MidTermSoln

# MidTermSoln - MATH 136 Winter 2008 Midterm Examination...

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MATH 136 Winter 2008 Midterm Examination: Solutions February 13, 2008 [10] Question 1: Mark each statement True or False. No justification necessary. 1 mark for each correct answer, - 1 2 for each wrong answer and 0 for no answer. (i) Let A be a 4 × 3 real matrix. Then Col A is a subspace of R 4 . True False (ii) If A is any 5 × 3 matrix and b is any vector in R 5 such that the equation Ax = b has a solution, then the solution is unique. True False (iii) Let V be a vector space over a field F . If ax = bx for a, b F and x V , then a = b . True False (iv) If f and g are polynomials of degree 100 over rationals, then f + g is a polynomial of degree 100. True False (v) The span of three vectors in a vector space V is a subspace of V . True False (vi) The set of real numbers is a vector space over the field of rational numbers. True False (vii) Let V = C 2 = { ( x, y ) : x, y C } and W = { ( x, y ) V : x 2 + y 2 = 0 } . Then W is a subspace of V . True False v 1 = (1 , i ) , v 2 = ( - 1 , i ) gives v 1 + v 2 = (0 , 2 i ) / W . (viii) Let T : R 2 R 2 be given by T ( u 1 , u 2 ) = ( u 1 , u 2 ) if u 1 = 0 (0 , 0) if u 1 = 0 . Then T is a linear transformation. True False v 1 = (0 , 1) , v 2 = (1 , 1) gives T ( v 1 + v 2 ) = T ( v 1 ) + T ( v 2 ). (ix) Let T : R 3 R be given by T ( u 1 , u 2 , u 3 ) = u 1 + u 2 + u 3 . Then T is a linear transformation. True False (x) Let T : R n R be such that T ( u ) =largest component of u . Then T is a linear transformation. True False v 1 = (1 , 2 , · · · , n ) , v 2 = ( - 1 , - 2 , · · · , - n ) gives T ( v 1 ) = n, T ( v 2 ) = - 1 but T ( v 1 + v 2 ) = 0.

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MATH 136 Winter 2008 Midterm Examination: Solutions Page 2 [6] Question 2: You are given M 2 × 2 ( R ), the set of all 2 × 2 matrices over R ; P 3 ( R ), the set of all polynomials of degree 3 over R together with the zero polynomial, and R 4 . Standard rules for matrix, polynomial and vector addition apply, as does the standard rule for scalar multiplication. Scalars are drawn from R . [2]( i ) Is A = 3 4 0 2 in Span B = 1 1 0 1 , C = 1 0 0 1 , D = 0 0 0 1 ? If yes, express A as a linear combi- nation of B, C and D . If not, explain why not. Yes . Let b, c, d be such that b 1 1 0 1 + c 1 0 0 1 + d 0 0 0 1 = 3 4 0 2 or b + c b 0 b + c + d = 3 4 0 2 . Then b = 4 , b + c = 3 , b + c + d = 2 giving b = 4 , c = d = - 1. Hence A = 4 B - C - D . [2]( ii ) Is a ( t ) = 3 + 4 t + 2 t 3 in Span { b ( t ) = 1 + t + t 3 , c ( t ) = 1 + t 3 , d ( t ) = t 3 } ? If yes, express a ( t ) as a linear combination of b ( t ) , c ( t ) and d ( t ). If not, explain why not. Yes . Let b, c, d be such that b (1 + t + t 3 ) + c (1 + t 3 ) + dt 3 = 3 + 4 t + 2 t 3 or ( b + c ) + bt + ( b + c + d ) t 3 = 3 + 4 t + 2 t 3 Then b = 4 , b + c = 3 , b + c + d = 2 giving b = 4 , c = d = - 1. Hence a ( t ) = 4 b ( t ) - c ( t ) - d ( t ).
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