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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
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.
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.
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.
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.
