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Unformatted text preview: 18.06 Problem Set 6 Solution Due Wednesday, 8 April 2009 at 4 pm in 2106. Total: 100 points. Problem 1: If A is a 7 × 7 matrix and det A = 17, what is det(3 A 2 )? Solution (10 points) For any square matrix M,N of the same size, we have det( MN ) = det( M )det( N ). Thus, det( A 2 ) = det( A )det( A ) = 17 2 . Note that we proved in class that multiplying a single row by 3 multiplies the determinant by 3. Multiplying the whole 7 × 7 matrix by 3 multiplies all 7 rows by 3, and hence multiplies the determinant by 3 7 . Hence, we have det(3 A 2 ) = 3 7 · 17 2 = 632043. (It is okay to leave it as 3 7 · 17 2 . Problem 2: The determinant of a 2 × 2 matrix A = a b c d ¶ is det A = ad bc . Assuming no row swaps are required, perform elimination on A and show explicitly that ad bc is the product of the pivots. Solution (5 points) We need to subtract c/a times row 1 from row 2, that leaves us a b c d ¶ ; a b d b · a/c ¶ The product of the pivots is a · ( d ab/c ) = ad bc = det A . REMARK: We proved that this is true for any n × n matrix in class (it is also in the book), not including row swaps (which flip the sign). Problem 3: If ~x and ~ y are vectors in R n ( n > 1), what is the determinant of ~x~ y T ? (This is not the dot product ~x T ~ y .) Hint: the rank of ~x~ y T is . Solution (5 points) The product A = ~x~ y T is an n × n matrix. But we know its column space C ( A ) ⊆ C ( ~x ) is at most 1dimensional (spanned by ~x if ~ y 6 = ~ 0). Hence A is singular because n > 1 ≥ rank( A ) = dim C ( A ). Therefore, det( A ) = 0. 1 An alternative way to see it is that all the rows are a multiple of ~ y T , and all the columns are multiples of ~x . Hence, the determinant must be zero because the determinant is zero when any rows or columns are linearly dependent. Problem 4: Does det( AB ) = det( BA ) in general? (a) True or false if A and B are square n × n matrices? (b) True or false if A is m × n and B is n × m , with m 6 = n ? For both (a) and (b), give a reason if true or a counterexample if false. Solution (10 points = 5+5) (a) True. This is because if A and B are square matrix, det( AB ) = det( A )det( B ) = det( B )det( A ) = det( BA ). (b) False. This is almost always not true. We may use the solution to problem 3 to make a counter example: det( xy T ) = 0, but det( x T y ) = x T y . For example, A = 1 1 ¶ , B = ( 1 1 ) Then, AB = 1 1 1 1 ¶ , BA = (2); det( AB ) = 0 , det( BA ) = 2 . REMARK: In fact, if m > n , det( AB ) = 0. This is because the column space C ( AB ) ⊆ C ( A ) has dimension ≤ n < m . So, AB , as an m × m matrix, is singular....
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This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Spring '08 term at MIT.
 Spring '08
 Strang

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