Question

Question 1 (1 point) Let A be the matrix defined below. -8 8 -8 1 -9 7 4 3 A= 7 6 -7 -9 4 9 5 5 -5 7 6 -7 -1 0 -7 -7 Suppose we know that ele 100 0 1 0 } RREFA= 10 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 Find a basis for the null space of A. O -87 6 5 O -9 9 -9 3 من و داده و حاد O 3 5 1 a 0 1 0 O

Answer 1

Sol 1: Given! -8 -8 A = 8 -9 HJ 3 7 7 -7 9 4 - 4 6 -9 La 6 41 -7 -7 7 -7 and row reduced echelon form of A is 513 213 413 1/3 0 RREF A 1 Ооо OOO find To null space of A Ax=0 : XERS 6 له Nul A since as RREFA is reduced echelon form form of A, solution set of AX=0 is same the solution set of (RREF A) X which gives, 4/3 RI - 513 213 0 1 0 23 - -1 sly 0 0 us 0 0

t 1) - " أتر Х 1 4 м, го хо sinh 2 3 О хч + (- 5 ) і. my Sқ free variables with Az and us ч ху - Уз 1x Cu() Б ) 3 2 х3 L 3 - X₂ К. free 24 Хr free - ч/з С. - S|3 - 2/3 Thus xt | -1/3 15 x2 ใน 1 1 he 1 Nr

Thus 513 -213 -413 -13 spans null space of A 1 0- vectors are two ALSO these as not multiple of each Linearly independent are other, they 1 5/3 -2/3 Hence basis for -113 is a is 1 A. null space of