Question

Q5. Assume that the following two matrices are row equivalent: A= -2 4 -2 4 2 -6 -3 -3 8 2 -3 1 B= 1 0 6 - 7 0 2 5 - 5 0 0 0 -4 Find bases for the column space and null space of A.

Answer 1

Given A= 2 4 -I 4 -3 1 3 9-3 basis for Column Space Multiply dow 1 by -1/2 ; i-e R,x-1 / 2 A- -2 1 - 2 -6 -3 +8 g -3 -3 Now perform this operation on Rowa (RO) Rg -RitRg A = - 5 5 2 -3 -3 f8 on Row 3 (R3) Now perfoom this operation R37 3 R14 R₂ 1 A= TE of or on -2-5 buy is Multiply dowa by -1/2 i . e RgX- A= -2 - Now 5/2-5/2 2 5-9 perform this operation on Row 3 ROW3 (R3) R3 - 2 Rg +R3

A= 59-519 o --4 Multiply dow3 by - i. e R₂X- Ч A - 5/2 -52 0 0 1. Ajow perform this operation on Row 2 R2 / / 3 R₂ + Rg (R3) As I 572 0 Now perform this operation of Row A (R1) R, QRgt RI As - [ AS 5/20 Now perform this operation on Row 1 (R1) Rit Q Rg+R, CO A leading one is first non zero-entry in a row. The columns Containing leading ones pivot Columns. 6 5/2 o are the

To obtain a basis for the Column space, we just use the pivot Columns from original matrix basis for Coloma space - RCA () * [ 5/9 'Basis for Null space :- we began with Given matrix : -9 y As 4 -9 410 2 -3 q O co -3 -3 multiply row 1 by at 2: R, Xhe As -6 8 -3 2 -3. -3 Now perform this operation on Row 2 (R2): Rg ~ -2R1 + R2 A= -3 -5 2 8 -3 Now perform this operation on Row 3 (R3) R37 3R + R3 -2 - A= -2 - 2 2 s 5 - 9 0

A = 1-2 Multiply dowa bya) -(fekete 1 +12 720 og 5-910 perform this oparation on Row 3 (B):4R37 - 2 R3tR3 -2 5/ -5/2 -4 Multiply dow by : - (R3x+1) = " - -2 5/25/2 Now paform this operation on dow 2 (R2) Rg - 5 R3 t Rg - A= 5/9 Now perform this operation on dow R, QR3+RI As 5/9 * 1 Now perform this operation on dow. 1 (R.) Rit Q Rg+RI 6 0 1 5/9 o оо A= O

convert the matrix equation back to an equivalent system x + 6x₂ = 0 X2 + 3 x = 0 Xy=0 Add an equation for each free variable Xi + 6xz so X2 + 5 x = 0 =Xz xyco solve variable in terms of free variables : x = -623 2q = -5/2dz x3 =23 1-683 Factor collect terms into vectors 23 for each xy=0 ai x2 w 2016 dy Factor out variables on right side 21 22 23 512 X3 xy Thus a basis for null space is -6 -5/9

basis for for matrix After calculating column space and noll space matrix , we can find column space of (A&B) (not equal to). and we can find null space of A= since A and B are row equident,