tut5 - and λ 2 =-2 (with algebraic multiplicity 1). λ 1 =...

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MATH 1104F - Solutions to Tutorial Assignment 5 March 9, 2010 Total: 10 points 1. Solve the system 4 x - y + 3 z = 1 6 x + 2 y - z = 0 3 x + 3 y + 2 z = - 1 by using Cramer’s rule. Solution: det A = det 4 - 1 3 6 2 - 1 3 3 2 = 79 x = det 1 - 1 3 0 2 - 1 - 1 3 2 79 = 12 79 y = det 4 1 3 6 0 - 1 3 - 1 2 79 = - 37 79 z = det 4 - 1 1 6 2 0 3 3 - 1 79 = - 2 79 2. Let A,B and C be 3 × 3 matrices such that det A = - 1, det B = 2 and det C = 4. What is det(2 A 3 BC T B - 1 )? Solution: det(2 A 3 BC T B - 1 ) = 2 3 (det A ) 3 (det B )(det C ) 1 det B = 8 · ( - 1) 3 · 2 · 4 · 1 2 = - 32 3. Let A = 3 0 0 0 1 - 2 1 0 1 and v = 2 - 1 1 . Show that v is an eigenvector of A and find the corresponding eigenvalue.
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Solution: Av = 3 0 0 0 1 - 2 1 0 1 · 2 - 1 1 = 6 - 3 3 = 3 · 2 - 1 1 = 3 v Therefore v is an eigenvector of A corresponding to the eigen- value λ = 3. 4. Let A = - 1 0 1 3 0 - 3 1 0 - 1 . Find the eigenvalues of A , their algebraic multiplicities, and a basis for each one of their corre- sponding eigenspaces. Solution: det( A - λI ) = det - 1 - λ 0 1 3 - λ - 3 1 0 - 1 - λ = - λ 2 ( λ + 2) The eigenvalues of A are λ 1 = 0 (with algebraic multiplicity 2)
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Unformatted text preview: and λ 2 =-2 (with algebraic multiplicity 1). λ 1 = 0 A-λ 1 I = A = -1 0 1 3 0-3 1 0-1 ∼ 1 0-1 0 0 0 0 x y z = z y z = y 1 + z 1 1 Therefore 1 , 1 1 is a basis of the eigenspace corre-sponding to λ 1 = 0. λ 2 =-2 A-λ 2 I = A +2 I = 1 0 1 3 2-3 1 0 1 ∼ 1 0 1 0 2-6 0 0 ∼ 1 0 1 0 1-3 0 0 x y z = -z 3 z z = z -1 3 1 Therefore -1 3 1 is a basis of the eigenspace correspond-ing to λ 2 =-2....
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This note was uploaded on 03/22/2010 for the course MATH 1104 taught by Professor Unknown during the Winter '10 term at Carleton.

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tut5 - and λ 2 =-2 (with algebraic multiplicity 1). λ 1 =...

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