Linear Algebra and Differential Equations Problem Solver
Eigenvalues and Eigenvectors
Theory and General Method
Characteristic Equation: det(A - λI) = 0. For a 3x3 matrix, the equation is: λ³ - S₁λ² + S₂λ - |A| = 0
- S₁: Trace of A (sum of diagonal elements).
- S₂: Sum of principal minors =
|a₂₂ a₂₃; a₃₂ a₃₃| + |a₁₁ a₁₃; a₃₁ a₃₃| + |a₁₁ a₁₂; a₂₁ a₂₂|. - |A|: Determinant of matrix A.
Step-by-Step Procedure
- Solve the cubic equation for roots
(λ₁, λ₂, λ₃). - Select the largest numerical value for
λ_max. - Solve
(A - λ_max * I) X = 0using row reduction or simultaneous equations to find the vectorX = [x₁, x₂, x₃]ᵀ.
Solution for Given Values
Given matrix: A = [1 1 -2; -1 2 1; 0 1 -1]
S₁ = 1 + 2 + (-1) = 2S₂ = |2 1; 1 -1| + |1 -2; 0 -1| + |1 1; -1 2| = -3 - 1 + 3 = -1|A| = 1(-3) - 1(1) - 2(-1) = -2
Equation: λ³ - 2λ² - λ + 2 = 0
Roots: λ = 2, 1, -1
Largest Eigenvalue: λ = 2
Eigenvector for λ = 2
Solve (A - 2I)X = 0:[-1 1 -2; -1 0 1; 0 1 -3] [x₁; x₂; x₃] = [0; 0; 0]
- From Equation 2:
-x₁ + x₃ = 0 => x₁ = x₃ - From Equation 3:
x₂ - 3x₃ = 0 => x₂ = 3x₃ - Let
x₃ = 1, thenx₁ = 1andx₂ = 3.
Eigenvector X: [1, 3, 1]ᵀ
Matrix of Quadratic Form
Theory and General Method
A quadratic form Q(x) = Σ(a_ij * x_i * x_j) can be written in symmetric matrix form Xᵀ A X.
- Diagonal elements (a_ii): Coefficient of
x_i². - Off-diagonal elements (a_ij = a_ji):
(Coefficient of x_i * x_j) / 2.
Solution for Given Values
Given: Q = x₁² + 2x₂² + 3x₃² + x₄² + 4x₁x₂ - 2x₁x₃ - 2x₁x₄ + 4x₂x₃ - 6x₂x₄ + 8x₃x₄
a₁₁ = 1, a₂₂ = 2, a₃₃ = 3, a₄₄ = 1a₁₂ = a₂₁ = 4/2 = 2a₁₃ = a₃₁ = -2/2 = -1a₁₄ = a₄₁ = -2/2 = -1a₂₃ = a₃₂ = 4/2 = 2a₂₄ = a₄₂ = -6/2 = -3a₃₄ = a₄₃ = 8/2 = 4
Matrix A:[ 1 2 -1 -1][ 2 2 2 -3][-1 2 3 4][-1 -3 4 1]
Linear Transformation Mapping
Theory and General Method
To find T(x,y) given T(v₁) = w₁ and T(v₂) = w₂:
- Write an arbitrary vector
(x,y)as a linear combination of basis vectors:(x,y) = a*v₁ + b*v₂. - Solve the linear system for coefficients
aandbin terms ofxandy. - Apply the linearity property:
T(x,y) = a*T(v₁) + b*T(v₂).
Solution for Given Values
Given: T(1,2) = (3,0) and T(2,1) = (1,2)
(x,y) = a(1,2) + b(2,1)x = a + 2b(Eq. 1)y = 2a + b(Eq. 2)- Solve for a and b: From Eq. 2,
b = y - 2a. Substitute into Eq. 1:x = a + 2(y - 2a) = -3a + 2y => a = (2y - x)/3
Substitute back:b = (2x - y)/3 - Apply Transformation:
T(x,y) = a*T(1,2) + b*T(2,1)T(x,y) = [(2y - x)/3]*(3,0) + [(2x - y)/3]*(1,2)T(x,y) = ( (6y - 3x + 2x - y)/3 , (0 + 4x - 2y)/3 )T(x,y) = ( (-x + 5y)/3 , (4x - 2y)/3 )
Nature of Quadratic Form
Theory and General Method
For a quadratic form f(x,y,z) with symmetric matrix A, calculate the Leading Principal Minors (D₁, D₂, D₃):
D₁ = |a₁₁|D₂ = |a₁₁ a₁₂; a₂₁ a₂₂|D₃ = det(A)
Classification Rules
- Positive Definite:
D₁ > 0, D₂ > 0, D₃ > 0 - Negative Definite:
D₁ < 0, D₂ > 0, D₃ < 0(alternating signs starting negative) - Indefinite: Minors do not follow these patterns and
det(A) ≠ 0 - Positive/Negative Semi-definite:
det(A) = 0with appropriate non-negative or non-positive signs.
Solution for Given Values
Given: f(x,y,z) = 3x² + 5y² + 3z² - 2xy + 2xz - 2yz
Matrix A: [3 -1 1; -1 5 -1; 1 -1 3]
D₁ = 3 > 0D₂ = |3 -1; -1 5| = 15 - 1 = 14 > 0D₃ = det(A) = 3(15-1) - (-1)(-3+1) + 1(1-5) = 42 - 2 - 4 = 36 > 0
Since D₁ > 0, D₂ > 0, D₃ > 0, the quadratic form is Positive Definite.
Gram-Schmidt Orthogonalization Process
Theory and General Method
Given an independent basis B = {v₁, v₂, v₃}, compute the orthogonal basis {u₁, u₂, u₃} using the inner product ⟨x,y⟩ = x₁y₁ + x₂y₂ + x₃y₃:
- Step 1:
u₁ = v₁ - Step 2:
u₂ = v₂ - [⟨v₂, u₁⟩ / ||u₁||²] * u₁ - Step 3:
u₃ = v₃ - [⟨v₃, u₁⟩ / ||u₁||²] * u₁ - [⟨v₃, u₂⟩ / ||u₂||²] * u₂
(Note: ||u||² = ⟨u, u⟩)
Solution for Given Values
Given: v₁ = (1,0,1), v₂ = (1,0,-1), v₃ = (0,3,4)
- u₁ = (1,0,1);
||u₁||² = 1² + 0² + 1² = 2 ⟨v₂, u₁⟩ = (1)(1) + (0)(0) + (-1)(1) = 0u₂ = v₂ - (0/2)*u₁
u₂ = (1,0,-1);||u₂||² = 1² + 0² + (-1)² = 2⟨v₃, u₁⟩ = (0)(1) + (3)(0) + (4)(1) = 4⟨v₃, u₂⟩ = (0)(1) + (3)(0) + (4)(-1) = -4u₃ = (0,3,4) - (4/2)*(1,0,1) - (-4/2)*(1,0,-1)u₃ = (0,3,4) - (2,0,2) + (2,0,-2)
u₃ = (0,3,0)
Orthogonal Basis: {(1,0,1), (1,0,-1), (0,3,0)}
Largest Eigenvalue and Eigenvector Analysis
Matrix A:[4 -1 6][2 1 6][2 -1 8]
Step 1: Characteristic Equation
S₁ = 4 + 1 + 8 = 13S₂ = |1 6; -1 8| + |4 6; 2 8| + |4 -1; 2 1| = 14 + 20 + 6 = 40|A| = 4(14) - (-1)(4) + 6(-4) = 56 + 4 - 24 = 36
Equation: λ³ - 13λ² + 40λ - 36 = 0
Trial root λ = 2: 8 - 52 + 80 - 36 = 0.
Factoring: (λ - 2)(λ² - 11λ + 18) = (λ - 2)(λ - 2)(λ - 9) = 0.
Eigenvalues: λ = 9, 2, 2
Largest Eigenvalue: λ = 9
Step 2: Eigenvector for λ = 9
Solve (A - 9I)X = 0:[-5 -1 6] [x₁] [0][ 2 -8 6] [x₂] = [0][ 2 -1 -1] [x₃] [0]
Using Cramer’s rule on Row 1 and Row 2:x₁ / (-6 - (-48)) = -x₂ / (-30 - 12) = x₃ / (40 - (-2))x₁ / 42 = x₂ / 42 = x₃ / 42 => x₁ / 1 = x₂ / 1 = x₃ / 1
Eigenvector X for λ = 9: [1, 1, 1]ᵀ
Basis of a Vector Space in R³
Theory and General Steps
For three vectors in R³, construct matrix A = [v₁ v₂ v₃].
- If
det(A) ≠ 0, the vectors are linearly independent and form a basis ofR³. - If
det(A) = 0, the vectors are linearly dependent and do not form a basis.
Set 1 Analysis
Vectors: v₁ = (1,1,0), v₂ = (2,2,3), v₃ = (1,1,-1)
Matrix A:[1 2 1][1 2 1][0 3 -1]
det(A) = 1(-2 - 3) - 2(-1 - 0) + 1(3 - 0) = -5 + 2 + 3 = 0
Since det(A) = 0, the vectors are linearly dependent.
Result: Set 1 does not form a basis of R³.
Set 2 Analysis
Vectors: v₁ = (1,1,0), v₂ = (2,1,1), v₃ = (1,-1,2)
Matrix A:[1 2 1][1 1 -1][0 1 2]
det(A) = 1(2 - (-1)) - 2(2 - 0) + 1(1 - 0) = 3 - 4 + 1 = 0
Since det(A) = 0, the vectors are linearly dependent.
Result: Set 2 does not form a basis of R³.
Differential Equations in Electrical Circuits
Analysis of R-C Circuits
Given: R = 20, C = 0.01, E(t) = 200e⁻⁵ᵗ, q(0) = 0
Differential Equation: R(dq/dt) + (1/C)q = E(t)20(dq/dt) + (1/0.01)q = 200e⁻⁵ᵗ20(dq/dt) + 100q = 200e⁻⁵ᵗ => dq/dt + 5q = 10e⁻⁵ᵗ
Standard Linear Form: dq/dt + Pq = Q (where P = 5, Q = 10e⁻⁵ᵗ)
Integrating Factor (IF): e^∫P dt = e⁵ᵗ
Solution:q * IF = ∫(Q * IF) dt + cq * e⁵ᵗ = ∫(10e⁻⁵ᵗ * e⁵ᵗ) dt + c = ∫10 dt + cq * e⁵ᵗ = 10t + c => q(t) = (10t + c)e⁻⁵ᵗ
At t = 0, q = 0:0 = (0 + c)(1) => c = 0
Charge Equation: q(t) = 10t e⁻⁵ᵗ Coulombs
Analysis of L-R Circuits
Given: V = 100, R = 20, L = 10, i(0) = 0. Find i(2).
Differential Equation: L(di/dt) + Ri = V10(di/dt) + 20i = 100 => di/dt + 2i = 10
Integrating Factor (IF): e^∫2 dt = e²ᵗi * e²ᵗ = ∫(10e²ᵗ) dt + c = 5e²ᵗ + ci(t) = 5 + ce⁻²ᵗ
At t = 0, i = 0:0 = 5 + c => c = -5i(t) = 5(1 - e⁻²ᵗ)
At t = 2 seconds:i(2) = 5(1 - e⁻⁴) Amperes