Consider $5$ functions $f_1, …, f_5$ defined on the interval $[0,5]$, and $\textbf{assume they have the following property:}$ given any numbers $b_1, …, b_5$ there always exist coefficients $c_1, ..., c_5$ such that
$\,\,\,\,\,\,\,\,\,\,$ $\displaystyle \sum_{j=1}^5 c_jf_j(i) = b_i$
Then, $\textbf{show}$ that under these circumstances that once the number $b_1, …, b_5$ are fixed the coefficients $c_1, ..., c_5$ that solve the above problem are uniquely determined, this will in particular mean there is a linear function that determines the $c_j's$ from the $b_i's$, which means there is a matrix $M_{ij}$ such that
$\,\,\,\,\,\,\,\,\,\,$ $\displaystyle c_j = \sum_{i=1}^5 M_{ij}b_i$
Without knowing anything else about $M$, say what is $\left(M^{-1}\right)_{ij}$.
I am not sure if I am setting this up correctly or taking the proper approach. First, I wrote out the first equation as a system of equations and got:
$c_1f_1(1)+c_2f_2(1)+c_3f_3(1)+c_4f_4(1)+c_5f_5(1)=b_1$
$c_1f_1(2)+c_2f_2(2)+c_3f_3(2)+c_4f_4(2)+c_5f_5(2)=b_2$
$\vdots$
$c_1f_1(5)+c_2f_2(5)+c_3f_3(5)+c_4f_4(5)+c_5f_5(5)=b_5$.
So, I set up a matrix out of these systems:
$\begin{pmatrix}
c_1f_1(1) & c_2f_2(1) & ... & c_5f_5(1)\\
\vdots \\
c_1f_1(5) & ... & ... & c_5f_5(5)
\end{pmatrix} = \begin{pmatrix}
b_1 \\
\vdots \\
b_5
\end{pmatrix}$ $\implies$
$\begin{pmatrix}
f_1(1) & f_2(1) & ... & f_5(1)\\
\vdots \\
f_1(5) & ... & ... & f_5(5)
\end{pmatrix}\begin{pmatrix}
c_1 \\
\vdots \\
c_5
\end{pmatrix} = \begin{pmatrix}
b_1 \\
\vdots \\
b_5
\end{pmatrix}$
Let $M = \begin{pmatrix} f_1(1) & f_2(1) & ... & f_5(1)\\ \vdots \\ f_1(5) & ... & ... & f_5(5) \end{pmatrix}$, $\overline{\textbf{c}} = \begin{pmatrix} c_1 \\ \vdots \\ c_5 \end{pmatrix}$, and $\overline{\textbf{b}} = \begin{pmatrix} b_1 \\ \vdots \\ b_5 \end{pmatrix}$.
So, we have $M \,\overline{\textbf{c}} = \overline{\textbf{b}}$.
**Here I wasn't too sure what to do next. I wanted to show that $M$ is invertible, but the only way I know is that the determinant is not zero. I didn't know how to do that with this $5 \times 5$ matrix. I wanted to show this becuase I figured it would help me show that $\overline{\textbf{c}} = M^{-1} \overline{\textbf{b}}$ which would mean that $\left(M^{-1}\right)_{ij} = M^T$. Or am I mistaken about the inverse being the transpose? As far as uniqueness goes for $\overline{\textbf{c}}$, I thought if we assume there is a $\overline{\textbf{c}}^{'}$, then we could obtain $\overline{\textbf{b}}$ which int turn means $\overline{\textbf{c}}^{'} = \overline{\textbf{c}}$.
Note that the matrix that you have called $M$ does the reverse of what is being requested: given the vector $\mathbf c$, you have found a matrix $A$ for which $A\mathbf c = \mathbf b$. What is required is a matrix "that determines the $c_j$'s from the $b_i$'s". In other words, $M$ should satisfy $\mathbf c = M \mathbf b$.
However, your work so far makes this easy. Note that $$ A \mathbf c = \mathbf b \implies \mathbf c = A^{-1} \mathbf b. $$ In other words, the desired matrix $M$ is the inverse of the matrix you found (which I have called $A$). Note that this requires the inverse of $M$ to exist, but we can conclude that this must hold because the $c_j$'s are uniquely determined from the $b_i$'s.
With that, the last part of the question is easy: the inverse of $M$ is simply $(A^{-1})^{-1} = A$. In other words, we have $$ (M^{-1})_{ij} = f_j(i). $$