Thus u t e 3 t v 1 3 e 5 t v 2 solves the given di ff

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Thus u ( t ) = e 3 t v 1 + 3 e 5 t v 2 solves the given di ff erential equation. Conse- quently, u (2) = 4 e 6 9 e 10 4 e 6 + 0 e 10 . 010 10.0 points Let u ( t ) satisfy d u dt = A u ( t ) , u (0) = 11 9 . Compute u (2) when A is a 2 × 2 matrix with eigenvalues 4 and 2 and corresponding eigen- vectors v 1 = 2 2 , v 2 = 1 1 . 1. u (2) = 10 e 8 e 4 10 e 8 + e 4 2. u (2) = 10 e 8 e 4 10 e 8 e 4 3. u (2) = 10 e 8 + e 4 10 e 8 + e 4 4. u (2) = 10 e 8 e 4 10 e 8 + e 4 correct 5. u (2) = 10 e 8 + e 4 10 e 8 e 4 6. u (2) = 10 e 8 + e 4 10 e 8 e 4 Explanation: Since v 1 and v 2 are eigenvectors corre- sponding to distinct eigenvalues of A , they form an eigenbasis for R 2 . Thus u (0) = c 1 v 1 + c 2 v 2 To compute c 1 and c 2 we apply row reduction to the augmented matrix [ v 1 v 2 u (0) ] = 2 1 11 2 1 9 1 0 5 0 1 1 . This shows that c 1 = 5, c 2 = 1 and u (0) = 5 v 1 + v 2 . Since v 1 and v 2 are eigenvectors correspond- ing to the eigenvalues 4 and 2 respectively, set u ( t ) = 5 e 4 t v 1 + e 2 t v 2 . Then u (0) is the given initial value and A u ( t ) = 5 e 4 t A v 1 + e 2 t A v 2 = 5 ( 4 e 4 t ) v 1 + ( 2 e 2 t ) v 2 = d u ( t ) dt . Thus u ( t ) = 5 e 4 t v 1 + e 2 t v 2
farias (df8272) – HW11 – gilbert – (56780) 8 solves the given di ff erential equation. Conse- quently, u (2) = 10 e 8 e 4 10 e 8 + e 4 . 011 10.0 points Find the area of the triangle ABC when A = (1 , 1 , 0) , B = (1 , 0 , 1) , and C = (0 , 1 , 2) . Hint: Check if ABC is right-angled . 1. area = 6 2. area = 3 3. area = 6 2 correct 4. area = 1 2 5. area = 3 2 Explanation: Set u = −−→ BA = (0 , 1 , 1) and v = −−→ BC = ( 1 , 1 , 1) . Then u · v = 0, so ABC is a right-angled triangle with side BA perpendicular to side BC . Thus area ABC = 1 2 base × height = u ∥∥ v = 2 3 2 Consequently, ABC has area = 6 2 . 012 10.0 points If v 1 , v 2 , . . ., v p span a subspace W of R n and if x is a vector in R n such that x · v j = 0 , j = 1 , . . . , p , then x is in W . True or False? 1. FALSE 2. TRUE correct Explanation: By definition W consists of all vectors in R n that are orthogonal to any set that spans W . So if Span { v 1 , v 2 , . . ., v p } = W and if x is a vector in R n such that x · v j = 0 for all 1 j p , then x is in W . Consequently, the statement is TRUE . 013 10.0 points Not every orthogonal set in R n is linearly independent. True or False? 1. FALSE 2. TRUE correct Explanation: Since the zero vector 0 is orthogonal to ev- ery vector in R n and any set containing 0 is linearly dependent, only orthogonal sets of non-zero vectors in R n are linearly indepen- dent. Consequently, the statement is TRUE . 014 10.0 points
farias (df8272) – HW11 – gilbert – (56780) 9 If a set S = { u 1 , ..., u p } has the property that u i · u j = 0 whenever i ̸ = j , then S is an orthonormal set. True or False? correct
Explanation: The above statement defines an orthogonal set. An orthonormal set is an orthogonal set of unit vectors.

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