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Unformatted text preview: Math 217, Linear Algebra, Fall 2002 Exam 2 Solutions 1.(8pts) Let A be the matrix A = 1 2 3 4 5 6 7 8 9 10 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 60 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 . It turns out that A x = is true for both x = 12 1 and x = 6580 5 4 3 2 1 . Does A x = b have a solution for all possible b ∈ R 9 ? Explain. (Hint: don’t use your calculator). We know that rank( A ) + dim(Nul( A )) = 10, the number of columns. Each of the given vectors is an element in Nul( A ) = { x ∈ R 10  A x = } , and they are clearly independent (because there are only two, we may check to see that they are not multiples of each other, which is true), so dim(Nul( A )) ≥ 2. This means that rank( A ) ≤ 8, and in particular that A can have at most 8 pivots. This means that in row reduced echelon form, A has a row of zeros, and thus the equation A x = b will not be consistent for all b ∈ R 9 . 2.(10pts) Suppose that the matrix B = 1 2 0 0 0 3 0 0 0 0 3 2 0 0 0 1 was obtained from A by the following row and column operations (performed in the order listed). 1. I replaced row 4 with row 4 plus 2 times row 3. 2. I replaced col 1 with col 1 minus 3 times col 4. 3. I swapped rows 2 and 4. 4. I multiplied row 3 row by the scalar 3. 5. I replaced row 4 with row 4 plus 4 times row 3. 6. Finally I swapped columns 3 and 1. What is det A ? Show your work....
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This note was uploaded on 04/13/2010 for the course ECON 330 taught by Professor Minetti during the Fall '08 term at Michigan State University.
 Fall '08
 MINETTI

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