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How to find "x" from this equation $$ \begin{bmatrix} a_1 & a_1^2 & \cdots & a_1^n \\ a_2 & a_2^2 & \cdots & a_2^n \\ \vdots & \vdots& \ddots & \vdots \\ a_m & a_m^2 & \cdots & a_m^n \\ \end{bmatrix} \begin{bmatrix} x \\ b_2\\ \vdots\\ b_m \\ \end{bmatrix} = \begin{bmatrix} c_1\\ c_2\\ \vdots\\ c_m \\ \end{bmatrix} $$

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2 Answers 2

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If $a_1 \ne 0$, just multiply the first row of the matrix by the vector $(x, b_2, \cdots, b_n)^T$ (you have a typo, it should be $b_n$, not $b_m$).

$$ a_1 x + a_1^2 b_2 + \cdots + a_1^n b_n = c_1 \Leftrightarrow x = \frac{1}{a_1} \left(c_1 -\sum_{i=2}^n a_1^i b_i \right) $$

Otherwise, just pick a row $k$ such that $a_k \ne 0$ and compute

$$ a_k x + a_k^2 b_2 + \cdots + a_k^n b_n = c_k \Leftrightarrow x = \frac{1}{a_k} \left(c_k -\sum_{i=2}^n a_k^i b_i \right) $$

If all coefficients $a_k$ are zero, then either all coefficients $c_k$ are zero and any $x$ will do, or some coefficient $c_k$ is not zero and there is no solution.

[EDIT] There is an issue that I did not address before but is also important. For every non zero coefficient $a_k$, you have an alternative expression for $x$. So, there will only be a solution if all these alternatives lead to the same value of $x$ (otherwise we will have incompatible equations).

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If $n \ne m$, then the given equation makes no sense.

So, let $n=m$ and $j \in\{1,2,...,m\}.$ Then we have

$$a_jx+a_j^2b_2+...+a_j^mb_m=c_j.$$

If $a_j \ne 0$ we get

$$x= \frac{c_j}{a_j}-(a_jb_2+...+a_{j}^{m-1}b_m).$$

If $a_1=a_2=...=a_m=0$, then you can't find $x$.

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