Question

4. A= -2 4 -2 2-6-3 8 2 B= 1 -3 1 0 6 - 7 0 2 5 -5 0 0 0 -4 bases for the column space and null space of A.

Answer 1

lo 4 -2 À 2 -6 - 3 -3 8 2 -3 Basis for Coloumn Space of A. Multiply Row by the Add -2 times Row I to Row 2 Soj A = 2 o -2 -5 2 ook m -3 8 Add 3 times Row And to Row 3 then multiply Row & by a by - 1 / 1 -2 Az -2 -512 O 512 5 0 حو -9 Add -2 times Row 2 to Row 3 them multiply Row 3 by - 4 - 2 Az -a -512 512

Add وا؟ times Row 3 to Row Az DO 512 0 Add 2 times Row 3 to Row and then Add 2 times Row 2 to Row I --> A - 0 6 0 ' 0 572 0 0 0 Leading One first is non zero entry in a Row To obtain a Basis for Column Space; we Pivot celoumns from Original use the Just Matrix A : Basis of Colaumn Space A 2 4 ) ) 2 -6 -3 -3

Now Basis for Null space of ay A 4 - 2 J 2 - 6 -3 -3 8 2 Converting into echlen form After from equation * we get A = O 0 0 512 o 0 Set AX=0 Where X 42 13 Hy -> Hit bug = Ug+ [43=0 Hy to H3=43

Solve for each Variable Hia -603 Har -43 Hz - H3 Hyco M и2 - 6 -512 из U3 hy thus basis for Null Space of A A is - 6 -512 0