Question

Find a basis for the row space of A. 1 -1 3 2 -3 8 A-0 1 -2 Find a basis for the null space of A. Verify that every vector in row(A) is orthogonal to every vector in null(A). Need Help? Submit Answer Save Progress Practice Another Version 17. -12 points PooleLinAlg4 5.2.009. Find a basis for the column space of A. My Notes Ask Your Tea 1-1 3 5 2 1 A- 012 T. Find a basis for the null space of A' for the given matrix. Verify that every vector in col(A) is orthogonal to every vector in null(A')

Answer 1

382 1 1313 1205

1	-1	3
2	-3	8
0	1	-2
5	-3	11

convert into  Reduced Row Eschelon Form...

Add (-2 * row1) to row2

1	-1	3
0	-1	2
0	1	-2
5	-3	11

Add (-5 * row1) to row4

1	-1	3
0	-1	2
0	1	-2
0	2	-4

Divide row2 by -1

1	-1	3
0	1	-2
0	1	-2
0	2	-4

Add (-1 * row2) to row3

1	-1	3
0	1	-2
0	0	0
0	2	-4

Add (-2 * row2) to row4

1	-1	3
0	1	-2
0	0	0
0	0	0

Add (1 * row2) to row1

1	0	1
0	1	-2
0	0	0
0	0	0

reduced matrix is

01 -2

there are 2 pivot entry at first and second column

so basis of row space are

(1 0 1)

(01-2

.

.

.

reduced system is

-200 0100 1000

$x+z=0................x=-z$

$y-2z=0................y=2z$

$z=z$.........free

.

general solution is

1

.

basis of null space is

-121

.

.

.

.

.

.

.

.

31-1 1 2 150-

1	-1	3
5	2	1
0	1	-2
-1	-1	1

convert into

Reduced Row Eschelon Form...

Add (-5 * row1) to row2

1	-1	3
0	7	-14
0	1	-2
-1	-1	1

Add (1 * row1) to row4

1	-1	3
0	7	-14
0	1	-2
0	-2	4

Divide row2 by 7

1	-1	3
0	1	-2
0	1	-2
0	-2	4

Add (-1 * row2) to row3

1	-1	3
0	1	-2
0	0	0
0	-2	4

Add (2 * row2) to row4

1	-1	3
0	1	-2
0	0	0
0	0	0

Add (1 * row2) to row1

1	0	1
0	1	-2
0	0	0
0	0	0

reduced matrix is

01 -2

there are 2 pivot entry at first and second column

so basis of column space are

1 150

.

.

.

.

take transpose

$A^T=\begin{pmatrix}1&5&0&-1\\ -1&2&1&-1\\ 3&1&-2&1\end{pmatrix}$

1	5	0	-1
-1	2	1	-1
3	1	-2	1

convert into

Reduced Row Eschelon Form...

Add (1 * row1) to row2

1	5	0	-1
0	7	1	-2
3	1	-2	1

Add (-3 * row1) to row3

1	5	0	-1
0	7	1	-2
0	-14	-2	4

Divide row2 by 7

1	5	0	-1
0	1	1/7	-2/7
0	-14	-2	4

Add (14 * row2) to row3

1	5	0	-1
0	1	1/7	-2/7
0	0	0	0

Add (-5 * row2) to row1

1	0	-5/7	3/7
0	1	1/7	-2/7
0	0	0	0

reduced system is

$\begin{pmatrix}1&0&-\frac{5}{7}&\frac{3}{7}\\ 0&1&\frac{1}{7}&-\frac{2}{7}\\ 0&0&0&0\end{pmatrix} \:\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=0$

$x-\frac{5}{7}z+\frac{3}{7}w=0...................x=\frac{5}{7}z-\frac{3}{7}w$

$y+\frac{1}{7}z-\frac{2}{7}w=0...................y=-\frac{1}{7}z+\frac{2}{7}w$

$y=y$.........free

$z=z$.........free

.

general solution is

$\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}\frac{5}{7}\\ -\frac{1}{7}\\ 1\\ 0\end{pmatrix}z+\begin{pmatrix}-\frac{3}{7}\\ \frac{2}{7}\\ 0\\ 1\end{pmatrix}w$

.

basis of null space are

${\color{Red} \begin{pmatrix}\frac{5}{7}\\ -\frac{1}{7}\\ 1\\ 0\end{pmatrix},\:\begin{pmatrix}-\frac{3}{7}\\ \frac{2}{7}\\ 0\\ 1\end{pmatrix}}$