Question

1 2 -3 1 -6 -2 5 2. 4. (10 points) Let A = (a) (5 points) Find a basis for col(A) and calculate rank(A). (b) (5 points) Find a basis for null(A) and calculate nullity(A).

Answer 1

Solution:

$\small A=\begin{bmatrix} 1 & -3 & 1 & 5\\ 2 & -6 & -2 &2 \\ -1 & 3 & 0 & -3 \end{bmatrix}$

Reduce the matrix $\small A$ using elementary row transformation.

$\small R_{2}-2R_{1},\: \: R_{3}+R_{1}$

$\small A=\begin{bmatrix} 1 & -3 & 1 & 5\\ 0 & 0 & -4 &-8 \\ 0 & 0 & 1 & 2 \end{bmatrix}$

$\small R_{3}+\left ( \frac{1}{4} \right )R_{2}$

$\small A=\begin{bmatrix} 1 & -3 & 1 & 5\\ 0 & 0 & -4 &-8 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

(a)

In the row reduced echelon form, the first and the third columns form a linearly independent columns.

So, the first and third columns of the original matrix form the basis for the $\small col\left ( A \right )$ .

Thus, the basis is:

$\small \beta =\left \{ \begin{bmatrix} 1\\ 2\\ -1\\ \end{bmatrix},\begin{bmatrix} 1\\ -2\\ 0\\ \end{bmatrix} \right \}$

$\small \therefore \mathrm{rank}\left ( A \right )=2$

(b)

$\small \mathrm{null}\left ( A \right )$ is the set  of solutions to $\small AX=0$ .

From the row reduced echelon form , we get

$\small -4c-8d=0\Rightarrow c=-2d$

$\small a-3b+c+5d=0\Rightarrow a=3b-3d$

$\small \therefore X=\begin{bmatrix} a\\ b\\ c\\ d\\ \end{bmatrix}=\begin{bmatrix} 3b-3d\\ b\\ -2d\\ d\\ \end{bmatrix}=b\begin{bmatrix} 3\\ 1\\ 0\\ 0\\ \end{bmatrix}+d\begin{bmatrix} -3\\ 0\\ -2\\ 1\\ \end{bmatrix}$

$\small \therefore \mathrm{null}\left ( A \right )=span\left \{ \begin{bmatrix} 3\\ 1\\ 0\\ 0\\ \end{bmatrix},\: \begin{bmatrix} -3\\ 0\\ -2\\ 1\\ \end{bmatrix} \right \}$

Also, this vectors are linearly independent and so they form a basis for  $\small \mathrm{null}\left ( A \right )$ .

$\small \therefore \gamma =\left \{ \begin{bmatrix} 3\\ 1\\ 0\\ 0\\ \end{bmatrix},\: \begin{bmatrix} -3\\ 0\\ -2\\ 1\\ \end{bmatrix} \right \}$

$\small \therefore \mathrm{nullity}\left ( A \right )=2$