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Practice 4 Solutions

# Practice 4 Solutions - 2 2 Exercise Set A(distinct roots...

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2 × 2 Exercise Set A (distinct roots), November , 2 × 2 Exercise Set A (distinct roots) Linear Algebra, Dave Bayer, November , [ 1 ] Find A n where A is the matrix A = 1 - 1 3 - 3 λ = - 2, 0 A n = (- 2 ) n 2 - 1 1 - 3 3 + 0 n 2 3 - 1 3 - 1 [ 2 ] Find A n where A is the matrix A = 1 - 1 - 3 - 1 λ = - 2, 2 A n = (- 2 ) n 4 1 1 3 3 + 2 n 4 3 - 1 - 3 1 [ 3 ] Find A n where A is the matrix A = 0 1 2 1 λ = - 1, 2 A n = (- 1 ) n 3 2 - 1 - 2 1 + 2 n 3 1 1 2 2 [ 4 ] Find A n where A is the matrix A = 1 2 3 0 λ = - 2, 3 A n = (- 2 ) n 5 2 - 2 - 3 3 + 3 n 5 3 2 3 2 [ 5 ] Find A n where A is the matrix A = 2 2 - 2 - 3 λ = - 2, 1 A n = (- 2 ) n 3 - 1 - 2 2 4 + 1 3 4 2 - 2 - 1 [ 6 ] Find A n where A is the matrix A = - 3 1 - 2 0

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2 × 2 Exercise Set A (distinct roots), November , λ = - 2, - 1 A n = (- 2 ) n 2 - 1 2 - 1 + (- 1 ) n - 1 1 - 2 2 [ 7 ] Find A n where A is the matrix A = 1 1 2 0 λ = - 1, 2 A n = (- 1 ) n 3 1 - 1 - 2 2 + 2 n 3 2 1 2 1 [ 8 ] Find A n where A is the matrix A = 0 1 - 2 - 3 λ = - 2, - 1 A n = (- 2 ) n - 1 - 1 2 2 + (- 1 ) n 2 1 - 2 - 1
2 × 2 Exercise Set B (distinct roots), November , 2 × 2 Exercise Set B (distinct roots) Linear Algebra, Dave Bayer, November , [ 1 ] Find e At where A is the matrix A = 3 1 - 3 - 1 λ = 0, 2 e At = 1 2 - 1 - 1 3 3 + e 2 t 2 3 1 - 3 - 1 [ 2 ] Find e At where A is the matrix A = - 3 2 - 3 2 λ = - 1, 0 e At = e - t 3 - 2 3 - 2 + - 2 2 - 3 3 [ 3 ] Find e At where A is the matrix A = - 1 - 3 - 2 0 λ = - 3, 2 e At = e - 3 t 5 3 3 2 2 + e 2 t 5 2 - 3 - 2 3 [ 4 ] Find e At where A is the matrix A = 3 3 - 1 - 1 λ = 0, 2 e At = 1 2 - 1 - 3 1 3 + e 2 t 2 3 3 - 1 - 1 [ 5 ] Find e At where A is the matrix A = 1 - 1 - 3 - 1 λ = - 2, 2 e At = e - 2 t 4 1 1 3 3 + e 2 t 4 3 - 1 - 3 1 [ 6 ] Find e At where A is the matrix A = - 2 2 - 3 3

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2 × 2 Exercise Set B (distinct roots), November , λ = 0, 1 e At = 3 - 2 3 - 2 + e t - 2 2 - 3 3 [ 7 ] Find e At where A is the matrix A = 2 - 2 3 - 3 λ = - 1, 0 e At = e - t - 2 2 - 3 3 + 3 - 2 3 - 2 [ 8 ] Find e At where A is the matrix A = 0 - 1 - 3 2 λ = - 1, 3 e At = e - t 4 3 1 3 1 + e 3 t 4 1 - 1 - 3 3
2 × 2 Exercise Set C (distinct roots), November , 2 × 2 Exercise Set C (distinct roots) Linear Algebra, Dave Bayer, November , [ 1 ] Solve the di erential equation y 0 = Ay where A = 1 1 2 2 , y ( 0 ) = 1 0 λ = 0, 3 e At = 1 3 2 - 1 - 2 1 + e 3 t 3 1 1 2 2 y = 1 3 2 - 2 + e 3 t 3 1 2 [ 2 ] Solve the di erential equation y 0 = Ay where A = 0 - 1 2 - 3 , y ( 0 ) = 1 2 λ = - 2, - 1 e At = e - 2 t - 1 1 - 2 2 + e - t 2 - 1 2 - 1 y = e - 2 t 1 2 + e - t 0 0 [ 3 ] Solve the di erential equation y 0 = Ay where A = 1 - 1 - 3 - 1 , y ( 0 ) = 1 1 λ = - 2, 2 e At = e - 2 t 4 1 1 3 3 + e 2 t 4 3 - 1 - 3 1 y = e - 2 t 4 2 6 + e 2 t 4 2 - 2 [ 4 ] Solve the di erential equation y 0 = Ay where A = - 1 1 2 - 2 , y ( 0 ) = 1 0 λ = - 3, 0 e At = e - 3 t 3 1 - 1 - 2 2 + 1 3 2 1 2 1 y = e - 3 t 3 1 - 2 + 1 3 2 2 [ 5 ] Solve the di erential equation y 0 = Ay where A = 1 1 2 0 , y ( 0 ) = 1 2 λ = - 1, 2 e At = e - t 3 1 - 1 - 2 2 + e 2 t 3 2 1 2 1 y = e - t 3 - 1 2 + e 2 t 3 4 4

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2 × 2 Exercise Set C (distinct roots), November , [ 6 ] Solve the di erential equation y 0 = Ay where A = - 3 2 - 3 2 , y ( 0 ) = 1 2 λ = - 1, 0 e At = e - t 3 - 2 3 - 2 + - 2 2 - 3 3 y = e - t - 1 - 1 + 2 3 [ 7 ] Solve the di erential equation y 0 = Ay where A = - 2 - 1 - 2 - 1 , y ( 0 ) = 1 0 λ = - 3, 0 e At = e - 3 t 3 2 1 2 1 + 1 3 1 - 1 - 2 2 y = e - 3 t 3 2 2 + 1 3 1 - 2 [ 8 ] Solve the di erential equation y 0 = Ay where A = - 1 - 2 1 2 , y ( 0 ) = 1 0 λ = 0, 1 e At = 2 2 - 1 - 1 + e t - 1 - 2 1 2 y = 2 - 1 + e t - 1 1
2 × 2 Exercise Set D (repeated roots), November , 2 × 2 Exercise Set D (repeated roots) Linear Algebra, Dave Bayer, November , [ 1 ] Find A n where A is the matrix A = 1 3 - 3 - 5 λ = - 2, - 2 A n = (- 2 ) n 1 0 0 1 + n (- 2 ) n - 1 3 3 - 3 - 3 [ 2 ] Find A n where A is the matrix A = 1 1 - 1 - 1 λ = 0, 0 A n = 0 n 1 0 0 1 + n 0 n - 1 1 1 - 1 - 1 [ 3 ] Find A n

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Practice 4 Solutions - 2 2 Exercise Set A(distinct roots...

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