#$&* course Mth 279 5/8 2 Query 28 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Find the algebraic and geometric multiplicity of each eigenvalue and, if possible, diagonalize the matrix [7, -2, 2; 8, -1, 4; 8, 4, -1 ]. The characteristic equation of this matrix is (lambda - 3)^2 ( lambda + 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For lambda = 3, algebraic mult. = 2 and geometric mult. = 2 For lambda = 1, algebraic mult. = 1 and geometric mult. = 1 T=[1,1,0;2,0,0;0,-2,0] ?????I can’t figure out what I’m doing incorrectly,I get the vectors,x1 =[1;2;0] and x2=[1;0;-2] for `lambda=3. For lambda=1 I get x3=[0;0;0]????? I think I see my problem, we can’t find a diagonalizable matrix because x3 isn’t linearly independent right????? I’m going to say that A = [7, -2, 2; 8, -1, 4; 8, 4, -1 ], cannot be diagonalize.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Find the algebraic and geometric multiplicity of each eigenvalue and, if possible, diagonalize the matrix [ 5, -1, 1; 14, -3, 6; 5, -2, 5 ]. The characteristic equation of this matrix is (lambda - 2)^2 ( lambda -3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For lambda = 2, algebraic mult. = 2 and geometric mult. = 1 For lambda = 1, algebraic mult. = 1 and geometric mult. = 1 A is not diagonalizable due to lambda=2 having geometric mult of 1. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Solve the system y ' = [ -4, -6; 3, 5 ] y + [e^(2 t) - 2 e^t; -e^(2 t) + e^t] YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First we diagonalize the matrix where, A= [ -4, -6; 3, 5 ] Has eigenvalues `lambda_1 =-1, `lambda_2 =2 Giving us he eigenvector x_1 = [2; -1], x_2 = [-1,1] So that T matrix equals, [2, -1;-1,1] Finding T^-1 [1,1;1,2] So that T^-1*(AT) = [1,1;1,2]*( [-4, -6; 3, 5]* [2, -1;-1,1]) =[1,1;1,2]* [-2, -2; 1,2] =[-1,0;0,2] Where [-1,0;0,2] is our Diagonalized matrix. Letting y(t) = T*z(t) we get d/dt[z1;z2] = [-1,0;0,2]*[z1;z2] + [e^(2 t) - 2 e^t; -e^(2 t) + e^t] Which we can transform into 2 uncoupled scalar equations z1’ = -z1 + e^(2 t) - 2 e^t z2’ = 2(z2) - e^(2 t) + e^t Giving us solutions z1(t) = -1-e^(t) + c1e^(-t) z2(t) = -(1/2)e^(4t)+e^(3t) + c2e^(2t) Since y(t) = T*z(t) y(t) = [2, -1;-1,1]*[ -1-e^(t) + c1e^(-t); -(1/2)e^(4t)+e^(3t) + c2e^(2t)]
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Solve x '' = [ 6, 7; -15, -16] x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Rewritten as x’’ +(-Ax)= 0 Where -A = [-6,-7;15,16] First we diagonalize the matrix where, A= [-6,-7;15,16] Has eigenvalues `lambda_1 =1, `lambda_2 =9 Giving us he eigenvector x_1 = [1; -1], x_2 = [-1,(15/7)] So that T matrix equals, [1,- 1;-1,(15/7)] Finding T^-1 [(15/8),( 7/8);(7/8),(7/8)] So that T^-1*(AT) = [(15/8),( 7/8);(7/8),(7/8)]*( [-6,-7;15,16]* [1, 1;-1,(15/7)]) =[1,1;1,2]* [-2, -2; 1,2] =[1,0;0,9] Where [1,0;0,9] is our Diagonalized matrix. We now rearrange into the form w’’ + Dw = 0 So when uncoupled we have w_1’’ + w_1 = 0 w_2’’ + 9w_2 = 0 Solving characteristic equation for the equations we get r = -i,i for w_1 and r= -3i and 3i for w_2. So we have w_1(t) = Acos(t) + Bsin(t) w_2(t) = Acos(3t) + Bsin(3t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Would you believe I have worked through this whole assignment before I just realized that the diagolzination is related directed to the lambda values. Is that a coincidence that that the lambda values for almost every problem I’ve worked out was the values that were on the diagonal or is that how it always works out???????? I feel SOOO stupid sometimes!!!!