2 cor rect 3 4 explanation if a can be diagonalized

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2. cor-rect3. 4. Explanation:If A can be diagonalized by ,then .Now A can be diagonalized if we can find an eigenbasis of R2 of eigenvectors v1, v2 corresponding to eigenvalues λ1, λ2, for then: .But ,i.e., λ1 = 2 and λ2 = 1. Corresponding eigenvectors are v, vso 1.d2 = 5, d3 = 4, 0 0 P= 0 1 1 3 TRUE of A ,.Consequently, 011 10.0 points Find a matrix P and d2, d3 so that ,is a diagonalization of the matrix .100
HW 10 gilbert 7 2.d2 = 5, d3 = 4, 1 0 P= 0 1 0 0 3 13.d2 = 4, d3 = 1 0 P= 0 1 0 3 correct4.d2 = 4, d3 = 0 0 P= 0 1 0 1 3 05.d2 = 5, d3 = 4, 1 0 P= 0 0 0 1 6.d2 = 4, d3 = 1 0 P= 0 0 0 1 Explanation:The entries 5, d2, d3 in the diagonal matrix are the respective eigenvalues λ1, λ2, λ3 of A. But So λ1 = 5, λ2 = 4, λ3 = Now let u1, u2, u3 be eigenvectors of corresponding to λ1, λ2, λ3 respectively. Since the eigenvalues are distinct, P = [u1 u2 uhas orthogonal columns. is a diagonalization of ATo determine u1 we row reduce AλI with 5: rref(.Thus u.To determine u2 we row reduce A λI with 4: rref(.
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HW 10 gilbert 8 Thus u.To determine u3 we row reduce A λI with = 5: rref(.Thus, finally, u.Consequently, d2 = 4, d3 = 012 3.4.5.6.Explanation:To begin, we must find the eigenvectors and eigenvalues of A. To do this, we will use the characteristic equation, det(A λI) = 0. That is, we will look for the zeros of the characteristic polynomial. det(A λI) = (2 λ)(5 λ) + 12 = λ2 + 3λ = (λ + 1)(λ + 2) = 0P= 1 0 0 1 0 3 .
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10.0 points Find a matrix P so that is a diagonalization of the matrix 1. correct2.
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.So .Now to find the eigenvectors of A, we will solve for the nontrivial solution of the characteristic equation by row reducing the related augmented matrices: while
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HW 10 gilbert 9 So, P = [u1 u2] and A = PDPis a diagonalization of AConsequently, .013 10.0 points Let A be a 2 × 2 matrix with eigenvalues 2 and 1 and corresponding eigenvectors v.Let {xk} be a solution of the difference equation x.Compute x1correct Explanation:To find x1 we must compute Ax0. Now, express x0 in terms of v1 and v2. That is, find c1 and c2 such that x0 = c1v1 + c2v2. This is certainly possible because the eigenvectors v1 and v2 are linearly independent (by inspection and also because correspond to distinct eigenvalues) and hence form a basis for R2. The row reduction [v1 v2 shows that x0 = v1 v2. Since v1 and eigenvectors (for the eigenvalues 2 and respectively): xConsequently, 014 10.0 points Let A be a 2 × 2 matrix with eigenvalues 5 and 4 and corresponding eigenvectors v.Determine the solution {xk} of the difference equation xk+1 = Axk, 1.xk = 8(5)k v1 6(4)k 2.xk = 8(5)k v1 + 3(4)k 3.xk = 8(5)k v1 3(4)k 4.xk = 4(5)k v1 + 6(4)k 5.xk = 4(5)k v1 6(4)k 6.xk = 4(5)k v1 3(4)k v2 correctD= 1 0 0 2 ,P= 4 x1= 2 3
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HW 10 gilbert 10 Explanation:Since v1 and v2 are eigenvectors

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