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hw9solutions - Math 215 HW#9 Solutions 1 Problem 4.4.12 If...

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Math 215 HW #9 Solutions 1. Problem 4.4.12. If A is a 5 by 5 matrix with all | a ij | ≤ 1, then det A . Volumes or the big formula or pivots should give some upper bound on the determinant. Answer: Let v i be the i th column of A . Then | v i | = a 2 1 i + a 2 2 i + a 2 3 i + a 2 4 i + a 2 5 i . Since | a ij | ≤ 1 for each i and j , it’s also true that each a 2 ij 1. Hence, the right hand side of the above equation is no bigger than 1 + 1 + 1 + 1 + 1 = 5, and so we see that | v i | ≤ 5 . This means that each edge of the 5-dimensional box spanned by the columns of A is no longer than 5, meaning that the volume of that box can be no bigger than 5 5 = 5 5 / 2 = 25 5 . Since | det A | is exactly equal to the volume of the box spanned by the columns of A , this reasoning implies that | det A | ≤ 25 5 55 . 9 . Another way to get an upper bound is to use the big formula for the determinant from p. 212. Notice that, for any 5 by 5 matrix, there are 5! = 120 terms in the sum. When all entries of A are smaller than 1 (in absolute value), it must be the case that each of the 120 terms in the sum is smaller than 1 (in absolute value). Therefore, the whole sum (which is just the determinant), must be no bigger than 120 in absolute value. 2. Problem 4.4.18. Find A - 1 from the cofactor formula C T / det A . Use symmetry in part (b): (a) A = 1 2 0 0 3 0 0 4 1 . (b) A = 2 - 1 0 - 1 2 - 1 0 - 1 2 . (a) We need to determine the various cofactors of A to find the cofactor matrix C : C 11 = ( - 1) 1+1 det A 11 = 3 0 4 1 = 3 C 12 = ( - 1) 1+2 det A 12 = - 0 0 0 1 = 0 C 13 = ( - 1) 1+3 det A 13 = 0 3 0 4 = 0 C 21 = ( - 1) 2+1 det A 21 = - 2 0 4 1 = - 2 1
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C 22 = ( - 1) 2+2 det A 22 = 1 0 0 1 = 1 C 23 = ( - 1) 2+3 det A 23 = - 1 2 0 4 = - 4 C 31 = ( - 1) 3+1 det A 31 = 2 0 3 0 = 0 C 32 = ( - 1) 3+2 det A 32 = - 1 0 0 0 = 0 C 33 = ( - 1) 3+3 det A 33 = 1 2 0 3 = 3 Therefore, C = C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 = 3 0 0 - 2 1 - 4 0 0 3 . Also, since we’ve already computed the relevant cofactors, it’s easy to find det A : det A = a 11 C 11 + a 12 C 12 + a 13 C 13 = 1 · 3 + 2 · 0 + 0 · 0 = 3 . Therefore, A - 1 = 1 det A C T = 1 3 3 - 2 0 0 1 0 0 - 4 3 = 1 - 2 / 3 0 0 1 / 3 0 0 - 4 / 3 1 (b) Again, we need to determine the various cofactors of A to find the matrix C , but now we can use the fact that A is symmetric, which implies that A ij = A T ji and so C ij = ( - 1) i + j det A ij = ( - 1) i + j det A T ji = ( - 1) j + i det A ji = C ji . Thus, we only need to compute the C ij for which i j : C 11 = ( - 1) 1+1 det A 11 = 2 - 1 - 1 2 = 3 C 12 = ( - 1) 1+2 det A 12 = - - 1 - 1 0 2 = 2 C 13 = ( - 1) 1+3 det A 13 = - 1 2 0 - 1 = 1 C 22 = ( - 1) 2+2 det A 22 = 2 0 0 2 = 4 C 23 = ( - 1) 2+3 det A 23 = - 2 - 1 0 - 1 = 2 C 33 = ( - 1) 3+3 det A 33 = 2 - 1 - 1 2 = 3 2
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Therefore, C = C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 = 3 2 1 2 4 2 1 2 3 . Also, det A = a 11 C 11 + a 12 C 12 + a 13 C 13 = 2 · 3 + ( - 1) · 2 + 0 · 1 = 4 , so we have that A - 1 = 1 det A C T = 1 4 3 2 1 2 4 2 1 2 3 = 3 / 4 1 / 2 1 / 4 1 / 2 1 1 / 2 1 / 4 1 / 2 3 / 4 . 3. Problem 4.4.24. If all entries of A are integers, and det A = 1 or - 1, prove that all entries of A - 1 are integers. Give a 2 by 2 example.
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