Query 28

#$&*

course MTH 279

4/20 11pm

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

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

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Your solution:

D = [-1 , 0 ; 0 , 6]

Using the fact that D is composed of the evals of the matrix, I found evals to be -1 and 6 using det[ 2- lam , 3 ; 3 , 3-lam]

@&

det[ 2- lam , 3 ; 3 , 3-lam] should be

det[ 2- lam , 3 ; 2, 3-lam]

det([2-lambda, 3; 2, 3-lambda]) = lambda^2 - 5 lambda.

Setting this equal to zero we get lambda = 0 or lambda = 5.

Corresponding eigenvectors would be

[2; -3] and [1; 1].

This gives you your T matrix for the similarity transformation; you then perform the similariy transformation to get the diagonalized matrix (i.e., the matrix relative to the eigenvector basis).

Specifically:

T := [2, 1; -3; 1])

T^-1 = [1/5, - 1/5; 3/5, 2/5]

For A = [2, 3; 2, 3] we get

T^-1 A T = [0. 0; 0, 5]

*@

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

[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:

lambda = 3 : algebraic multiplicity 2, geometric multiplicity 1

lambda = -1; algebraic multiplicity 1, geometric multiplicity 1

The matrix is not diagonalizable.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I am confident in the algrbraic multiplicity of both evals and the geometric multiplicity of the second eval. I was a bit confused on how to get the geometric multiplicity of the first eval. I used the fact that there was only 1 free parameter in the row reduced matrix A - lam*I. And 1 free parameter means 1 evec, which gives geometric multiplicity 1.

@&

That reasoning is correct.

*@

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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:

lambda = 2 : algebraic multiplicity 2, geometric mulltiplicity 1

lambda = 3 algebraic multiplicity 1, geometric multiplicity 1

Not diagonalizable.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Same issue here.

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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:

y= [.25e^2t + Ce^-2t -2te^t - 2Ke^t ; -.25e^2t + Ce^-2t + te^t + Ke^t]

I started out by finding my evals to be 2 and -1. My corresponding evecs were [-1 ; 1] and [-2 ; 1]. This gave me my T matrix to be [-1 , -2 ; 1 , 1] and T_inv to be [ 1 , 2 ; -1 , -1].

I then computed T*A*T_inv to find D = [-2 , 0 ; 0 , 1]

I came up with equations z_1' = -2z_1 -e^2t and z_2' = z_2 +e^t, which had solutions:

z= [-.25e^t + Ce^-2t ; te^t + Ke^t]

to recover y, I multiplied Tz to find my final solution.

confidence rating #$&*:

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Question: Solve

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

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Your solution:

x= [ -7A/15cos3t - 7B/15sin3t -Ccost + Dsint ; Acos3t + Bsin3t + Ccost + Dsint]

I found evals of -9 and -1 and evecs of [-7/15 ; 1] and [-1 ; 1]

T = [-7/15 , -1 ; 1 , 1] T_inv= [15/8 , 15/8 ; -15/8 , -7/8] D = [-9 , 0 ; 0 , -1]

Which led to equations of z_1'' = -9z_1 and z_2'' = -z_1,

which had solutions z= [Acos3t + Bsin3t ; Ccost + Dsint]

multiplying through by T led me to my solution.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK"

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