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136A4S

# 136A4S - ASSIGNMENT 4 for SECTION 001 Solutions Questions...

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ASSIGNMENT 4 for SECTION 001 Solutions Questions from the textbook Section 3-3: B1 [3 marks] , B3 (c) [1 mark] , B4 [4 marks] , B5 (a) [2 marks] , B6 (a) [3 marks] , D4 [3 marks] and D6 [2 marks] In addition to the questions listed here, do the MATLAB question from Assignment 4 for the other sections; this may be found under Content > Assignments > Assignment 4 [2 marks] Other questions 1. [6 marks] Given an angle θ , let R θ denote the linear transformation reflecting vectors across the line formed by rotating the x -axis counterclockwise by θ . Find all conditions on α and β such that R α and R β commute. R θ reflects vectors across the line x sin θ - y cos θ = 0; that is, across the vector r = cos θ sin θ . Let e 1 and e 2 denote the columns of the 2 × 2 identity matrix. We have R θ ( e 1 ) = e 1 - 2 ( e 1 - proj r e 1 ) = 2proj r e 1 - e 1 = 2 e 1 · r r 2 r - e 1 = 2 cos 2 θ - 1 2 cos θ sin θ = cos(2 θ ) sin(2 θ ) . Similarly, R θ ( e 1 ) = sin(2 θ ) - cos(2 θ ) . Thus R θ has standard matrix cos(2 θ ) sin(2 θ ) sin(2 θ ) - cos(2 θ ) . Let A and B denote the standard matrices for R α and R β , respectively. Then AB = cos(2 α ) cos(2 β ) + sin(2 α ) sin(2 β ) cos(2 α ) sin(2 β ) - sin(2 α ) cos(2 β ) sin(2 α
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Unformatted text preview: β )-cos(2 α ) sin(2 β ) cos(2 α ) cos(2 β ) + sin(2 α ) sin(2 β ) ² = ± cos (2( α-β ))-sin (2( α-β )) sin (2( α-β )) cos (2( α-β )) ² . Thus R α ◦ R β is a rotation by 2( α-β ). Similarly, R β ◦ R α is a rotation by 2( β-α ). These rotations are equal when α and β are congruent modulo π/ 2. 2. [4 marks] Let K : R m → R n and L : R n → R m be linear transformations, and let m < n . Prove that the reduced row-echelon form of the standard matrix of K ◦ L is not an identity matrix. Let K and L have standard matrices A and B , respectively. AB , the standard matrix of K ◦ L , is m × m . Now the reduced row-echelon form of AB is the m × m identity matrix if, and only if, Nul( AB ) = { } . Since m < n , B has at least one free variable. Hence there exists a nonzero vector x ∈ R m such that B x = , and AB x = . Thus Nul( AB ) 6 = { } ....
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