2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Who is "we", and in what context you need this? What are $A$ and $A_j$? $\endgroup$ – Andrés E. Caicedo May 7 '13 at 5:40
  • $\begingroup$ I don't see how a solution is obviously obtained by Cramer's rule, but for any matrix, we have $\det(A) = \det(A^{T})$. $\endgroup$ – Alex Wertheim May 7 '13 at 5:41
  • 2
    $\begingroup$ 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. $\endgroup$ – Raskolnikov May 7 '13 at 5:42
  • $\begingroup$ The problem is $det(A_j)$ when $A_j$ is the matrix we get by replacing column $j$ with $b = \begin{pmatrix} 1 \\ b_1 \\ b^2 \\ .. \\ b^{n-1}\end{pmatrix}$ $\endgroup$ – TheNotMe May 7 '13 at 5:48
  • $\begingroup$ Why is it a problem? Do you know the Vandermonde determinant formula? Just replace the $a_j$'s with $b$'s. $\endgroup$ – Raskolnikov May 7 '13 at 6:27
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Cannot several factors be cancelled out in the expression for $x_k$? That could give a simpler expression. $\endgroup$ – Martin Sleziak May 7 '13 at 8:04
  • $\begingroup$ Sure, all the factors not containing $a_k$'s below will cancel with respect to the same factors above. $\endgroup$ – Raskolnikov May 7 '13 at 8:05
  • $\begingroup$ 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? $\endgroup$ – TheNotMe May 7 '13 at 12:11
  • $\begingroup$ There's no index to $b$. Multiplication goes over all values of $i \in \{1,\ldots,n\}\setminus\{k\}$. $\endgroup$ – Raskolnikov May 7 '13 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.