Module 8: Practice Problems
Introduction to Matrices
About These Practice Problems
This set contains 20 comprehensive problems covering all topics in Module 8. Work through each problem carefully, then check your solution. Each problem includes detailed step-by-step explanations.
Problems 1-5: Matrix Operations
Problem 1: Matrix Addition
Given A = [2 -3; 1 4] and B = [5 1; -2 6], find A + B.
Solution:
Step 1: Check dimensions
Both matrices are 2×2, so addition is defined.
Both matrices are 2×2, so addition is defined.
Step 2: Add corresponding elements
A + B = [2+5 -3+1; 1+(-2) 4+6]
A + B = [2+5 -3+1; 1+(-2) 4+6]
Step 3: Simplify
A + B = [7 -2; -1 10]
A + B = [7 -2; -1 10]
Final Answer: [7 -2; -1 10]
Problem 2: Scalar Multiplication
Given C = [3 -1 2; 0 4 -5], find 3C.
Solution:
Step 1: Multiply each element by the scalar 3
3C = [3(3) 3(-1) 3(2); 3(0) 3(4) 3(-5)]
3C = [3(3) 3(-1) 3(2); 3(0) 3(4) 3(-5)]
Step 2: Simplify
3C = [9 -3 6; 0 12 -15]
3C = [9 -3 6; 0 12 -15]
Final Answer: [9 -3 6; 0 12 -15]
Problem 3: Matrix Subtraction
Given D = [6 2; -1 5] and E = [3 -4; 2 1], find 2D - E.
Solution:
Step 1: Calculate 2D
2D = [2(6) 2(2); 2(-1) 2(5)] = [12 4; -2 10]
2D = [2(6) 2(2); 2(-1) 2(5)] = [12 4; -2 10]
Step 2: Subtract E from 2D
2D - E = [12-3 4-(-4); -2-2 10-1]
2D - E = [12-3 4-(-4); -2-2 10-1]
Step 3: Simplify
2D - E = [9 8; -4 9]
2D - E = [9 8; -4 9]
Final Answer: [9 8; -4 9]
Problem 4: Matrix Multiplication (2×2)
Given A = [1 3; 2 -1] and B = [4 0; -2 5], find AB.
Solution:
Step 1: Check dimensions
A is 2×2, B is 2×2. Since columns of A (2) = rows of B (2), multiplication is defined.
Result will be 2×2.
A is 2×2, B is 2×2. Since columns of A (2) = rows of B (2), multiplication is defined.
Result will be 2×2.
Step 2: Calculate element (1,1)
Row 1 of A · Column 1 of B: (1)(4) + (3)(-2) = 4 - 6 = -2
Row 1 of A · Column 1 of B: (1)(4) + (3)(-2) = 4 - 6 = -2
Step 3: Calculate element (1,2)
Row 1 of A · Column 2 of B: (1)(0) + (3)(5) = 0 + 15 = 15
Row 1 of A · Column 2 of B: (1)(0) + (3)(5) = 0 + 15 = 15
Step 4: Calculate element (2,1)
Row 2 of A · Column 1 of B: (2)(4) + (-1)(-2) = 8 + 2 = 10
Row 2 of A · Column 1 of B: (2)(4) + (-1)(-2) = 8 + 2 = 10
Step 5: Calculate element (2,2)
Row 2 of A · Column 2 of B: (2)(0) + (-1)(5) = 0 - 5 = -5
Row 2 of A · Column 2 of B: (2)(0) + (-1)(5) = 0 - 5 = -5
Step 6: Write the result
AB = [-2 15; 10 -5]
AB = [-2 15; 10 -5]
Final Answer: [-2 15; 10 -5]
Problem 5: Matrix Multiplication (Different Dimensions)
Given P = [2 -1 3; 0 4 1] and Q = [1 5; -2 0; 3 -1], find PQ.
Solution:
Step 1: Check dimensions
P is 2×3, Q is 3×2. Since columns of P (3) = rows of Q (3), multiplication is defined.
Result will be 2×2.
P is 2×3, Q is 3×2. Since columns of P (3) = rows of Q (3), multiplication is defined.
Result will be 2×2.
Step 2: Calculate element (1,1)
Row 1 of P · Column 1 of Q: (2)(1) + (-1)(-2) + (3)(3) = 2 + 2 + 9 = 13
Row 1 of P · Column 1 of Q: (2)(1) + (-1)(-2) + (3)(3) = 2 + 2 + 9 = 13
Step 3: Calculate element (1,2)
Row 1 of P · Column 2 of Q: (2)(5) + (-1)(0) + (3)(-1) = 10 + 0 - 3 = 7
Row 1 of P · Column 2 of Q: (2)(5) + (-1)(0) + (3)(-1) = 10 + 0 - 3 = 7
Step 4: Calculate element (2,1)
Row 2 of P · Column 1 of Q: (0)(1) + (4)(-2) + (1)(3) = 0 - 8 + 3 = -5
Row 2 of P · Column 1 of Q: (0)(1) + (4)(-2) + (1)(3) = 0 - 8 + 3 = -5
Step 5: Calculate element (2,2)
Row 2 of P · Column 2 of Q: (0)(5) + (4)(0) + (1)(-1) = 0 + 0 - 1 = -1
Row 2 of P · Column 2 of Q: (0)(5) + (4)(0) + (1)(-1) = 0 + 0 - 1 = -1
Step 6: Write the result
PQ = [13 7; -5 -1]
PQ = [13 7; -5 -1]
Final Answer: [13 7; -5 -1]
Problems 6-9: Determinants
Problem 6: Determinant of 2×2 Matrix
Find the determinant of A = [3 5; -2 4].
Solution:
Step 1: Use the 2×2 determinant formula
For A = [a b; c d], det(A) = ad - bc
For A = [a b; c d], det(A) = ad - bc
Step 2: Identify elements
a = 3, b = 5, c = -2, d = 4
a = 3, b = 5, c = -2, d = 4
Step 3: Calculate
det(A) = (3)(4) - (5)(-2)
det(A) = 12 - (-10)
det(A) = 12 + 10 = 22
det(A) = (3)(4) - (5)(-2)
det(A) = 12 - (-10)
det(A) = 12 + 10 = 22
Final Answer: det(A) = 22
Problem 7: Determinant of 3×3 Matrix (Method 1)
Find the determinant of B = [2 1 3; 0 -1 4; 5 2 -2].
Solution:
Step 1: Use cofactor expansion along row 1
det(B) = a11C11 + a12C12 + a13C13
det(B) = a11C11 + a12C12 + a13C13
Step 2: Calculate C11
C11 = (-1)1+1 · det([-1 4; 2 -2])
C11 = 1 · [(-1)(-2) - (4)(2)]
C11 = 1 · [2 - 8] = -6
C11 = (-1)1+1 · det([-1 4; 2 -2])
C11 = 1 · [(-1)(-2) - (4)(2)]
C11 = 1 · [2 - 8] = -6
Step 3: Calculate C12
C12 = (-1)1+2 · det([0 4; 5 -2])
C12 = -1 · [(0)(-2) - (4)(5)]
C12 = -1 · [0 - 20] = 20
C12 = (-1)1+2 · det([0 4; 5 -2])
C12 = -1 · [(0)(-2) - (4)(5)]
C12 = -1 · [0 - 20] = 20
Step 4: Calculate C13
C13 = (-1)1+3 · det([0 -1; 5 2])
C13 = 1 · [(0)(2) - (-1)(5)]
C13 = 1 · [0 + 5] = 5
C13 = (-1)1+3 · det([0 -1; 5 2])
C13 = 1 · [(0)(2) - (-1)(5)]
C13 = 1 · [0 + 5] = 5
Step 5: Combine results
det(B) = (2)(-6) + (1)(20) + (3)(5)
det(B) = -12 + 20 + 15
det(B) = 23
det(B) = (2)(-6) + (1)(20) + (3)(5)
det(B) = -12 + 20 + 15
det(B) = 23
Final Answer: det(B) = 23
Problem 8: Determinant with Row Operations
Find the determinant of C = [1 2 3; 2 5 7; 3 7 9] using row operations.
Solution:
Step 1: Original matrix
[1 2 3; 2 5 7; 3 7 9]
[1 2 3; 2 5 7; 3 7 9]
Step 2: R2 - 2R1 → R2
[1 2 3; 0 1 1; 3 7 9]
(Row operations don't change the determinant)
[1 2 3; 0 1 1; 3 7 9]
(Row operations don't change the determinant)
Step 3: R3 - 3R1 → R3
[1 2 3; 0 1 1; 0 1 0]
[1 2 3; 0 1 1; 0 1 0]
Step 4: R3 - R2 → R3
[1 2 3; 0 1 1; 0 0 -1]
[1 2 3; 0 1 1; 0 0 -1]
Step 5: Calculate determinant (upper triangular)
For upper triangular matrix, det = product of diagonal elements
det(C) = (1)(1)(-1) = -1
For upper triangular matrix, det = product of diagonal elements
det(C) = (1)(1)(-1) = -1
Final Answer: det(C) = -1
Problem 9: Determinant Properties
If det(A) = 5 for a 3×3 matrix A, find det(2A).
Solution:
Step 1: Recall the scalar multiplication property
For an n×n matrix A and scalar k: det(kA) = kn · det(A)
For an n×n matrix A and scalar k: det(kA) = kn · det(A)
Step 2: Identify values
k = 2, n = 3 (since A is 3×3), det(A) = 5
k = 2, n = 3 (since A is 3×3), det(A) = 5
Step 3: Apply the formula
det(2A) = 23 · det(A)
det(2A) = 8 · 5
det(2A) = 40
det(2A) = 23 · det(A)
det(2A) = 8 · 5
det(2A) = 40
Important: The scalar multiplies each element, so for n×n matrix, the determinant is multiplied by kn.
Final Answer: det(2A) = 40
Problems 10-14: Inverse Matrices
Problem 10: Inverse of 2×2 Matrix (Formula Method)
Find the inverse of A = [3 1; 5 2] using the 2×2 formula.
Solution:
Step 1: Calculate the determinant
det(A) = (3)(2) - (1)(5) = 6 - 5 = 1
det(A) = (3)(2) - (1)(5) = 6 - 5 = 1
Step 2: Check if inverse exists
Since det(A) = 1 ≠ 0, the inverse exists.
Since det(A) = 1 ≠ 0, the inverse exists.
Step 3: Use the 2×2 inverse formula
For A = [a b; c d], A-1 = (1/det(A)) · [d -b; -c a]
For A = [a b; c d], A-1 = (1/det(A)) · [d -b; -c a]
Step 4: Apply the formula
A-1 = (1/1) · [2 -1; -5 3]
A-1 = [2 -1; -5 3]
A-1 = (1/1) · [2 -1; -5 3]
A-1 = [2 -1; -5 3]
Step 5: Verify (optional)
AA-1 = [3 1; 5 2][2 -1; -5 3] = [6-5 -3+3; 10-10 -5+6] = [1 0; 0 1]
AA-1 = [3 1; 5 2][2 -1; -5 3] = [6-5 -3+3; 10-10 -5+6] = [1 0; 0 1]
Final Answer: A-1 = [2 -1; -5 3]
Problem 11: Inverse Using Gauss-Jordan (2×2)
Find the inverse of B = [2 3; 1 4] using the Gauss-Jordan method.
Solution:
Step 1: Set up augmented matrix [B | I]
[2 3 | 1 0; 1 4 | 0 1]
[2 3 | 1 0; 1 4 | 0 1]
Step 2: Get 1 in position (1,1) by swapping rows
R1 ↔ R2
[1 4 | 0 1; 2 3 | 1 0]
R1 ↔ R2
[1 4 | 0 1; 2 3 | 1 0]
Step 3: Eliminate below pivot
R2 - 2R1 → R2
[1 4 | 0 1; 0 -5 | 1 -2]
R2 - 2R1 → R2
[1 4 | 0 1; 0 -5 | 1 -2]
Step 4: Get 1 in position (2,2)
(-1/5)R2 → R2
[1 4 | 0 1; 0 1 | -1/5 2/5]
(-1/5)R2 → R2
[1 4 | 0 1; 0 1 | -1/5 2/5]
Step 5: Eliminate above pivot
R1 - 4R2 → R1
[1 0 | 4/5 -3/5; 0 1 | -1/5 2/5]
R1 - 4R2 → R1
[1 0 | 4/5 -3/5; 0 1 | -1/5 2/5]
Step 6: Read the inverse from right side
B-1 = [4/5 -3/5; -1/5 2/5]
B-1 = [4/5 -3/5; -1/5 2/5]
Final Answer: B-1 = [4/5 -3/5; -1/5 2/5] or [0.8 -0.6; -0.2 0.4]
Problem 12: Inverse of 3×3 Matrix
Find the inverse of C = [1 0 2; 0 1 3; 1 2 0] using the Gauss-Jordan method.
Solution:
Step 1: Set up augmented matrix [C | I]
[1 0 2 | 1 0 0; 0 1 3 | 0 1 0; 1 2 0 | 0 0 1]
[1 0 2 | 1 0 0; 0 1 3 | 0 1 0; 1 2 0 | 0 0 1]
Step 2: Eliminate below first pivot
R3 - R1 → R3
[1 0 2 | 1 0 0; 0 1 3 | 0 1 0; 0 2 -2 | -1 0 1]
R3 - R1 → R3
[1 0 2 | 1 0 0; 0 1 3 | 0 1 0; 0 2 -2 | -1 0 1]
Step 3: Eliminate below second pivot
R3 - 2R2 → R3
[1 0 2 | 1 0 0; 0 1 3 | 0 1 0; 0 0 -8 | -1 -2 1]
R3 - 2R2 → R3
[1 0 2 | 1 0 0; 0 1 3 | 0 1 0; 0 0 -8 | -1 -2 1]
Step 4: Get 1 in position (3,3)
(-1/8)R3 → R3
[1 0 2 | 1 0 0; 0 1 3 | 0 1 0; 0 0 1 | 1/8 1/4 -1/8]
(-1/8)R3 → R3
[1 0 2 | 1 0 0; 0 1 3 | 0 1 0; 0 0 1 | 1/8 1/4 -1/8]
Step 5: Eliminate above third pivot
R1 - 2R3 → R1 and R2 - 3R3 → R2
[1 0 0 | 3/4 -1/2 1/4; 0 1 0 | -3/8 1/4 3/8; 0 0 1 | 1/8 1/4 -1/8]
R1 - 2R3 → R1 and R2 - 3R3 → R2
[1 0 0 | 3/4 -1/2 1/4; 0 1 0 | -3/8 1/4 3/8; 0 0 1 | 1/8 1/4 -1/8]
Step 6: Read the inverse
C-1 = [3/4 -1/2 1/4; -3/8 1/4 3/8; 1/8 1/4 -1/8]
C-1 = [3/4 -1/2 1/4; -3/8 1/4 3/8; 1/8 1/4 -1/8]
Final Answer: C-1 = [3/4 -1/2 1/4; -3/8 1/4 3/8; 1/8 1/4 -1/8]
Problem 13: Determine if Matrix is Invertible
Determine if D = [2 4; 3 6] is invertible. If so, find its inverse.
Solution:
Step 1: Calculate the determinant
det(D) = (2)(6) - (4)(3) = 12 - 12 = 0
det(D) = (2)(6) - (4)(3) = 12 - 12 = 0
Step 2: Check invertibility condition
A matrix is invertible if and only if its determinant is non-zero.
Since det(D) = 0, the matrix is NOT invertible.
A matrix is invertible if and only if its determinant is non-zero.
Since det(D) = 0, the matrix is NOT invertible.
Notice that the second row is 1.5 times the first row, making the rows linearly dependent. This is why the determinant is zero.
Final Answer: D is NOT invertible (singular matrix) because det(D) = 0
Problem 14: Verify Inverse
Verify that B = [1 -3; -2 7] and B-1 = [7 3; 2 1] are inverses.
Solution:
Step 1: Calculate BB-1
BB-1 = [1 -3; -2 7][7 3; 2 1]
BB-1 = [1 -3; -2 7][7 3; 2 1]
Step 2: Compute element (1,1)
(1)(7) + (-3)(2) = 7 - 6 = 1
(1)(7) + (-3)(2) = 7 - 6 = 1
Step 3: Compute element (1,2)
(1)(3) + (-3)(1) = 3 - 3 = 0
(1)(3) + (-3)(1) = 3 - 3 = 0
Step 4: Compute element (2,1)
(-2)(7) + (7)(2) = -14 + 14 = 0
(-2)(7) + (7)(2) = -14 + 14 = 0
Step 5: Compute element (2,2)
(-2)(3) + (7)(1) = -6 + 7 = 1
(-2)(3) + (7)(1) = -6 + 7 = 1
Step 6: Write result
BB-1 = [1 0; 0 1] = I
BB-1 = [1 0; 0 1] = I
Since BB-1 = I, we have verified that B-1 is indeed the inverse of B.
Final Answer: Verified. BB-1 = I, so the matrices are inverses.
Problems 15-20: Matrix Equations and Applications
Problem 15: Solve Matrix Equation AX = B
Solve for X if [2 1; 3 2]X = [5; 7].
Solution:
Step 1: Identify A and B
A = [2 1; 3 2], B = [5; 7]
We need to solve AX = B
A = [2 1; 3 2], B = [5; 7]
We need to solve AX = B
Step 2: Find A-1
det(A) = (2)(2) - (1)(3) = 4 - 3 = 1
A-1 = (1/1)[2 -1; -3 2] = [2 -1; -3 2]
det(A) = (2)(2) - (1)(3) = 4 - 3 = 1
A-1 = (1/1)[2 -1; -3 2] = [2 -1; -3 2]
Step 3: Multiply both sides by A-1
A-1AX = A-1B
X = A-1B
A-1AX = A-1B
X = A-1B
Step 4: Calculate A-1B
X = [2 -1; -3 2][5; 7]
X = [(2)(5) + (-1)(7); (-3)(5) + (2)(7)]
X = [10 - 7; -15 + 14]
X = [3; -1]
X = [2 -1; -3 2][5; 7]
X = [(2)(5) + (-1)(7); (-3)(5) + (2)(7)]
X = [10 - 7; -15 + 14]
X = [3; -1]
Final Answer: X = [3; -1]
Problem 16: System of Equations Using Matrices
Solve the system using matrices: 2x + 3y = 7, x - y = -1.
Solution:
Step 1: Write in matrix form AX = B
[2 3; 1 -1][x; y] = [7; -1]
[2 3; 1 -1][x; y] = [7; -1]
Step 2: Find A-1
A = [2 3; 1 -1]
det(A) = (2)(-1) - (3)(1) = -2 - 3 = -5
A-1 = (1/-5)[-1 -3; -1 2] = [1/5 3/5; 1/5 -2/5]
A = [2 3; 1 -1]
det(A) = (2)(-1) - (3)(1) = -2 - 3 = -5
A-1 = (1/-5)[-1 -3; -1 2] = [1/5 3/5; 1/5 -2/5]
Step 3: Multiply X = A-1B
[x; y] = [1/5 3/5; 1/5 -2/5][7; -1]
[x; y] = [1/5 3/5; 1/5 -2/5][7; -1]
Step 4: Calculate
x = (1/5)(7) + (3/5)(-1) = 7/5 - 3/5 = 4/5
y = (1/5)(7) + (-2/5)(-1) = 7/5 + 2/5 = 9/5
x = (1/5)(7) + (3/5)(-1) = 7/5 - 3/5 = 4/5
y = (1/5)(7) + (-2/5)(-1) = 7/5 + 2/5 = 9/5
Step 5: Verify
2(4/5) + 3(9/5) = 8/5 + 27/5 = 35/5 = 7
4/5 - 9/5 = -5/5 = -1
2(4/5) + 3(9/5) = 8/5 + 27/5 = 35/5 = 7
4/5 - 9/5 = -5/5 = -1
Final Answer: x = 4/5, y = 9/5
Problem 17: Cramer's Rule (2×2 System)
Use Cramer's Rule to solve: 3x + 2y = 8, x - 4y = -6.
Solution:
Step 1: Calculate det(A)
A = [3 2; 1 -4]
det(A) = (3)(-4) - (2)(1) = -12 - 2 = -14
A = [3 2; 1 -4]
det(A) = (3)(-4) - (2)(1) = -12 - 2 = -14
Step 2: Calculate det(Ax)
Replace first column with constants:
Ax = [8 2; -6 -4]
det(Ax) = (8)(-4) - (2)(-6) = -32 + 12 = -20
Replace first column with constants:
Ax = [8 2; -6 -4]
det(Ax) = (8)(-4) - (2)(-6) = -32 + 12 = -20
Step 3: Calculate det(Ay)
Replace second column with constants:
Ay = [3 8; 1 -6]
det(Ay) = (3)(-6) - (8)(1) = -18 - 8 = -26
Replace second column with constants:
Ay = [3 8; 1 -6]
det(Ay) = (3)(-6) - (8)(1) = -18 - 8 = -26
Step 4: Apply Cramer's Rule
x = det(Ax)/det(A) = -20/-14 = 10/7
y = det(Ay)/det(A) = -26/-14 = 13/7
x = det(Ax)/det(A) = -20/-14 = 10/7
y = det(Ay)/det(A) = -26/-14 = 13/7
Final Answer: x = 10/7, y = 13/7
Problem 18: Cramer's Rule (3×3 System)
Use Cramer's Rule to solve for x only: x + 2y + z = 6, 2x + y - z = 1, x - y + 2z = 5.
Solution:
Step 1: Calculate det(A)
A = [1 2 1; 2 1 -1; 1 -1 2]
Using cofactor expansion along row 1:
det(A) = 1·det([1 -1; -1 2]) - 2·det([2 -1; 1 2]) + 1·det([2 1; 1 -1])
= 1(2-1) - 2(4-(-1)) + 1(-2-1)
= 1(1) - 2(5) + 1(-3)
= 1 - 10 - 3 = -12
A = [1 2 1; 2 1 -1; 1 -1 2]
Using cofactor expansion along row 1:
det(A) = 1·det([1 -1; -1 2]) - 2·det([2 -1; 1 2]) + 1·det([2 1; 1 -1])
= 1(2-1) - 2(4-(-1)) + 1(-2-1)
= 1(1) - 2(5) + 1(-3)
= 1 - 10 - 3 = -12
Step 2: Calculate det(Ax)
Replace first column with constants:
Ax = [6 2 1; 1 1 -1; 5 -1 2]
det(Ax) = 6·det([1 -1; -1 2]) - 2·det([1 -1; 5 2]) + 1·det([1 1; 5 -1])
= 6(2-1) - 2(2-(-5)) + 1(-1-5)
= 6(1) - 2(7) + 1(-6)
= 6 - 14 - 6 = -14
Replace first column with constants:
Ax = [6 2 1; 1 1 -1; 5 -1 2]
det(Ax) = 6·det([1 -1; -1 2]) - 2·det([1 -1; 5 2]) + 1·det([1 1; 5 -1])
= 6(2-1) - 2(2-(-5)) + 1(-1-5)
= 6(1) - 2(7) + 1(-6)
= 6 - 14 - 6 = -14
Step 3: Apply Cramer's Rule for x
x = det(Ax)/det(A) = -14/-12 = 7/6
x = det(Ax)/det(A) = -14/-12 = 7/6
Final Answer: x = 7/6
Problem 19: Application - Encryption
A simple encryption uses the matrix A = [1 2; 1 3] to encode the message vector [5; 3]. What is the encoded message? How would you decode it?
Solution:
Step 1: Encode the message
Encoded = A · Message
Encoded = [1 2; 1 3][5; 3]
Encoded = A · Message
Encoded = [1 2; 1 3][5; 3]
Step 2: Calculate the product
Encoded = [(1)(5) + (2)(3); (1)(5) + (3)(3)]
Encoded = [5 + 6; 5 + 9]
Encoded = [11; 14]
Encoded = [(1)(5) + (2)(3); (1)(5) + (3)(3)]
Encoded = [5 + 6; 5 + 9]
Encoded = [11; 14]
Step 3: To decode, find A-1
det(A) = (1)(3) - (2)(1) = 3 - 2 = 1
A-1 = (1/1)[3 -2; -1 1] = [3 -2; -1 1]
det(A) = (1)(3) - (2)(1) = 3 - 2 = 1
A-1 = (1/1)[3 -2; -1 1] = [3 -2; -1 1]
Step 4: Decode by multiplying by A-1
Original = A-1 · Encoded
Original = [3 -2; -1 1][11; 14]
Original = [(3)(11) + (-2)(14); (-1)(11) + (1)(14)]
Original = [33 - 28; -11 + 14]
Original = [5; 3]
Original = A-1 · Encoded
Original = [3 -2; -1 1][11; 14]
Original = [(3)(11) + (-2)(14); (-1)(11) + (1)(14)]
Original = [33 - 28; -11 + 14]
Original = [5; 3]
Final Answer: Encoded message is [11; 14]. To decode, multiply by A-1 = [3 -2; -1 1].
Problem 20: Application - Network Flow
A simple network has three nodes. Flow equations are: x + y = 10 (into node 1), y + z = 12 (into node 2), x + z = 14 (into node 3). Find all flows x, y, z using matrices.
Solution:
Step 1: Write system in matrix form
[1 1 0; 0 1 1; 1 0 1][x; y; z] = [10; 12; 14]
[1 1 0; 0 1 1; 1 0 1][x; y; z] = [10; 12; 14]
Step 2: Set up augmented matrix
[1 1 0 | 10; 0 1 1 | 12; 1 0 1 | 14]
[1 1 0 | 10; 0 1 1 | 12; 1 0 1 | 14]
Step 3: Row reduce
R3 - R1 → R3
[1 1 0 | 10; 0 1 1 | 12; 0 -1 1 | 4]
R3 - R1 → R3
[1 1 0 | 10; 0 1 1 | 12; 0 -1 1 | 4]
Step 4: Continue reducing
R3 + R2 → R3
[1 1 0 | 10; 0 1 1 | 12; 0 0 2 | 16]
R3 + R2 → R3
[1 1 0 | 10; 0 1 1 | 12; 0 0 2 | 16]
Step 5: Solve by back substitution
From row 3: 2z = 16, so z = 8
From row 2: y + 8 = 12, so y = 4
From row 1: x + 4 = 10, so x = 6
From row 3: 2z = 16, so z = 8
From row 2: y + 8 = 12, so y = 4
From row 1: x + 4 = 10, so x = 6
Step 6: Verify
x + y = 6 + 4 = 10
y + z = 4 + 8 = 12
x + z = 6 + 8 = 14
x + y = 6 + 4 = 10
y + z = 4 + 8 = 12
x + z = 6 + 8 = 14
Final Answer: x = 6, y = 4, z = 8 (units of flow)
Ready to Test Your Knowledge?
Now that you have practiced these problems, take the Module 8 Quiz to demonstrate your mastery!