Question

2) Let (1 3 15 7 -20 A= 2 4 22 8 3 1 2 7 34 17 -1 3 be given (a)( 10 pts.) Find the reduced echelon form of A. (b)(5 pts.) Find a basis for the Row(A). (c)( 5 pts.) Find a basis for the Col(A). (d) (5 pts.) Find a basis for the Null(A). (e)( 5 pts.) What are the rank and nullity of A?

Answer 1

O-M in A: 12ů 22 8 3 3 5 -2 l El 2 7 34 17 - 3 Reduced Echelon form R₂ = -2R,+ R2 si 3 15 7 -2 on 2 7 34 17 - 3 R3 = -2R, tR3 Ti 3 15 7 -2 og 10 -2 -8 6 711 Lo 4 3 3 3 1 3 15 7 - 20 -> +1 +4 +3 - -! Lol 4 3 3 3 J Rz=-R₂ + R₂ si 3 is 7- 20 7 Loooo 137, R3 = 3 Rz Eri 3 is 7-2 o u 3 --), Am I R₂ = 7 R₂R [ 3 15 o 7 -2 21 Loo oo z 13 R = 2 R₂ + R 13 15 o 142 ou 3018 Loooo 7 R = -3R + R. Sio 3 -2 o 407 Is It is in Reduced lo 4 30 12 Echelon form. ooo 0 72 J s leading entry in each a It Irow is only non-rend l in the column. I

(6) Row oWCA - ) 7, -2,0), (2,4, 22, 8, 31), (2, 7, 34, 17,-1,32 VA Now, to check basis all three vectors should be linearly independent : av, tbv + cvz = 0 a+26+ 2c co7 a-6c+20= 0 ) a=4C - 3 at ub+7c = 0 150 +226 +34c = 0 7a18b+ 170=0 - 2a + 3b-cool 6+30 = 0 6 = -3C = . Put a&b in (2. oooo 3 (4c) +42-30) 47 c = 20 =0 & CEO So, abec:0 Only bossible solution for a=6=c=0 So, V, V₂, U3 are linearly indebent Hence, basis for Row (A) = $(1, 3, 15, 7, -2,0), (2, 4, 22,8,31) ? -(2,7, 34,17, -1,3) Col(A) = 0,2,3), (,4,7), (5,92,34), 0,4,17), 63, 3-1), v va uz (0.153) 75 for basis, we need to find set of lineare in vectors from COLCA). So, avtovat CV₂ + avut erst fuo ; if only ones o lincerly independent borazbac-de-fio then indehendent -

no 3-2 o 40 Lol4 3 0 18 Loooo! Row-reduced form a +38-2d-40 f = 0 OOOOOO 1 I c, d, f are free 6+40+3d +18 f = 0 e + 7 f = 0 a = 31-2d-40 of 6=-4c-3d - 18 € e = -7 f GA N(A) = N(rired (A)) = seuruet ax cooow i & N(A)= span/ (-376-3 MJoooo OOO pool

For Column space, au, + buat cuz t ovat eust fy=0 Now set free variable C = 0, d=0, f = 1 62-18 ant from breviously calculated equations L 6 B ce 7 a=30-2d- of e-7f je b= +4C-3d-18 of Ae so v aut bv devist ters]; vois lebendas rector so Now, check for v2, vs, Vy, us aut butt vzt dut evsco but cond=lf has removed, So remove it a = 3c-2d =-2 / Non-zero solution for a, b, c, d, e. 6 = -3 e-o so, Vu - Lavt buat cua te vor] ; Vu is defendent lector Remove it from the set & a from the epation. Now, check for U,,U2, U3, vs. avt buat Waters. Now, c is free variable but cal a= 32 so vsa vtbe te vol Va is dependent vator. So Remove it from set. & cfrom free variable. Now Set become su, V2, VZ & No free variable a=07 so, Column (A) = sban ( e vog bzo e=0

span (ColumA) = f (1, 2, 3, 6, 4,7) (-2, 3, -1)} na 17,0 = (No too vectors are parallel as well). (Noomber of elements in Nullsbace). Nullity (A) = 3 Rank of A = 3 (Number of element in columnsbace) Sot
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