Question

1 3 -2 -5 2 11 1. Let A= 3 9 -5 -13 6 3 1 -2 -6 8 18 -1 -1 (a) Find a basis for the row space of A, i.e. Row(A). (b) Find a basis for the column space of A, i.e. Col(A). (c) Find a basis for the null space of A, i.e. Null(A). (d) Determine rankA and dim(Null(A)).

Answer 1

- Solution - Acsi 3 3 9 - -2 -6 -2 -5 8 -5 -13 18 2 17 G 31 -4-id 3 R2-R2 - 32 | R3 R3+2R, lo lo -2 o I 4 -5 2 8 2 17 ool 0 ] o R34 R3-4R2 [13 - -2 1 -5 2 2 0 o R,+R, -R3 - 0 0 - 0 oow 0 o woow. 0 - 0 0 ° : N - as 2 Noon oo 0 - 00 R, R,+2R2 1 0 0 To 0 -1 2 0 0 row.echelon from a) The non- zero rows in the reduced . are a basis for row space. s [ 1,3,0,-1,2,0] ] (A) = 1 [0,0,1, 2,0,0]'? [0,0,0,0,0,1) Row b) Recall that a leading one is the first entry in row. The columns containing lealing ones are the pivot columns To obtain basis for column space, we just use pirot columns from original matrix. Col(A)= ( 11) -2 ] solve the system c) To ☆ find basis for null space, ri 30 -1 207 0 0 1 2 0 0 0 0 0 0

leading variables 21,713, H6 free variables 22,24,25 24,+372 - 244 +275 = 0 23 +224 :o 216 -O 22=t,tEIR an=s ,SEIR, 25=u , UGIR → 23 --2014 --25 x = -3x2 +24-225 <-3t+S-24 F-35 + 5-2017 Ot+1-25 + +ť a uso oooo 0 - 0 0 on oo- .: Basis for null space of A is Nul(A) Oooo. 0-000 d) rank A=3 .... No. of non-zero rows in reduced now echelon form ... No. of vectors in dim (Nullian) = 3 Nell (A)