Question

Find bases for Cola, RowA and Nula, for A=) 2 - 2 2 - 2 (4-4-4-4) 3 -3 30

Answer 1

1 Given matrix ? Solution - PR • R2 A- admt -I -2 2 -2 -3 3 - 4 4 -4 R4 4x4 Firstly we find Echelon form of A. ܬ -l 1 To -17 An R₂ R₂-2R, R₂ R₂-3R, 3 O o RA Re-the he I va OOO gowy R₂ GR al I I A R2 & Re c) Row (A) = span { Ri, R1 R3, R3} 1 Basis for Row (A) 2 non zero :: B1A) a no. of: = no of elements in basis of Row (A) = 2. columne A) span -2 3 -4 from row reduced Echelon form of matrix A, we have only two pivot elements corresponding to first and fourth coleman so first and fourth column is linearly independent.

i Bosis for column (A) = 2 -2 3 Bus Further Null(A) {xegt: : Ax=0} X= (22.22,23,24) ER*.st let AXO cie 1 x 2 -2 2 x2 3 -3 -4. . 3 4 -4 23 Exa using row reduced form of from eguna 1 - ㅗ 2 X, 0 0 24 or) x,-X2 +83 xq=0 2470 *,*248320 (**420) %= %,+83 -: = $_272 No. of independent 8019 = no SIA) choose х,а , *3= ß iky a OB. "X = B 6-161-0-00 {[!103 i Bosis for Null (A) = (a) = { Answer