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Unformatted text preview: Linear Algebra Review Problems for First Exam–Solutions 1. (12 points) Find all solutions to the following system of linear equations 1 2 1 2 1 1 7 5 2 x y z =  3 1 Solution: Represent as a matrix and row reduce: 1 2 1 3 2 1 1 1 7 5 2 1 2 1 3 3 3 7 9 9 21 1 2 1 3 0 1 1 7 3 0 0 1 0 1 5 3 0 1 1 7 3 0 0 So the solution set is { 5 3 s 7 3 + s s ; s ∈ R } . 2. (10 points) Let V be the set of all 2 × 2 matrices. (a) Explain why V forms a vector space, describe a basis for V and find dim V . (b) Let S be the set of all matrices a b c d such that a + b + c = 0 and a + d = 0. Show that S is a subspace of V , find its dimension, and find a basis for S . Solution: (a) V forms a vector space since we have addition and scalar multiplication. All the vector space axioms hold since they hold for matrix algebra. V has the basis { 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 } so V has dimension 4. (b) Let a b c d , e f g h ∈ S . Consider a + e b + f c + g d + h . Then ( a + e ) + ( b + f ) + ( c + g ) = ( a + b + c ) + ( e + f + g ) = 0 + 0 = 0 and ( a + e ) + ( d + h ) = ( a + d ) + ( e + h ) = 0 + 0 = 0 so that a b c d + e f g h ∈ S . Similarly, if k ∈ R then consider k a b c d = ka kb kc kd . This matrix satisfies ka + kb + kc = k ( a + b + c ) = k 0 = 0 and ka + kd = k ( a + d ) = k 0 = 0 so that k a b c d ∈ S . Since 0 0 0 0 ∈ S , S is nonempty so that S forms a subspace of V . Notice that any element in S may be written as s s t t s = s 1 1 1 + t 1 1 so the set T = { 1 1 , 1 1 } spans S . T is obviously linearly independent, so that T is a basis for S . Since S has 2 elements, it follows that dim S = 2....
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This note was uploaded on 10/18/2011 for the course S 2010D taught by Professor Peters during the Spring '10 term at Columbia.
 Spring '10
 Peters

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