2 solve x 2 1 6 2 5 2 x solution we begin by finding

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2. Solve x 0 = 2 1 6 0 2 5 0 0 2 x . Solution. We begin by finding the eigenvalues. Because our matrix is an upper triangular matrix, the eigenvalues are the diagonal entries, hence λ = 2 mult. 3. We now find the eigenvectors for λ = 2: 2 - 2 1 6 0 0 2 - 2 5 0 0 0 2 - 2 0 0 1 6 0 0 0 5 0 0 0 0 0 The second row tells us that x 3 = 0, and the first row tells us that x 2 = - 6 x 3 . But this implies that x 2 = 0. x 1 is our free variable, so we can choose it to be 1, and we get the eigenvector v 1 = 1 0 0 We only found one eigenvector, which is less than the multiplicity of our eigenvalue. This means two things: we must find more eigenvectors, and we must have a t 2 e 2 t and te 2 t in our solution (i.e. we must “bump up” our solution). We find the next eigenvector by solving the equation ( A - ) v 2 = v 1 : 0 1 6 1 0 0 5 0 0 0 0 0 The second row tells us x 3 = 0, and the first row tells us that x 2 = - 6 x 3 +1, which means that x 2 = 1. x 1 is free, so if we choose it to be 0 then the second eigenvector is v 2 = 0 1 0 We need to find another eigenvector, so we solve the equation ( A - ) v 3 = v 2 : 0 1 6 0 0 0 5 1 0 0 0 0 The second row tells us that x 3 = 1 5 , the first row tells us that x 2 = - 6 x 3 = - 6 5 . x 1 is free, so if we choose it to be 0 then the third eigenvector is v 3 = 0 - 6 5 1 5 . Therefore the solution is x = c 1 1 0 0 e 2 t + c 2 1 0 0 te 2 t + 0 1 0 e 2 t + c 3 1 0 0 t 2 2 e 2 t + 0 1 0 te 2 t + 0 - 6 5 1 5 e 2 t . 3
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3. Solve x 0 = 4 0 1 0 6 0 - 4 0 4 x . Solution. We begin by finding the eigenvalues: 4 - λ 0 1 0 6 - λ 0 - 4 0 4 - λ = (4 - λ ) 6 - λ 0 0 4 - λ - 0 + 1 0 6 - λ - 4 0 = (4 - λ )(6 - λ )(4 - λ ) + 4(6 - λ ) = (6 - λ )[(4 - λ )(4 - λ ) + 4] = (6 - λ )( λ 2 - 8 λ + 20) We use the quadratic formula to find the eigenvalues of λ 2 - 8 λ + 20: λ = 8 ± p 64 - 4(1)(20) 2 = 8 ± - 16 2 = 8 ± 4 i 2 = 4 ± 2 i Therefore the eigenvalues are λ 1 = 6 , λ 2 = 4 + 2 i, λ 3 = 4 - 2 i . When λ 1 = 6: 4 - 6 0 1 0 0 6 - 6 0 0 - 4 0 4 - 6 0 - 2 0 1 0 0 0 0 0 - 4 0 - 2 0 - 2 0 1 0 0 0 0 0 0 0 - 4 0 The last row tells us that x 3 = 0, the first row tells us that 2 x 1 = x 3 which implies that x 1 = 0. x 2 is free, so if we choose it to be 1 then we have the eigenvector v 1 = 0 1 0 When λ 2 = 4 + 2 i : 4 - (4 + 2 i ) 0 1 0 0 6 - (4 + 2 i ) 0 0 - 4 0 4 - (4 + 2 i ) 0 - 2 i 0 1 0 0 2 - 2 i 0 0 - 4 0 - 2 i 0 The second row tells us that (2 - 2 i ) x 2 = 0, or that x 2 = 0. The first row tells us that 2 ix 1 = x 3 . If we choose x 1 = 1, then we have the eigenvector v 2 = 1 0 2 i . This implies that the third eigenvector is v 3 = 1 0 - 2 i . Therefore the solution to the system is x = c 1 0 1 0 e 6 t + c 2 1 0 2 i e (4+2 i ) t + c 3 1 0 - 2 i e (4 - 2 i ) t , 4
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however we want to write our solution as a real solution. Therefore we must expand v 2 e λ 2 t (you can also do the same for v 3 e λ 3 t , but it is enough to just do one expansion): 1 0 2 i e 4 t e i 2 t = 1 0 2 i e 4 t (cos 2 t + i sin 2 t ) = e 4 t cos 2 t + i sin 2 t 0 2 i cos 2 t - 2 sin 2 t = e 4 t cos 2 t 0 - 2 sin 2 t
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