Linear Algebra and Its Applications
6th Edition -- Pearson -- ISBN: 978-0-321-98238-4
1,680 solutions
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10 chapters
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98% solved
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★ 4.8 (2,420 ratings)
Chapters
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Chapter 1: Linear Equations
186 solutions
1.1 Systems of Linear Equations24
1.2 Row Reduction28
1.3 Vector Equations22
Chapter 2: Matrix Algebra
204 solutions
Chapter 3: Determinants
142 solutions
Chapter 4: Vector Spaces
186 solutions
Chapter 5: Eigenvalues
168 solutions
5.1 Eigenvectors32
5.2 Characteristic Equation28
5.3 Diagonalization36
Chapter 6: Orthogonality
156 solutions
Chapter 7: Symmetric Matrices
124 solutions
Section 5.1: Eigenvectors & Eigenvalues
32 solved problems -- All step-by-step
Section 5.1 -- Problem 12
Find the eigenvalues and corresponding eigenvectors of A = [[3, 1], [1, 3]].
Step 1 Set up the characteristic equation
det(A - lambda*I) = 0
Subtract lambda from each diagonal entry and take the determinant.
Step 2 Compute the determinant
(3 - lambda)^2 - 1 = 0
Expanding: lambda^2 - 6*lambda + 9 - 1 = lambda^2 - 6*lambda + 8 = 0
Step 3 Factor
(lambda - 4)(lambda - 2) = 0
Eigenvalues: lambda_1 = 4, lambda_2 = 2
Step 4 Find eigenvector for lambda = 4
(A - 4I)v = 0 => v_1 = [1, 1]
Solving [[-1, 1], [1, -1]][x1, x2] = [0, 0] gives x1 = x2. Choose v = [1, 1].
Step 5 Find eigenvector for lambda = 2
(A - 2I)v = 0 => v_2 = [1, -1]
Solving [[1, 1], [1, 1]][x1, x2] = [0, 0] gives x1 = -x2. Choose v = [1, -1].
lambda_1 = 4, v_1 = [1, 1] | lambda_2 = 2, v_2 = [1, -1]
Section 5.1 -- Problem 13
Is lambda = 0 an eigenvalue of [[4, -6], [2, -3]]? Explain.
Step 1 Check if A is singular
det(A) = (4)(-3) - (-6)(2) = -12 + 12 = 0
Since det(A) = 0, the matrix is singular. Lambda = 0 IS an eigenvalue because Ax = 0x = 0 has a nontrivial solution.
Yes -- lambda = 0 is an eigenvalue (det(A) = 0)