06aut-m2sols

06aut-m2sols - Math 51 Midterm 2 Solutions PROBLEM 1(a). A...

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Math 51 Midterm 2 Solutions PROBLEM 1(a). A one-to one transformation must have null space { 0 } . The columns must be independent, there are pivots in every column, and no free variables, so the answer is rank ( A ) = n . PROBLEM 1(b). Onto means the column space is all of R m . So the answer is rank ( B ) = m . PROBLEM 1(c). 0 0 1 0 1 - 3 1 2 2 ± ± ± 1 0 0 0 1 0 0 0 1 7→ 0 0 1 0 1 0 1 2 0 ± ± ± 1 0 0 3 1 0 - 2 0 1 7→ 0 0 1 0 1 0 1 0 0 ± ± ± 1 0 0 3 1 0 - 8 - 2 1 7→ 1 0 0 0 1 0 0 0 1 ± ± ± - 8 - 2 1 3 1 0 1 0 0 . The answer is 0 0 1 0 1 - 3 1 2 2 - 1 = - 8 - 2 1 3 1 0 1 0 0 . PROBLEM 2(a). Since u 1 = 1 0 2 2 and u 2 = 0 1 - 2 2 are already orthogo- nal, we first just need to replace the third vector w = 0 - 7 5 4 by the vector orthogonal to both u 1 and u 2 given by u 3 = w - w · u 1 u 1 · u 1 u 1 - w · u 2 u 2 · u 2 u 2 . Calculating gives u 3 = - 2 - 6 - 1 2 , which you should double check is orthogonal to u 1 and u 2 . Finally we need to normalize, that is, divide u 1 , u 2 , and u 3 by their lengths to get an orthonormal basis. The answer is 1 / 3 0 2 / 3 2 / 3 , 0 1 / 3 - 2 / 3 2 / 3 , - 2 / 45 - 6 / 45 - 1 / 45 2 / 45 . PROBLEM 2(b). Since we assume w span { n 1 , n 2 , n 3 } , we can write w = c 1 n 1 + c 2 n 2 + c 3 n 3 . We need to determine the constants c j . This is very easy, just take the dot product of both sides of the equation with n j . For example, 1
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w · n 1 = c 1 n 1 · n 1 + c 2 n 2 · n 1 + c 3 n 3 · n 1 = ( c 1 )1 + ( c 2 )0 + ( c 3 )0 = c 1 , since the n 0 s are orthonormal. In the same way, w · n 2 = c 2 and w · n 3 = c 3 . Thus w = ( w · n 1 ) n 1 + ( w · n 2 ) n 2 + ( w · n 3 ) n 3 . PROBLEM 3(a). B = 4 0 0 0 2 2 0 0 - 3 1 3 0 1 - 5 - 7 1 . (For example, the first column in B gives the coordinates of T ( v 1 )= 4 v 1 +2 v 2 - 3 v 3 + v 4 in the { v j } basis.) PROBLEM 3(b). C is just the matrix with the { v j } as its columns. That is, C = 7 4 0 0 5 3 0 0 0 0 6 5 0 0 5 4 . PROBLEM 3(c). Since B is triangular, det(B) = (4)(2)(3)(1) = 24 is the product of the diagonal entries. (Think about how many ways you can circle one entry from each row and each column and get a non-zero product. One way, that’s how many, by circling the diagonal entries.) By properties of determinants, det(A) = det(C)det(B)det( C - 1 ) = det(C)det(B)(1/det(C)) = det(B) = 24.
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This note was uploaded on 01/12/2010 for the course MATH 51 at Stanford.

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06aut-m2sols - Math 51 Midterm 2 Solutions PROBLEM 1(a). A...

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