HW5 - p.301, problem 6 1 a 1 a 1 1 a 1 1 1 a 1 → 0 a −...

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Unformatted text preview: p.301, problem 6 1 a 1 a 1 1 a 1 1 1 a 1 → 0 a − 1 1 − a → 0 a − 1 1− a a 1 1 0 1 − a 1 − a 2 0 0 2 − a − a2 2 2 − a − a = 0 ⇒ a = 1,−2 Rank = 3 if a≠ 1or -2. Rank = 2 if a=-2. Rank = 1 if a=1. Page 301, problem 11 4 8 16 4 8 16 2 4 8 16 2 2 16 8 4 2 0 − 24 − 60 − 126 0 − 24 − 60 − 126 → → 4 8 16 2 0 0 0 0 − 30 0 0 − 30 0 − 12 0 − 30 − 75 2 16 8 4 0 12 0 4 8 16 2 0 − 24 − 60 − 126 → 0 0 − 30 − 75 0 0 − 30 0 Rank = 4. Page 301, problem 15 6 0 3 0 −1 2 12 3 0 6 0 → 0 − 1 0 0 1 4 2 3 1 4 2 6 0 7 0 5 → 0 − 1 2 7 0 5 0 3 − 6 − 21 0 − 15 − 19 8 − 11 3 1 4 2 2 7 0 5 0 0 0 0 Therefore linearly dependent. Page 314, problem 8 14 2 5 2 0 8 = 14( 0 − 8 ⋅ 8) − 2( − 2 ⋅ 2 − 8 ⋅ 5) + 5( 2 ⋅ 8 − 0) = −728 5 8 −2 Page 322, problem 4 2 / 3 1/ 3 2 / 3 1 0 0 2 / 3 1/ 3 2 / 3 1 1 1 1 − 2 / 3 2 / 3 1/ 3 0 1 0 → 0 1/ 3 2 / 3 − 2 / 3 0 0 1 0 1/ 2 − 1 − 1/ 2 2 / 3 1/ 3 2 / 3 1 0 0 1 1/ 2 1 3 / 2 → 0 1 11 1 0 → 0 1 1 1 0 0 − 3 / 2 − 1 − 1/ 2 1 0 0 1 2 / 3 0 0 1 0 0 1 0 0 1 0 1/ 3 − 2 / 3 1 1/ 2 0 5 / 6 − 1/ 3 2 / 3 1 0 0 2 / 3 − 2 / 3 1/ 3 → 0 1 0 1/ 3 2 / 3 2 / 3 → 0 1 0 1/ 3 2 / 3 2/3 0 0 1 2 / 3 1/ 3 − 2 / 3 0 0 1 2 / 3 1/ 3 − 2 / 3 2 / 3 1/ 3 2 / 3 ∴ − 2 / 3 2 / 3 1/ 3 1 / 3 2 / 3 − 2 / 3 −1 2 / 3 − 2 / 3 1/ 3 = 1/ 3 2 / 3 2/3 2 / 3 1 / 3 − 2 / 3 Page 322, problem 5 3 −1 1 1 0 0 3 −1 1 1 1 0 5 − 15 6 − 5 0 1 0 → 0 5 0 − 1/ 3 1/ 3 − 5 / 3 − 2 2 0 0 1 3 − 1 1 1 0 0 1 − 1/ 3 1/ 3 1/ 3 → 0 1 0 5 1 0 → 0 1 05 0 0 1/ 3 0 1/ 3 1 0 0 10 0 0 1 0 0 1 0 0 1 0 1 3 1 − 1 / 3 0 1 / 3 − 1 / 3 − 1 1 0 0 2 0 − 1 → 0 1 05 1 0 → 0 1 0 5 1 0 0 0 10 1 3 0 0 1 0 1 3 −1 −1 1 3 2 0 − 1 ∴ − 15 6 − 5 = 5 1 0 5 0 1 3 −2 2 Page 338, problem 19: 13 − λ 5 2 3 2 7 − λ − 8 = −λ3 + 27λ2 − 243λ + 729 = −( λ − 9 ) = 0 ⇒ λ = 9,9,9. 5 4 7−λ 2 x1 0 4 5 2 − 2 − 8 x 2 = 0 5 4 − 2 x 0 3 Let x3 = 1 ⇒ 4 x1 + 5 x 2 = −2, 2 x1 − 2 x 2 = 8 ⇒ x1 = 2, x 2 = −2 2 ⇒ − 2 1 is an eigenvector. Page 338, problem 25: −1− λ 0 0 0 0 −1− λ 0 0 12 0 −1− λ −4 0 12 2 2 = ( − 1 − λ ) λ 2 + 2λ − 15 = ( λ + 1) ( λ + 5) ( λ − 3) = 0 −4 −1− λ ( ) ⇒ λ = −1, − 1, − 5, 3 λ = −1 : 0 0 0 0 0 12 0 x 1 0 0 12 x 2 0 0 − 4 x 3 0 − 4 0 x 4 λ = −5 : 0 1 0 0 0 1 = 0 ⇒ x 3 = 0, x 2 = 0, x 1 and x 2 arbitrary ⇒ 0 and 0 eigenvectors 0 0 0 0 12 0 x 1 4 0 12 x 2 0 4 − 4 x 3 0 − 4 4 x 4 λ =3: 0 − 3 0 − 3 = 0 Let x 3 = 1 ⇒ x 1 = −3, x 4 = 1, x 2 = −3 ⇒ 1 is an eigenvector 1 0 4 0 0 0 − 4 0 12 0 x 1 0 − 4 0 12 x 2 0 0 − 4 − 4 x 3 0 0 − 4 − 4 x 4 0 3 0 − 3 = 0 Let x 3 = 1 ⇒ x 1 = 3, x 4 = −1, x 2 = −3 ⇒ 1 is an eigenvector − 1 0 ...
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