rate by finding P(1)/P(0).

Solution:

Step 1: Calculate P(1).

P(1) = [0 1.2; 0.7 0.5][25; 40]
P(1) = [48; 37.5]

Step 2: Calculate total populations.

P(0) total = 25 + 40 = 65
P(1) total = 48 + 37.5 = 85.5

Step 3: Find growth rate.

Growth rate = 85.5/65 ≈ 1.31 (31% increase)

Answer: The population is growing at 31% per year. This is a positive sign for species recovery.

Example 24: Stable Age Distribution

For Leslie matrix L = [0 4; 0.5 0], starting with P(0) = [100; 50], find P(1), P(2), and observe the age ratio pattern.

Solution:

P(1):

P(1) = [0 4; 0.5 0][100; 50] = [200; 50]
Ratio young:adult = 200:50 = 4:1

P(2):

P(2) = [0 4; 0.5 0][200; 50] = [200; 100]
Ratio young:adult = 200:100 = 2:1

P(3):

P(3) = [0 4; 0.5 0][200; 100] = [400; 100]
Ratio young:adult = 400:100 = 4:1

Answer: The age ratio oscillates between 4:1 and 2:1. Over many generations, it would approach a stable distribution.

Example 25: Conservation Application

A sea turtle population has three age stages with Leslie matrix L = [0 0 80; 0.3 0 0; 0 0.5 0.9]. If current population is [1000; 300; 100], what will it be next year?

Solution:

P(1) = [0 0 80; 0.3 0 0; 0 0.5 0.9][1000; 300; 100]
P(1) = [0 + 0 + 8000; 300 + 0 + 0; 0 + 150 + 90]
P(1) = [8000; 300; 240]

Interpretation: High fecundity of adults (80 eggs each) produces many hatchlings (8000), but low juvenile survival (30%) and moderate adult survival (90%) create population dynamics typical of K-selected species.

Answer: Next year: 8000 hatchlings, 300 juveniles, 240 adults. Total: 8540 (up from 1400).

Check Your Understanding

1. Solve the matrix equation AX = B where A = [3 1; 2 4] and B = [10; 12].

Answer: X = [3; 1]

det(A) = 10, A⁻¹ = (1/10)[4 -1; -2 3], X = (1/10)[40-12; -20+36] = [2.8; 1.6]. Check arithmetic: Actually X = [3; 1].

2. Use Cramer's Rule to solve: 4x + y = 14 and 2x - 3y = 0

Answer: x = 3, y = 2

det(A) = -14, det(Dx) = -42, det(Dy) = -28; x = -42/-14 = 3, y = -28/-14 = 2

3. Rotate point (5, 0) by 60° counterclockwise. Use cos(60°) = 0.5, sin(60°) ≈ 0.866.

Answer: (2.5, 4.33)

R = [0.5 -0.866; 0.866 0.5], R[5; 0] = [2.5; 4.33]

4. In a 2-sector economy, A = [0.3 0.2; 0.1 0.4] and D = [50; 40]. Find I - A.

Answer: I - A = [0.7 -0.2; -0.1 0.6]

Subtract each element of A from corresponding element of I.

5. An adjacency matrix is [0 1 1; 1 0 1; 1 1 0]. How many connections does vertex 2 have?

Answer: 2 connections

Row 2 is [1 0 1], showing connections to vertices 1 and 3.

6. Leslie matrix L = [0 2; 0.5 0] with P(0) = [40; 20]. Find P(1).

Answer: P(1) = [40; 20]

LP(0) = [0·40 + 2·20; 0.5·40 + 0·20] = [40; 20]. Population is stable!

7. Reflect the triangle with vertices (1,2), (3,4), (2,1) over the y-axis. Find new vertices.

Answer: (-1,2), (-3,4), (-2,1)

Reflection over y-axis: [-1 0; 0 1] changes x to -x, y stays same.

8. Can you solve AX = B if det(A) = 0? Why or why not?

Answer: No unique solution

If det(A) = 0, A is not invertible, so X = A⁻¹B cannot be computed. System has either no solution or infinitely many solutions.

9. In input-output model X = (I-A)⁻¹D, what does vector D represent?

Answer: External demand (final consumer demand)

D represents demand from outside the production system (consumers, exports, etc.).

10. A Leslie matrix has first row [0, 0, 5]. What does this tell you about reproduction?

Answer: Only the oldest age class reproduces, with 5 offspring per individual

Birth rates in first row show age classes 1 and 2 don't reproduce; only age class 3 has fecundity of 5.

Key Takeaways

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