Determine unkown scalars with given vectors Given the vectors $v_1$, $v_2$, $v_3$ and $u$ how do I determine the scalars $\mu_1$, $\mu_2$, $\mu_3$ in the following equation:
$$\mu_1\cdot v_1 + \mu_2\cdot v_2 + \mu_3\cdot v_3 = u$$
Can you give me a point to start from? I'm not really sure how to tackle this problem
 A: You get a linear system of 3 equations which may or may not have solution. If the vectors $v1, v2, v3$ are linearly independent then there is always an unique solution $\mu_1, \mu_2, \mu_3$. 
If the vectors $v1, v2, v3$ are linearly dependent then still you may be able to find a solution but not always. 
Even more, sometimes there will be infinite solutions.
For the case when $v1, v2, v3$ are linearly independent you have an explicit formula:
$$\left[\begin{array}{c}
\mu_{1}\\
\mu_{2}\\
\mu_{3}
\end{array}\right]=\left[\begin{array}{ccc}
v_{11} & v_{21} & v_{31}\\
v_{12} & v_{22} & v_{32}\\
v_{13} & v_{23} & v_{33}
\end{array}\right]^{-1}\left[\begin{array}{c}
u_{1}\\
u_{2}\\
u_{3}
\end{array}\right]$$
However it's easier to solve the system via Gauss method than calculating the inverse of a matrix.
A: Hint: Write out the vectors explicitly. Assuming that the vectors are 3d, you have
$$
\mu_1\left(\begin{matrix}v_{11}\\v_{12}\\v_{13}\end{matrix}\right)+\mu_2\left(\begin{matrix}v_{21}\\v_{22}\\v_{23}\end{matrix}\right)+\mu_3\left(\begin{matrix}v_{31}\\v_{32}\\v_{33}\end{matrix}\right)=\left(\begin{matrix}u_1\\u_2\\u_3\end{matrix}\right)
$$
Now, convert this into three equations, and then express as a matrix system.
A: Write the vectors as a linear combination, and if you have $3\times 1$ vectors, you will have three equations in three unknowns. We'll write $\vec v_1 = \vec x,\; \vec v_2 = \vec y,\;\vec v_3 = \vec z$
$$
\mu_1\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix} +\mu_2
\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}+\mu_3
\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}
=\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}
$$
The system of equations represented by the above linear combination of vectors can be expressed by augmented the coefficient matrix 
$\begin{pmatrix} \mu_1v_1 & \mu_2v_2 & \mu_3v_3 \mid & u \end{pmatrix}:$
$$
\begin{pmatrix} \mu_1 x_1 & \mu_2y_1 & \mu_3z_1\mid & u_1\\
\mu_1x_2 & \mu_2y_2 & \mu_3z_2 \mid & u_2 \\ 
\mu_1x_3 & \mu_2y_3 & \mu_3z_3 \mid & u_3 \\
\end{pmatrix}
$$
Now, Knowing each of $\vec x, \vec y, \vec z$ and $\vec u$, you can solve for $\mu_1, \mu_2, \mu_3$ quite readily, if solutions exist.
A: If you think about a matrix, $A$, with columns $v_1$, $v_2$, $v_3$, and the vector, $v$, with entries $\mu_1$, $\mu_2$ and $\mu_3$.
You can rewrite the equation you've been given in terms of $A$, $v$, and $u$.

 $$Av = u$$

You can then rearrange that to find $v$ explicitly.

$$v = A^{-1}u$$

Then solve, and use the fact that $v_i = \mu_i$ to get the answer.
