winter 2007assign1-sol

# winter 2007assign1-sol - MATH235 Winter 2007 SOLUTIONS TO...

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Unformatted text preview: MATH235 - Winter 2007. SOLUTIONS TO ASSIGNMENT 1. 1. Section 3.1 In Exercises 34 and 36, verify that det( EA ) = (det E )(det A ), where E is the elementary matrix shown and A = a b c d . 34. E = 1 k . Solution: EA = 1 k a b c d = a b kc kd . det EA = akd- bkc = k ( ad- bc ) det E = 1 k- 0 = k det A = ad- bc Therefore, det EA = (det E )(det A ) is verified. 36. E = 1 k 1 . Solution: EA = 1 k 1 a b c d = a b ka + c kb + d . det EA = a ( kb + d )- b ( ka + c ) = ad- bc det E = 1 det A = ad- bc Therefore, det EA = (det E )(det A ) is confirmed. Section 3.2 8. Find the determinant of the following matrix by row reduction to echelon form. Solution: A = 1 3 3- 4 1 2- 5 2 5 4- 3- 3- 7- 5 2 ( R 3- 2 R 1 ) → R 3 ( R 4 + 3 R 1 ) → R 4-→ 1 3 3- 4 1 2- 5- 1- 2 5 2 4- 10 1 ( R 3 + R 2 ) → R 3 ( R 4- 2 R 2 → R 4-→ 1 3 3- 4 1 2- 5 The operations that we use to reduce the matrix to echelon form do not change the determinant, thus D ( A ) = D 1 3 3- 4 1 2- 5 = 0 (seen by alternating property, or by linearity of D in the third or fourth row). 26. Use determinant to decide if the following set of vectors is linearly inde- pendent. 3 5- 6 4 , 2- 6 7 , - 2- 1 3 , - 3 Solution: Let A = 3 2- 2 5- 6- 1- 6 3 4 7- 3 , then the above set of vectors is linearly...
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winter 2007assign1-sol - MATH235 Winter 2007 SOLUTIONS TO...

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