MHB 10.3 Determine if A is in the span B

[karush]

Determine if $A=\begin{bmatrix}
1\\3\\2
\end{bmatrix}$ is in the span $B=\left\{\begin{bmatrix}
2\\1\\0
\end{bmatrix}
\cdot
\begin{bmatrix}
1\\1\\1
\end{bmatrix}\right\}$

ok I added A and B to this for the OP
but from examples it looks like this can be answered by scalors so if

$c_1\begin{bmatrix}
2\\1\\0
\end{bmatrix}
+
c_2\begin{bmatrix}
1\\1\\1
\end{bmatrix}=\begin{bmatrix}
1\\3\\2
\end{bmatrix}$

 

[Olinguito]

Hi karush.

So you have
$$\begin{eqnarray}2c_1 &+& c_2 &=& 1 \\ c_1 &+& c_2 &=& 3 \\ {} & {} & c_2 &=& 2.\end{eqnarray}$$
If you substitute $c_2=2$ from the last equation into the first two equations, you get two different values for $c_1$. Hence the above set of equations is inconsistent (has no solutions) showing that $\mathbf A\notin\mathrm{span}B$.

 

[karush]

Lets try this one... if $A=
\begin{bmatrix}
1\\3\\2
\end{bmatrix}$ is in the span $B=\left\{\begin{bmatrix}
2\\1\\0
\end{bmatrix}
\cdot
\begin{bmatrix}
1\\1\\1
\end{bmatrix}
\cdot
\begin{bmatrix}
0\\1\\1
\end{bmatrix}\right\}$
then
$\begin{array}{rrrrr}
2c_1 &+ c_2 & & =1 \\
c_1 &+ c_2 & +c_3 & =3 \\
& c_2 & +c_3 & =2
\end{array}$
Solving $c_1=1, c_2=−1, c_3=3$
so $A\in\mathrm{span}B$

 

[HallsofIvy]

You are missing some "+" signs, aren't you?

Yes, the definition of "span" requires that
$2c_1+ c_2= 1$
$c_1+ c_2+ c_3= 3$ and
$c_2+ c_3= 2$