1997solut1

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Unformatted text preview: Physical Sciences Division University of Toronto at Scarborough MATA04Y Solutions to Term Test I 1997/98 Part A 1. For all real numbers r; s 2 R and all vectors u; v; w 2 R the following eight conditions hold A1. (u + v) + w = u + (v + w) A2. u + v = v + u A3. 0 + v = v A4. v + (?v) = 0 S1. S2. S3. S4. r(u + v) = ru + rv (r + s)u = ru + su r(sv)) = (rs)v 1v = v 2. For all u; v 2 R we have ju vj jujjvj with equality if and only if u and v are parallel. 3. (a) The rst kind of row operations switches two rows of a matrix. The second kind multiplies a row by a non-zero scalar. The third kind replaces a row by its sum with a multiple of another row. (b) It is a matrix obtained from the identity matrix by applying a single elementary row operation. 4. We have (AB )(B ?1A?1 ) = ABB ?1 A?1 = AI A?1 = AA?1 = I . Similarly (or by appealing to a theorem proved in class) (B ?1A?1)(AB ) = I , so (AB )?1 = (B ?1A?1 ). 5. Let p be any solution of Ax = b (i.e., we are assuming the system in consistent). Then p + h is also a solution whenever h is a solution of the associated homogeneous system Ax = 0. Moreover every solution of Ax = b has this form. 6. (a) A subspace of R n is a subset W R n which is i. nonempty ii. closed under vector addition iii. closed under scalar multiplication (b) Let W be a subspace of R n . The vectors b1; : : : ; bk 2 W form a basis of W if and only if each vector in W is a unique linear combination of b1; : : : ; bk . An alternate de nition requires: MATA04Y page 2 7. (a) (b) (c) (d) (e) (f) (g) i. sp(b1 ; : : : ; bk) = W ii. The only solution to the equation x1 b1 + + xk bk = 0 is the trivial one x1 = 0; : : : ; xk = 0. False (The system may be inconsistent) True (The system is consistent and will have at least one free variable) True True False (The system Ax = b may be inconsistent) False (This is true when the system is homogeneous) False (this is true when the line passes through the origin) only only Part B 1. (a) The r...
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