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Given this:

$$\begin{pmatrix} 1 & 1 & 1 & ... & 1 \\ a_1 & a_2 & a_3 & ... & a_n \\ a_1^2 & a_2^2 & a_3^2 & ... & a_n^2 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & ... & a_n^{n-1}\end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n\end{pmatrix} = \begin{pmatrix} 1 \\ b \\ b^2 \\ \vdots \\ b^{n-1} \end{pmatrix} $$

We need to solve this. Now it is obviously by Cramer's rule, but how do we calculate $\det(A)$ and $\det(A_j)$? It is related to Vandermonde because it looks like it except that it is transposed.

And help is appreciated! thank you!

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Who is "we", and in what context you need this? What are $A$ and $A_j$? – Andres Caicedo May 7 at 5:40
I don't see how a solution is obviously obtained by Cramer's rule, but for any matrix, we have $\det(A) = \det(A^{T})$. – AWertheim May 7 at 5:41
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If it's by Cramer's rule, then it literally is an application of the Vandermonde determinant. I frankly don't see what more there is to tell if you already know this. – Raskolnikov May 7 at 5:42
The problem is $det(A_j)$ when $A_j$ is the matrix we get by replacing column $j$ with $b = \begin{pmatrix} 1 \\ b_1 \\ b^2 \\ .. \\ b^{n-1}\end{pmatrix}$ – TheNotMe May 7 at 5:48
Why is it a problem? Do you know the Vandermonde determinant formula? Just replace the $a_j$'s with $b$'s. – Raskolnikov May 7 at 6:27

1 Answer

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The Vandermonde determinant is equal to

$$\det \begin{pmatrix} 1 & 1 & 1 & ... & 1 \\ a_1 & a_2 & a_3 & ... & a_n \\ a_1^2 & a_2^2 & a_3^2 & ... & a_n^2 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & ... & a_n^{n-1}\end{pmatrix} = \prod_{i<j}(a_i-a_j)$$

Therefore, by Cramer's rule the solution of your linear system is

$$x_k = \frac{\prod_{\substack{i<j \\ k \leftrightarrow b}}(a_i-a_j)}{\prod_{i<j}(a_i-a_j)}$$

in which I introduced some notation to indicate that in the upper product, the $a_k$'s have to be switched for the $b$'s.

As suggested by Martin, this simplifies further to

$$x_k = \prod_{i\neq k}\frac{a_i-b}{a_i-a_k}$$

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Cannot several factors be cancelled out in the expression for $x_k$? That could give a simpler expression. – Martin Sleziak May 7 at 8:04
Sure, all the factors not containing $a_k$'s below will cancel with respect to the same factors above. – Raskolnikov May 7 at 8:05
In the simplified one, what is the index of the $b$? also, the multiplication goes with respect to what variable and from where to where? – TheNotMe May 7 at 12:11
There's no index to $b$. Multiplication goes over all values of $i \in \{1,\ldots,n\}\setminus\{k\}$. – Raskolnikov May 7 at 12:49

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