Answer to Question #222325 in Linear Algebra for andrew

Question

Determine the dimension and hence the basis for the vector space spanned by the vectors (-1,1,3),(2,3,4),(3,0,-5) and (-2,1,0)

Answer 1

$$\begin{pmatrix} -1 & 2 & 3 & -2 \\ 1 & 3 & 0 & 1 \\ 3 & 4 & -5 & 0 \\ \end{pmatrix}$$

$R_1=-R_1$

$$\begin{pmatrix} 1 & -2 & -3 & 2 \\ 1 & 3 & 0 & 1 \\ 3 & 4 & -5 & 0 \\ \end{pmatrix}$$

$R_2=R_2-R_1$

$$\begin{pmatrix} 1 & -2 & -3 & 2 \\ 0 & 5 & 3 & -1 \\ 3 & 4 & -5 & 0 \\ \end{pmatrix}$$

$R_3=R_3-3R_1$

$$\begin{pmatrix} 1 & -2 & -3 & 2 \\ 0 & 5 & 3 & -1 \\ 0 & 10 & 4 & -6 \\ \end{pmatrix}$$

$R_2=R_2/5$

$$\begin{pmatrix} 1 & -2 & -3 & 2 \\ 0 & 1 & 3/5 & -1/5 \\ 0 & 10 & 4 & -6 \\ \end{pmatrix}$$

$R_1=R_1+2R_2$

$$\begin{pmatrix} 1 & 0 & -9/5 & 8/5 \\ 0 & 1 & 3/5 & -1/5 \\ 0 & 10 & 4 & -6 \\ \end{pmatrix}$$

$R_3=R_3-10R_2$

$$\begin{pmatrix} 1 & 0 & -9/5 & 8/5 \\ 0 & 1 & 3/5 & -1/5 \\ 0 & 0 & -2 & -4 \\ \end{pmatrix}$$

$R_3=R_3/(-2)$

$$\begin{pmatrix} 1 & 0 & -9/5 & 8/5 \\ 0 & 1 & 3/5 & -1/5 \\ 0 & 0 & 1 & 2 \\ \end{pmatrix}$$

$R_1=R_1+9R_3/5$

$$\begin{pmatrix} 1 & 0 & 0 & 26/5 \\ 0 & 1 & 3/5 & -1/5 \\ 0 & 0 & 1 & 2 \\ \end{pmatrix}$$

$R_2=R_2-3R_3/5$

$$\begin{pmatrix} 1 & 0 & 0 & 26/5 \\ 0 & 1 & 0 & -7/5 \\ 0 & 0 & 1 & 2 \\ \end{pmatrix}$$

$rank A=3=>$ the dimension is 3.

the basis for the vector space is $\begin{bmatrix} -1 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix},\begin{bmatrix} 3 \\ 0 \\ -5 \end{bmatrix}$