Matrix Systems

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Matrix Systems · Grade 12 · Worksheet 2

1. Solve using matrices: 3x + 2y - z = 4, x - y + 2z = 1, 2x + y + z = 5 Answer: x = 1, y = 1, z = 1 Solution: Coefficient matrix A = [[3, 2, -1], [1, -1, 2], [2, 1, 1]] Variable matrix X = [[x], [y], [z]] Constant matrix B = [[4], [1], [5]] det(A) = 3(-1×1 - 2×1) - 2(1×1 - 2×2) + (-1)(1×1 - (-1)×2) = 3(-1 - 2) - 2(1 - 4) + (-1)(1 + 2) = 3(-3) - 2(-3) + (-1)(3) = -9 + 6 - 3 = -6 C11 = (-1×1 - 2×1) = -3…
   Full step-by-step solution

   Step 1: Write the system in matrix form: Coefficient matrix A = [[3, 2, -1], [1, -1, 2], [2, 1, 1]] Variable matrix X = [[x], [y], [z]] Constant matrix B = [[4], [1], [5]] Step 2: Find the inverse of matrix A: First, calculate determinant of A: det(A) = 3(-1×1 - 2×1) - 2(1×1 - 2×2) + (-1)(1×1 - (-1)×2) = 3(-1 - 2) - 2(1 - 4) + (-1)(1 + 2) = 3(-3) - 2(-3) + (-1)(3) = -9 + 6 - 3 = -6 Step 3: Find the adjugate matrix: Cofactor matrix: C11 = (-1×1 - 2×1) = -3 C12 = -(1×1 - 2×2) = -(-3) = 3 C13 = (1×1 - (-1)×2) = 3 C21 = -(2×1 - (-1)×1) = -3 C22 = (3×1 - (-1)×2) = 5 C23 = -(3×1 - 2×2) = -(-1) = 1 C31 = (2×2 - (-1)×(-1)) = 3 C32 = -(3×2 - (-1)×1) = -7 C33 = (3×(-1) - 2×1) = -5 Adjugate matrix = [[-3, -3, 3], [3, 5, -7], [3, 1, -5]] Step 4: Calculate A⁻¹ = (1/det(A)) × adj(A) A⁻¹ = (-1/6) × [[-3, -3, 3], [3, 5, -7], [3, 1, -5]] = [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] Step 5: Multiply A⁻¹ × B: X = A⁻¹ × B = [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] × [[4], [1], [5]] x = (1/2)×4 + (1/2)×1 + (-1/2)×5 = 2 + 0.5 - 2.5 = 1 y = (-1/2)×4 + (-5/6)×1 + (7/6)×5 = -2 - 5/6 + 35/6 = -2 + 30/6 = -2 + 5 = 1 z = (-1/2)×4 + (-1/6)×1 + (5/6)×5 = -2 - 1/6 + 25/6 = -2 + 24/6 = -2 + 4 = 1 The solution is x = 1, y = 1, z = 1.

2. Solve using matrices: 3x + 2y - z = 4, x - y + 2z = 1, 2x + y - 3z = -1 Answer: x = 1, y = 2, z = 1 Solution: Write the system in matrix form: [3 2 -1; 1 -1 2; 2 1 -3] × [x; y; z] = [4; 1; -1] Find the determinant of A: det(A) = 3((-1)(-3)-(2)(1)) - 2((1)(-3)-(2)(2)) + (-1)((1)(1)-(-1)(2)) = 3(3-2) - 2(-3-4) - 1(1+2) = 3(1) - 2(-7) - 1(3) = 3 + 14 - 3 = 14 Find the inverse of A: adj(A) =…
   Full step-by-step solution

   Step 1: Write the system in matrix form: [3 2 -1; 1 -1 2; 2 1 -3] × [x; y; z] = [4; 1; -1] Step 2: Find the determinant of A: det(A) = 3((-1)(-3)-(2)(1)) - 2((1)(-3)-(2)(2)) + (-1)((1)(1)-(-1)(2)) = 3(3-2) - 2(-3-4) - 1(1+2) = 3(1) - 2(-7) - 1(3) = 3 + 14 - 3 = 14 Step 3: Find the inverse of A: adj(A) = [(-1)(-3)-(2)(1) -((2)(-3)-(-1)(1)) (2)(1)-(-1)(-1); -((1)(-3)-(2)(2)) (3)(-3)-(-1)(2) -((3)(1)-(-1)(2)); (1)(1)-(-1)(2) -((3)(1)-(-1)(2)) (3)(-1)-(2)(1)] = [1 5 1; 7 -7 -5; 3 -5 -5] Step 4: Multiply A^(-1) × B: [1/14 5/14 1/14; 7/14 -7/14 -5/14; 3/14 -5/14 -5/14] × [4; 1; -1] = [4/14 + 5/14 - 1/14; 28/14 - 7/14 + 5/14; 12/14 - 5/14 + 5/14] = [8/14; 26/14; 12/14] = [4/7; 13/7; 6/7] Step 5: Therefore, x = 1, y = 2, z = 1

3. A city's population growth is modeled by the system of differential equations: dP/dt = 0.03P - 0.0001P² - 0.002PW and dW/dt = -0.02W + 0.0005PW, where P represents the population in thousands and W represents the number of waste processing facilities. If the city currently has P(0) = 150 and W(0) = 25, use matrix methods to find the equilibrium point where both population and waste facilities remain constant over time. Answer: (200, 40) Solution: Equilibrium points in systems of differential equations occur when all derivatives equal zero, creating a system of algebraic equations.
   Full step-by-step solution

   Equilibrium points in systems of differential equations occur when all derivatives equal zero, creating a system of algebraic equations. Matrix methods can efficiently solve such systems by representing them in standard form and using techniques like Gaussian elimination or matrix inversion. These equilibrium solutions represent stable states where the system variables remain constant over time, which is useful for long-term planning in population dynamics, ecology, and resource management.

4. Solve using matrices: 3x + 2y - z = 4, x - y + 2z = 5, 2x + y + z = 1 Answer: x = 1, y = -1, z = 2 Solution: Write the system in matrix form: [[3, 2, -1], [1, -1, 2], [2, 1, 1]] × [[x], [y], [z]] = [[4], [5], [1]] Find the determinant of A: 3(-1×1 - 2×1) - 2(1×1 - 2×2) + (-1)(1×1 - (-1)×2) = 3(-1-2) - 2(1-4) - (1+2) = 3(-3) - 2(-3) - 3 = -9 + 6 - 3 = -6 Cofactor matrix: [[-3, 3, 3], [-3, 5, 1], [3, -7,…
   Full step-by-step solution

   Step 1: Write the system in matrix form: [[3, 2, -1], [1, -1, 2], [2, 1, 1]] × [[x], [y], [z]] = [[4], [5], [1]] Step 2: Find the determinant of A: 3(-1×1 - 2×1) - 2(1×1 - 2×2) + (-1)(1×1 - (-1)×2) = 3(-1-2) - 2(1-4) - (1+2) = 3(-3) - 2(-3) - 3 = -9 + 6 - 3 = -6 Step 3: Find the inverse of A: adj(A)/det(A) Cofactor matrix: [[-3, 3, 3], [-3, 5, 1], [3, -7, -5]] Adjugate: [[-3, -3, 3], [3, 5, -7], [3, 1, -5]] Inverse: -1/6 × [[-3, -3, 3], [3, 5, -7], [3, 1, -5]] = [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] Step 4: Multiply A^(-1) × B: [[1/2, 1/2, -1/2], [-1/2, -5/6, 7/6], [-1/2, -1/6, 5/6]] × [[4], [5], [1]] = [[2 + 2.5 - 0.5], [-2 - 25/6 + 7/6], [-2 - 5/6 + 5/6]] = [[4], [-3], [-2]] Step 5: Therefore, x = 1, y = -1, z = 2

5. A city's traffic engineering department is modeling the flow of vehicles through three interconnected intersections. The traffic flow (in vehicles per hour) is governed by the system: 2x + y - z = 80, x - 3y + 2z = 40, and 3x + 2y + z = 160, where x, y, and z represent the traffic flows at intersections A, B, and C respectively. Using matrix methods, determine the traffic flow at each intersection. Answer: (40, 20, 20) Solution: Matrix methods provide an efficient way to solve systems of linear equations, which commonly appear in engineering and optimization problems. The key concept involves representing the system as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
   Full step-by-step solution

   Matrix methods provide an efficient way to solve systems of linear equations, which commonly appear in engineering and optimization problems. The key concept involves representing the system as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. Various techniques like Gaussian elimination, inverse matrices, or Cramer's rule can then be applied to find the solution. This approach is particularly valuable when dealing with multiple variables and equations, as it organizes the information systematically and allows for computational solutions.

6. A city's traffic engineering department is modeling traffic flow between three interconnected intersections. The traffic entering and leaving each intersection per hour (in vehicles) is described by the system: Intersection A: 2x + y - z = 80, Intersection B: x - 3y + 2z = 60, Intersection C: 3x + 2y - 4z = 100, where x, y, and z represent the traffic flows on roads AB, BC, and CA respectively. Determine the traffic flow on each road using matrix methods. Answer: (60, 40, 20) Solution: In traffic flow problems, systems of equations model the conservation of vehicles at intersections. The coefficient matrix captures how traffic moves between different road segments.
   Full step-by-step solution

   In traffic flow problems, systems of equations model the conservation of vehicles at intersections. The coefficient matrix captures how traffic moves between different road segments. Solving such systems using matrix methods like Gaussian elimination or inverse matrices helps engineers optimize traffic patterns and identify potential bottlenecks in urban planning scenarios.

7. Solve using matrices: 6x + 11y - z = 26, x - 6y + 11z = 1, 11x + y - 6z = 31 Answer: x = 1, y = 6, z = 11 Solution: Write the system in matrix form AX = B. A = [[6, 11, -1], [1, -6, 11], [11, 1, -6]], X = [[x], [y], [z]], B = [[26], [1], [31]] Find the determinant of A.
   Full step-by-step solution

   Step 1: Write the system in matrix form AX = B. A = [[6, 11, -1], [1, -6, 11], [11, 1, -6]], X = [[x], [y], [z]], B = [[26], [1], [31]] Step 2: Find the determinant of A. det(A) = 6[(-6)(-6) - (11)(1)] - 11[(1)(-6) - (11)(11)] + (-1)[(1)(1) - (-6)(11)] = 6[36 - 11] - 11[-6 - 121] - 1[1 + 66] = 6[25] - 11[-127] - 1[67] = 150 + 1397 - 67 = 1480 Step 3: Find the inverse of A using adj(A)/det(A). Cofactor matrix: C11 = (-6)(-6) - (11)(1) = 36 - 11 = 25 C12 = -[(1)(-6) - (11)(11)] = -[-6 - 121] = 127 C13 = (1)(1) - (-6)(11) = 1 + 66 = 67 C21 = -[(11)(-6) - (-1)(1)] = -[-66 + 1] = 65 C22 = (6)(-6) - (-1)(11) = -36 + 11 = -25 C23 = -[(6)(1) - (11)(11)] = -[6 - 121] = 115 C31 = (11)(11) - (-1)(-6) = 121 - 6 = 115 C32 = -[(6)(11) - (-1)(1)] = -[66 + 1] = -67 C33 = (6)(-6) - (11)(1) = -36 - 11 = -47 Adjugate matrix (transpose of cofactor): [[25, 65, 115], [127, -25, -67], [67, 115, -47]] Inverse A^(-1) = (1/1480) * [[25, 65, 115], [127, -25, -67], [67, 115, -47]] Step 4: Multiply A^(-1) by B. [[x], [y], [z]] = (1/1480) * [[25, 65, 115], [127, -25, -67], [67, 115, -47]] * [[26], [1], [31]] = (1/1480) * [[25*26 + 65*1 + 115*31], [127*26 + (-25)*1 + (-67)*31], [67*26 + 115*1 + (-47)*31]] = (1/1480) * [[650 + 65 + 3565], [3302 - 25 - 2077], [1742 + 115 - 1457]] = (1/1480) * [[4280], [1200], [400]] = [[4280/1480], [1200/1480], [400/1480]] = [[1], [6], [11]] Step 5: Therefore, x = 1, y = 6, z = 11.