<s>[INST] Linear Algebra - Linear transformation question Let $
b \in \mathbb{R}^4, \space A\in M_{4\times4} (\mathbb{R}). 
$ Suppose
$$
\begin{bmatrix}  0  \\ 1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}  1  \\ 0 \\ 1 \\ 1\end{bmatrix},\begin{bmatrix}  1  \\ 1 \\ 0 \\ 1\end{bmatrix},\begin{bmatrix}  1  \\ 1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}  1  \\ 1 \\ 1 \\ 1\end{bmatrix}$$
are all solutions of the equation $Ax=b$. Prove that $$b = \begin{bmatrix}  0  \\ 0 \\ 0 \\ 0\end{bmatrix}.$$
This question appeared at my recent algebra exam. I had an intuiton that the given vectors are linear-dependent since there are five of them and the dim of the vector space is four. The professor solved that question after the exam using linear transformation which seemed much easier and clean solution however he didn't explain the intuition behind. I  would like to get an explanation regarding that. Thanks
 [/INST] $$\begin{bmatrix}  0  \\ 1 \\ 1 \\ 1\end{bmatrix} + \begin{bmatrix}  1  \\ 0 \\ 1 \\ 1\end{bmatrix} + \begin{bmatrix}  1  \\ 1 \\ 0 \\ 1\end{bmatrix} + \begin{bmatrix}  1  \\ 1 \\ 1 \\ 0\end{bmatrix} - 3\begin{bmatrix}  1  \\ 1 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix}  0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$
$$A\begin{bmatrix}  0  \\ 1 \\ 1 \\ 1\end{bmatrix} + A\begin{bmatrix}  1  \\ 0 \\ 1 \\ 1\end{bmatrix} + A\begin{bmatrix}  1  \\ 1 \\ 0 \\ 1\end{bmatrix} + A\begin{bmatrix}  1  \\ 1 \\ 1 \\ 0\end{bmatrix} - 3A\begin{bmatrix}  1  \\ 1 \\ 1 \\ 1\end{bmatrix} = A\begin{bmatrix}  0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$
$$b + b + b + b -3b = 0$$
$$b=0$$
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