Query 28

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course MTH 279

Query 28 Differential Equations*********************************************

Question: Diagonalize the matrix [2, 3; 2, 3].

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Your solution:

det([2-λ, 3; 2, 3-λ]) = 0;

λ^2 - 5λ = 0,

λ = 0,5.

For λ = 0:

[2-0, 3; 2, 3-0] = [2, 3; 2, 3]. Using row reduction, we get:

[2, 3; 0, 0]. Interpreted algebraically, we get 2x + 3y = 0. Translating this into vector form: v_1 = [3; -2].

For λ = 5:

[2-5, 3; 2, 3-5] = [-3, 3; 2, -2]. Using row reduction, we get:

[-1, 1; 0, 0]. Interpreted algebraically, we get x = y. Translating this into vector form: v_2 = [1;1].

Our diagonalized matrix is:

[0, 0; 0, 5].

Confidence rating: 3

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Given Solution:

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Self-critique (if necessary): OK

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Question: Find the algebraic and geometric multiplicity of each eigenvalue and, if possible, diagonalize the matrix

[7, -2, 2; 8, -1, 4; 8, 4, -1 ].

The characteristic equation of this matrix is (lambda - 3)^2 ( lambda + 1).

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Your solution:

The characteristic equation of this matrix is not (λ-3)^2(λ+1), it is -(λ-7)(λ-3)(λ+5). Therefore, we cannot continue with this problem as stated. The algebraic multiplicity of the given λ = 3 is 2, and of λ = -1 is 1. The geometric multiplicity cannot be found.

The algebraic multiplicity of the actual eigenvalues is 1 for each λ = -5, λ = 3, λ = 7. The geometric multiplicity of each of the λ is 1 as well. Since the algebraic and geometric multiplicities of each λ are equal, the matrix can be diagonalized. The diagonalized matrix is:

[-5, 0, 0; 0, 3, 0; 0, 0, 7].

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Question: Find the algebraic and geometric multiplicity of each eigenvalue and, if possible, diagonalize the matrix

[ 5, -1, 1; 14, -3, 6; 5, -2, 5 ].

The characteristic equation of this matrix is (lambda - 2)^2 ( lambda -3).

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Your solution:

The algebraic multiplicity for λ = 2 is 2, and the algebraic multiplicity for λ = 3 is 1. We begin our search for eigenvectors with λ = 2, since we know that in order for this matrix to be diagonalizable its geometric multiplicity must be 2.

For λ = 2:

[3, -1, 1; 14, -5, 6; 5, -2, 5]. We make this an augmented matrix and use row reduction methods to determine [1, 0, -1; 0, 1, -4; 0, 0, 0]. We translate this into algebra and find that x = z, and y = 4z. Translated into vector form, we get [1; 4; 1]. This is our first eigenvector, but we discover that there is no more nonzero vectors that satisfy [A-λI]x = 0. This means that the geometric multiplicity of λ = 2 is just 1. Since the geometric and algebraic multiplicities don’t equate, we know that the matrix is not diagonalizable. The geometric multiplicity of λ = 3 is 1.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Question: Solve the system

y ' = [ -4, -6; 3, 5 ] y + [e^(2 t) - 2 e^t; -e^(2 t) + e^t]

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Your solution:

A = [ -4, -6; 3, 5 ].

det(A-λI) = det([-4-λ, -6; 3, 5-λ]) = 0,

λ = -1, 2.

For λ = 2:

[-4-2, -6; 3, 5-2] = [-6, -6; 3, 3]. Using row reduction methods, we get [-1, -1; 0, 0]. Algebraically, this is y = -x. In vector form, this is v_1 = [1; -1].

For λ = -1:

[-4+1, -6; 3, 5+1] = [-3, -6; 3, 6]. Using row reduction methods, we get [-1, -2; 0, 0]. Algebraically, this is y = -x/2. In vector form, this is v_2 = [2; -1].

y_c = c_1e^(2t)[1;-1] + c_2e^(-t)[2;-1] = [e^(2t), 2e^(-t); -e^(2t), -e^(-t)][c_1;c_2].

Now we tackle the task of finding our particular solution, y_p.

g(t) = [e^(2 t) - 2 e^t; -e^(2 t) + e^t] can be broken down into e^(2t)[1;-1] + e^(t)[-2;1].

We assume that the first particular solution will come in the form u = e^(2t)a with a being a constant 2x1 matrix. u’ = 2e^(2t)a.

We form the new differential equation: u’ = Au + g(t). A is still the matrix from our original DE, and g(t) is our new nonhomogeneity = e^(2t) [1;-1].

2e^(2t)a = Ae^(2t)a + e^(2t)g,

2a = Aa + g,

2a-Aa = g,

(2I-A)a = g,

[6, 6; -3,-3]a = g.

I’m a bit stumped, since 2 is one of our original eigenvalues, I’m not sure how we can go from here. The (2I-A) above gives us the eigenvector for eigenvalue 2.

We assume the second particular solution takes the form v = e^(t)a. v’ = e^(t)a. g is [-2;1]. Plugging in, we get:

e^(t)a = Ae^(t)a + e^(t)g,

a = Aa + g,

a-Aa = g,

(I-A)a = g

[5,6; -3, -4]a = [-2; 1].

a = [-1; ½].

y_p = ??? + e^(t) [-1; ½]

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I’m a little confused because I came up with a particular solution that was a repeat of the eigenpair used earlier. I’m not sure where I should have gone from there.

@&

The similarity transformation T^-1 A T can still be applied to both sides, where T is the matrix of eigenvectors T = [ 1, 2; -1, -1].

This transformation diagonalizes the A matrix, and applied to the column vector `g = [e^(2 t) - 2 e^t; -e^(2 t) + e^t] yields a column vector `h.

You end up with the equations

z_i ' = lambda_i z_i + h_i

where lambda_1 = 2 and labmda_1 = -1. For each value of i (in this case i = 1 and i = 2) you have first-order linear nonhomogeneous equation is z_i, so you easily obtain solutions z_1 and z_2.

Applying T^-1 to the column vector `z = [z_1; z_2], you get your solution vector [y_1; y_2].

This will work despite the fact that part of the nonhomogeneous part of the original solution is a solution to the homogeneous equation.

*@

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Question: Solve

x '' = [ 6, 7; -15, -16] x

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Your solution:

We need to uncouple the DEs, so we start by finding the eigenpairs of A.

det(A-λI) = 0,

det([6-λ, 7; -15, -16-λ]) = 0,

λ^2 + 10λ + 9 = 0,

λ = -9, -1.

For λ = -9:

[6+9, 7; -15, -16+9] = [15, 7; -15, -7] [15, 7; 0, 0]. This translates into y=c, 15x+7y=0. Solving, we get v_1 = c[-7;15].

For λ = -1:

[6+1, 7; -15, -16+1] = [7, 7; -15, -15] [1,1; 0,0]. This translates into y=c, x+y = 0. Solving, we get v_2 = c[-1,1].

This makes our T matrix T=[v_1,v_2]=[-7,-1; 15, 1]. Our T^(-1)=[1/8, 1/8; -15/8, -7/8].

We make the change of variable x=Tz, x’=Tz’, x’’=Tz’’. Substituting in, we get:

Tz’’=ATz. This simplifies to z’’=T^(-1)ATz. We know that T^(-1)AT is the diagonalized matrix of A. We compute this, and get D=[-9,0;0,-1].

We now have:

[z_1’’;z_2’’] = [-9z_1; -z_2]. Our DEs are successfully decoupled. We solve each second order DE separately, and get:

z_1 = (A+B)cos(3t)+(B-A)sin(3t) and z_2 = (A+B)cos(t)+(B-A)sin(t). We change our variables back through x=Tz.

x = Tz = [-7,-1; 15, 1]*[z_1;z_2] = [-7((A+B)cos(3t)+(B-A)sin(3t))-((A+B)cos(t)+(B-A)sin(t)); 15((A+B)cos(3t)+(B-A)sin(3t))+ (A+B)cos(t)+(B-A)sin(t)].

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK "

Self-critique (if necessary):

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Self-critique rating:

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&#Good responses. See my notes and let me know if you have questions. &#