Question

1 1. The matrix A and it reduced echelon form B are given below. 1 -2 9 5 4 1 0 3 0 0 -1 6 5 -3 0 1 -3 0 -7 A= ~B= -2 0 -6 1 -2 0 0 1 -2 4 9 1 -9 0 0 0 0 0 (a) Find p, q, r s.t Nul A, Col A, Row A is a subspace of RP, R9, R”, respectively o 1 Answer. p = a = re (b) Find a basis for Nul A (c) Find a basis for Col A (d) Find a basis for Row A (e) Is Col A = Col B? Justify (f) Fill in the spaces, and justify dim Col A dim Nul A rank A rank AT Nullity AT Answer:

Answer 1

If you have any doubts in the solution please ask me in comments here i used basic definition of all parts...

* * The matrise A and it seduced row echion form B ore given Below 9 5. u 3 03 :-2 - A = 5 -3 6 0 - -3 ܝܙ -1 B = - 2 0 0 1 -2 - - -2 0 - 6 -9 1 4 9 1 Xe Du0 A a) subspace. of IRP Sot. is NUI A A is GX5 matrum Ax=0 Ax is possible iff Xe (RS Null A is subspace of IR 5

there are 5 column Here in matrix А Row A is subspace of IR 5. there are 4-row in matrix A COLA is Subspace of IR 4. P=5 4 * 7 = 5 Answer .4 (6) NOW to zind Basis for NULLA 20 Null A Ax = 0 XEIRS 1 - 2 9 s 4 X 6 S -3 R2 - -2 - 6 -2 4 9 -g xy 25 Now reduced 20w echion form of A 16 B SO AX = 0 is same as B x = 0 System is Homogenous) 3 0 0 0 -3 O 22 -7 ***** 0 0 1 -2 O 0 22-373 745 = 0 , xu - 2x5 =O, 22, +3X3 =0, → 24 = 225, 24 = -383., 22 = 3x3 + 7H5 X) → - 3х3 3x₃ +785 13 -3 3 +7 Hz + 13 O 0 Ru 225 0 - US 25 Basis of Nul A = Answer

(c) TO Bind Basis of column A. Solve? 1 9 5 4 A = 1 6 s - 3 -2 O -6 - 2 - 4 1 9 -9 CT 2 cu 65 NOW deduced Dow echlon borm. of A 0 3 o 0 0 - 3 0 -7 pivot is only on 1st, 2nd, wth column in reduced row echion form of . A. -2 O C1, C2, Cu column are Linearly inden pendent S Basis of column A = Answe (D) To Bind Row A Basis reduced row echlon form only non in A matux Zero row form а Basis of Row - So Basis is collection Non zero row of Row A of an of B Basis of ROW A = -2]}. & [10300];[01-3 0 -7], [o oo { I 3 > Basis. of Row A= 0]; -7]; -3 0 Answer [o [o i o -2] }

e) Is Column A col B ? (Justify) Solution 001 A col B. Basis of col A = Because S All Νοω in matrix B it is alseady in reduced how echion form SO the Basis of col 3 = [:].6.1.1:13 COIA - = span ) colb= span s Col(A). 1:12:03:01 ColB never have the sith entry non Zew. COLA have Luth entry Non-zew. But Many time COIA Col B Answe (6). is the No. of total element dim of any iOR? Subspace the Basis of subspace in oth OIA dom (Got A ): Basis have 3 - element 2 NullA Basis have 2- element clim (NUTA) dim (CoIA) = rank (A) = 3. 3 Νοω rank (A) = rank (AT) = No of column in AT = 4. matrux 5xu AT is Null ATt rank AT 4 By Jank - Nullity theorom NurlL AT = 4-3 = 1 Nullity AT 1.

3; dim COLA 2. dim Nul(A) = 3 rank (AT) Ansul rank A 3 1 Answer Nullity AT