Midterm_Review_Solutions

Midterm_Review_Solutions - Math 115 Midterm Review...

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Unformatted text preview: Math 115 Midterm Review Solutions 1. (a) x y z = 1 + 3 t 1- 5 t t (b) x y z = 3 t- 5 t t 2. ( A- 1 BC )( C- 1 B- 1 A ) = A- 1 B ( CC- 1 ) B- 1 A ) = A- 1 BIB- 1 A = A- 1 ( BB- 1 ) A = A- 1 IA = A- 1 A = I Similarly, ( C- 1 B- 1 A )( A- 1 BC ) = I . 3. det 1 c 2 c c- 1 1 = det 1 2- 2 c c c- 1- c 2 1 = det- 2 c c- 1- c 2 1 = c ( c 2- 1) = c ( c- 1)( c + 1) . Hence the matrix is invertible if c 6 = 0 , 1 or- 1 . 4. Using the matrix inversion algorithm we have 1- 1- 2- 1 1 2 1 - 1 = 1 2 1- 2- 4- 1 1 3 1 . The equations are A~x = ~ b in matrix form where ~ b = [3 0 1] T , so the solution is ~x = A- 1 ~ b 1 2 1- 2- 4- 1 1 3 1 3 1 = 4- 7 4 . 5. The standard matrix for T is 1 1 cos( / 2)- sin( / 2) sin( / 2) cos( / 2) = 1- 1 , and T 1 2 = 1- 1 1 2 = 1- 2 6. (a) the standard matrix for T A T B is AB = 1 1 1- 1 3...
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This note was uploaded on 09/17/2011 for the course MATH 115 taught by Professor Dunbar during the Fall '07 term at Waterloo.

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Midterm_Review_Solutions - Math 115 Midterm Review...

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