For which vector is there no solution?

Question

We have the vector $$\vec{v}_1=\begin{pmatrix}3 \\ 2 \\ -1 \\ 5\end{pmatrix} , \ \ \vec{v}_2=\begin{pmatrix}0 \\ 6 \\ 2 \\ 7\end{pmatrix}$$ I want to find a vector $\vec{w}$ for that there are no $a,b\in \mathbb{R}$ such that $a\vec{v}_1+b\vec{v}_2=\vec{w}$.

For that do we calculate the rank for $A=\begin{pmatrix}3 & 0 \\ 2 & 6 \\ -1 & 2 \\ 5 & 7\end{pmatrix}$ and for $(A\mid \vec{w})=\begin{pmatrix}3 & 0 & w_1 \\ 2 & 6 & w_2 \\ -1 & 2 & w_3\\ 5 & 7 & w_4\end{pmatrix}$ and get the conditio that these two results have to be different?

Or is there also an other way?

Answer 1

Let consider by RREF

$$\begin{pmatrix}3 &2&-1&5\\ 0 & 6 & 2 & 7\end{pmatrix}\to \begin{pmatrix}9 &0&-5&8\\ 0 & 6 & 2 & 7\end{pmatrix}$$

therefore $w_1=(0,0,1,0)$ or $w_2=(0,0,0,1)$ or any $w=aw_1+bw_2$ are good choice.

Answer 2

If you only want to find quickly a vector that is not a linear combination of $\vec{v}_1$ and $\vec{v}_2$, then you can just construct a vector which is orthogonal to $\vec{v}_1$ and $\vec{v}_2$.

With $$\vec{v}_1=\begin{pmatrix}3 \\ 2 \\ -1 \\ 5\end{pmatrix} , \ \ \vec{v}_2=\begin{pmatrix}0 \\ 6 \\ 2 \\ 7\end{pmatrix}$$

Looking at $\vec{v}_2$, you can take, for example, $\vec{w}=\begin{pmatrix}x \\ -1 \\ 3 \\ 0\end{pmatrix}$. Use $\vec{v}_1$ to find $x$: $$3x + 2(-1) + (-1)3 + 0\cdot 5 =0 \Leftrightarrow x = \frac{5}{3}$$ $$\Rightarrow \vec{w}=\begin{pmatrix}\frac 53 \\ -1 \\ 3 \\ 0\end{pmatrix} \mbox{ is such a vector.}$$

Answer 3

We can show that $$(A\mid \vec{w})=\begin{bmatrix}3 & 0 & w_1 \\ 2 & 6 & w_2 \\ -1 & 2 & w_3\\ 5 & 7 & w_4\end{bmatrix} \xrightarrow{\text{row reduced }} \begin{bmatrix}3 & 0 & w_1 \\ 0 & 6 & w_2-\frac{2}{3}w_1 \\ 0 & 0 & w_3-\frac{1}{3}w_2+\frac{5}{9}w_1\\ 0 & 0 & w_4-\frac{7}{6}w_2-\frac{8}{9}w_1\end{bmatrix}.$$ So for inconsistency, you want at least one of the bottom two entries in the right hand column to be non-zero, i.e. $$\text{either } \quad w_3-\frac{1}{3}w_2+\frac{5}{9}w_1 \neq 0 \quad \text{ or } \quad w_4-\frac{7}{6}w_2-\frac{8}{9}w_1 \neq 0.$$ For example we can have $\vec{w}=\begin{bmatrix}1\\0\\0\\1\end{bmatrix}$