Question

For the given matrix B=

1	1	1
3	2	-2
4	3	-1
6	5	1

a.) Find a basis for the row space of matrix B.

b.) Find a basis for the column space of matrix B.

c.)Find a basis for the null space of matrix B.

d.) Find the rank and nullity of the matrix B.

Answer 1

matrix B is

1	1	1
3	2	-2
4	3	-1
6	5	1

convert into Reduced Row Eschelon Form...

Add (-3 * row1) to row2

1	1	1
0	-1	-5
4	3	-1
6	5	1

Add (-4 * row1) to row3

1	1	1
0	-1	-5
0	-1	-5
6	5	1

Add (-6 * row1) to row4

1	1	1
0	-1	-5
0	-1	-5
0	-1	-5

Divide row2 by -1

1	1	1
0	1	5
0	-1	-5
0	-1	-5

Add (1 * row2) to row3

1	1	1
0	1	5
0	0	0
0	-1	-5

Add (1 * row2) to row4

1	1	1
0	1	5
0	0	0
0	0	0

Add (-1 * row2) to row1

1	0	-4
0	1	5
0	0	0
0	0	0

reduced matrix is

there are 2 pivot entry at first two columns

hence rank is 2

so basis of column space are

.

basis of row space are  

.

.

and there are no pivot entry at third column hence nullity is 1

.

.reduced system is

...................free variable

.

general solution is

basis of null space is