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# 4.4 - Math 2940 Solutions Fall 2011 Section 4.4 100 10 10...

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Math 2940 Solutions, Fall 2011 Section 4 . 4 4 )(a) Q = 1 0 0 1 0 0 so QQ T = 1 0 0 1 0 0 1 0 0 0 1 0 = 1 0 0 0 1 0 0 0 0 . A square matrix Q will never work. (b) For this problem one of the two vectors must be 0. Choose any nonzero vector for the other. (c) 1 / 3(1 , 1 , 1) is our first vector. The other two must lie in the plane x + y + z = 0 (why?) So use 1 / 2(1 , 0 , - 1) for the second vector. If the third is ( a, b, c ), then to be perpen- dicular to the first, we need a + b + c = 0 and to be perpendicular to the second we need a - c = 0 so (1 , - 2 , 1) suggest itself. We normalize to get 1 / 6(1 , - 2 , 1). 5 ) We take (1 , - 1 , 0) as our first, and if ( a, b, c ) is the second, we need a + b + 2 c = 0 and a = b so (1 , 1 , - 1) works. Normalizing we get 1 / 2(1 , - 1 , 0) and 1 / 3(1 , 1 , - 1). 10 ) (a) Dotting the equation c 1 q 1 + c 2 q 2 + c 3 q 3 = 0 with q i gives c i = 0. This holds for i = 1 , 2 , 3. (b) Say Q is m × n with m n . Then Q T Q = I n , the n × n identity matrix. To solve Qx = 0 we multiply both side by Q T on the left to get Q T Qx = Ix = x = 0 is the only solution.

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4.4 - Math 2940 Solutions Fall 2011 Section 4.4 100 10 10...

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