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2. 34. The set of all polynomials of the form at3 + bt2 + (b − a), where a and b are any real numbers. n 35. We show that {cv1 , v2 , . . , vk } is also a set of k = dim V vectors which spans V . If v = vector in V , then ai vi is a i=1 n v= a1 ai vi . (cv1 ) + c i=2 36. Let d = max{d1 , d2 , . . , dk }. The polynomial td+1 + td + · · · + t + 1 cannot be written as a linear combination of polynomials of degrees ≤ d. 37. If dim V = n, then V has a basis consisting of n vectors. 10 then implies the result.
Use matrix multiplication cA where c is a row vector containing the coefficients and matrix A has rows that are the vectors from Rn . 4, p. 215 2. (a) 1 does not belong to span S. (b) Span S consists of all vectors of the form is not in span S. a 0 , where a is any real number. Thus, the vector 0 1 (c) Span S consists of all vectors of M22 of the form Thus, the vector 4. (a) Yes. (b) Yes. a b , where a and b are any real numbers. b a 1 2 is not in span S. 3 4 (c) No. (d) No. 6. (d). 8. (a) and (c).
Thus xh is in fact a solution to Ax = 0. 30. (a) 3x2 + 2 32. 3 2 2x (b) 2x2 − x − 1 − x + 12 . 34. (a) x = 0, y = 0 (b) x = 5, y = −7 36. r = 5, r2 = 5. 37. The GPS receiver is located at the tangent point where the two circles intersect. 38. 4Fe + 3O2 → 2Fe2 O3 40. x = 1 4 0 . − 14 i 42. No solution. 3, p. 124 1. The elementary matrix E which results from In by a type I interchange of the ith and jth row differs from In by having 1’s in the (i, j) and (j, i) positions and 0’s in the (i, i) and (j, j) positions.