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Unformatted text preview: MATH 51 MIDTERM 1 SOLUTIONS February 2, 2006 1. Find all solutions of the following system: x 1 + 2 x 2 + x 3 + x 4 = 7 x 1 + 2 x 2 + 2 x 3 + x 4 = 12 2 x 1 + 4 x 2 + 6 x 4 = 4 Solution: 1 2 1 1  7 1 2 2 1  12 2 4 6  4 → 1 2 1 1  7 1 2  5 2 4  10 → 1 2 3  2 1 2  5  . The last augmented matrix corresponds to the system x 1 + 2 x 2 + 3 x 4 = 2 x 3 2 x 4 = 5 = . So: x 1 = 2 2 x 2 3 x 4 x 3 = 5 + 2 x 4 (where x 2 and x 4 can take any values), or x 1 x 2 x 3 x 4 = 2 5 +  2 1 x 2 +  3 2 1 x 4 2(a). What three conditions must a set V of vectors in R n have in order to be a linear subspace? Solution: It must contain the zero vector, it must be closed under addition, and it must be closed under scalar multiplication. 2(b,c) . Determine which of those three properties each of the following sets has: 1 2(b) . The set V of vectors x in R n such that each coordinate of x is an integer. (The integers are the numbers 0, 1, 1, 2, 2, etc.) Solution: Contains , closed under addition, but not closed under scalar multipli cation. (For example, e 1 ∈ V , but (1 / 2) e 1 / ∈ V .) 2(c) . The set W of vectors v = x y z in R 3 such that xyz = 0. Solution: It contains the zero vector. Also, it is closed under scalar multiplication. But it is not closed under addition. (For example, 1 1 and 1 are both in W , but 1 1 + 1 = 1 1 1 is not in W .) 3. Consider the system: 1 2 3 3 7 10 2 5 7 1 3 4...
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 '07
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 Math, Linear Algebra, Algebra, Differential Calculus, Vector Space, scalar multiplication

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