Question

1. Consider the following matrix and its reduced row echelon form [1 0 3 3 5 187 [1 0 3 3 0 37 1 1 5 4 1 10 0 1 2 1 0 - A=1 4 1 0 3 3 -1 0 rref(A) = 10 0 0 0 1 3 2 0 6 6 -1 3 | 0 0 0 0 0 0 (a) Find a basis of row(A), the row space of A. (b) What is the dimension of row(A)? Is row(A) = R6? Explain. (c) Find a basis of col(A), the column space of A. (d) Find a basis of null(A), the nullspace of A.

Answer 1

A = 1 له - 0 1 o 0 3 5 3 6 3 4 3 6 5 1 -1 - 18 10 0 3 [2 ] rrel (A) = 1 o 0 0 1 0 lo NW 3 2 0 3 1 0 0 0 1 لاگ با 3 4 3 a) Basis of rowspare. All non zero rows in RREF forms a basis of row space. To 3 3 0 371 2 To I 2 I o 47 > Loooo I 3] 1 Dimension of now(A) = 3 Yow (A) + because there are only three non-zero rows.

c) - - column space. The columns containing leading ones are pivot columns. To obtain a baris of column space, we just use prot column from original matrix . T5 oo Null space rvef(A) = 2, + 3x2 + 3x + 3x = 0 x₂ +223 + xy + 40 to as + 32,20 x3, xu, a are free variable 2,5 - 3313 - 34 2 = -223 - y - 3116 - 426 y = as a - 326

+ 26 .: Basis of will spac. - wo - w