Question

1 2 0 1 10. Let A = 2 3 1 1 3 5 1 2 (a). Find the reduced row echelon form of A. (b). Using the answer for (a), find rank(A), and find a basis for Col(A). (c). Using the answer for (a), find a basis for Nul(A).

Answer 1

Solution (10) 1 2 A = 2. -- 3 - 3 5 (a) A R₂ - 2R, R₂ R₃-3R, 1 2 O - - -T7 R₂ R₃ - R₂ 0 0 FTD R₂ (-1) R₂ 2 O 2. O - -U O O R, RI 2R₂ 0 0 - 0 2 2 O O This is the reduced 9co echelon foem of A. (6) 48 the non-zere а row Basis for Colemn space A leading one firest entry in The columns containing leading ones the pivot columns, To obtain basis for the column space, we just wee the pivot columns focom the original matrix. Gece a

Barts for Column Space 98 2. 2 3 3 5 Rank of A dimension of column space of A Rank (A) = 2 the null space, solue c) To matrix equation find the O 2 - 21 O O X3 O xu O Since x3 axu + asce free variables Let xz=t > xy = 8 x + 2x₂ - xy = 0 U x = -2x3 + xy = x=-2t+8 xz - x3 + xy = 0 -) x₂ = x3 - xy = x, = t-s -2tts -2 t t-s + 8-1 O x3 t xu 8 Basis for null space is ១