Answer to Question #238438 in Linear Algebra for Zizan

Question

Find the basies of the column space and null space of the matrix.
A =

1 −2 5 −3 4
2 3 1 −4 2
.
• Find the the rank(A) and Nullity(A)
• If the matrix is augmented by a column what would be the rank and nullity of the
matrix?

Answer 1

rref(A)= |1 0 17/7 -17/7 16/7|

| 0 1 -9/7 2/7 -6/7|

Basis of column space are 1st and second column

Column space=

$\begin{bmatrix} 1 \\ 2 \end{bmatrix}$$\begin{bmatrix} -2 \\ 3 \end{bmatrix}$

Rank of A= 2

Nullity of A=3

Basis of Null space

From rref of (A)

X1 +17/7X3 -17/7X4 +16/7X5 =0

X2- 9/7X3+ 2/7X4 -6/7X5=0

Let X3, X4 and X5 = 1

X1 -17/7 17/7 -16/7

X2 9/7 -2/7 6/7

X3 = X3 1 + X4 0 + X5 0

X4 0 1 0

X5 0 0 1

Null space has basis formed from

-17/7 17/7 -16/7

9/7 -2/7 6/7

[ 1 , 0 , 0 ]

0 1 0

0 0 1

Augment A by a column e.g $\begin{bmatrix} 7 \\ 2 \end{bmatrix}$

(A|B)= [ 1 -2 5 -3 4 |7 ]

[ 2 3 1 -4 2 |2 ]

rref(A|B)= [1 0 17/7 -17/7 16/7 |25/7]

[ 0 1 -9/7 2/7 -6/7 |-12/7]

Rank of Augmented A =2

Nullity of Augmented A=4