Question

1. This question is intended to demonstrate your understanding of basic concepts. (a) Circle all rank 2 matrices below. Give a brief justification. [1 3 6 0] 1 3 6 0] (1 3 6 0 1 1 3 6 0] (i). 0 2 9 1 (i). 0 2 9 1 (iii) 0 2 9 1 (iv) 0 2 9 1 LO 0 0 0 0 1 0 0 1 1 -3 -1 0 0 1 3 (b) Circle all matrices with trivial null space (Nul(A) = {0}). You don't need to justify. (c) Circle all linearly independent vectors. You don't need to justify. (i).{t+1,3t+1,2} (ii). {t +1, 2t +3,4t +5} ( 1) (11) { [D]> 0: 0]} iv). 10, 11, 12 131 2116 [a b c] (d) Let A= d e f and B be another 3 x 3 matrix. We assume det A= 3, det B = -2. 9 h i Compute the following determinant. (i). det (2A7 B-1) [b 3a+c c] (ii). dete 3d + ff h 3g+i i (e) Write down the cofactor expansion of det A with respect to column 2. Do NOT simplify. A= 13 2 5] 0 1 0 (1 7 3

Answer 1

1 (a) i li 3 607 A=1029 1 R₂ = 2 2 ti 3 60 10 12 % ooool Hence matrix A has rank 2 (by row operation ) (ii) [ 3607 R₂ = 12 B = To 291 o oo 1 3 6 07 Tol9/2 % To o o Rz' = R3-fz 1 3 6 0 7 lo i 1 ½ R3 = - Rz To 0 / -1/2 15 [ 3 6 o Tol / / Too I / Hence matrix B has rank 3 r ri 3 6 07 = 1029 1 R₂ = R₃-R II -3 - Ti 3 60 10 2 9 1 Tor2-9- 1 R2 = ₂ R2 Ti 3 bon o 1/1/21 lol / ₂ | R₃ = R₂ +2 R2 0-2 -9 1 Hence matrix has rank a (iv) Similarly . 1 3 607 matrix Tozgi | has rank 3 Lo o 13 Hence for (a) correct answer is 0,

(6 by rank-nulity theorem we get [oo] has Nul (A) = {0} since rank = 2 Toil has nul (A) = {0} since rank = 1 lol has nul (A) = {0} since rank = 2 173 – 47 rank = 1 has nul (A) & 03 since & wronksian of {2+1, 3+1, 2} is wa Itti 3t+1 tl 3 27 - 2 ( 37+3 - 3t-17 Io o 21 - 470 1 Hence {++1, 3t+1, 2} linearly independent (1) since 2(t+1) + 2t+3 = 42+5 Hence {t+1, 2++3, 4445} linearly depedent GT) [:]+[*] = U, hence {10,0), 41] are linearly dependent nce 1 Since lol 2170 hence (iv) since loll 132 6 are linearly independent Hence Hence correct answer O D correct answer

(d) i) det (2AT8') - 2 det (AT) det (B-1) = 2.3 ( ² ) = -3 (1) 16 3atc cle e 3d + f f 1h 3gti il = 6{i (3246) - f (39 + i)} - pate) { ei-th} + ese (39+ i) - h (3h+f)} - 6{3d it if- afga if} - (39+ C) { ei-f} teszge tei-32h-hf} = 3 dib-3f9b - 3a ei +3 afh a cic-zaei + 3 gec teci-3d he ahfc = -3 (aei-ath-bdi + f g h + dhe-gec} -zaei sie ohfe = -343-3a e irhfc =-9-Ba ei-hfe r since det (A)=+37