Question

Show that the vector v = (2, −5, 3) ∈ R 3

cannot be expressed as a linear combination of the

vectors v1= (1, −3, 2), v2= (2, −4, −1), v3= (1, −5, 7).

Answer 1

Given,

The vector $\vec v=<2,-5,3>\ \ \in \R^3$

$\vec v_1=<1,-3,2>\\\vec v_2=<2,-4,-1>\\\vec v_3=<1,-5,7>$

Suppose $\vec v$ can be expressed as linear combination of $\vec v_1,\vec v_2\ and \ \vec v_3$

Then,

$\vec v=x_1\vec v_1+x_2\vec v_2+x_2\vec v_3\\\vec v=x_1\begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}+x_2\begin{pmatrix} 2\\ -4\\-1 \end{pmatrix}+x_3\begin{pmatrix} 1\\ -5\\7 \end{pmatrix}=\begin{pmatrix} 2 \\ -5\\3 \end{pmatrix}$

Augmented matrix of the above system is

$\begin{bmatrix} 1&2&\ \ 1\ |\ \ \ \ \ 2 \\ -3 & -4& -5\ |\ -5\\2&-1&7\ |\ \ \ 3 \end{bmatrix}$

Apply elementary row operation

$R_2\rightarrow R_2+3R_1\\R_3\rightarrow R_3-2R_1$

$=\begin{bmatrix} 1&2&1&|2\\ 0&2&-2&|1\\0&-5&5&\ \ \ \ |-1 \end{bmatrix}$

Now,

$R_2\rightarrow \frac{1}{2}R_2\\\ \\R_3\rightarrow \frac{1}{5}R_3$

$=\begin{bmatrix} 1&2&1&|2 \\ 0&1&-1&\ \ \ \ |1/2\\0&-1&1&\ \ \ \ \ \ \ \ |-1/5 \end{bmatrix}$

Also, $R_3\rightarrow R_3+R_2$

$=\begin{bmatrix} 1&2&1&|2 \\ 0&1&-1&\ \ \ \ |1/2\\0&0&0&|3/10 \end{bmatrix}$

From above, we can conclude that there is pivot position in the augmented column.

Hence given system has no solution.

So, Vector $\vec v$ can't be expressed as linear combination of $\vec v_1,\vec v_2\ and \ \vec v_3$