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Calculate n-order determinant (n>=2):

$\left|\begin{array}{ccccc}x_{1}-a_{1} & x_{2} & x_{3} & \cdots & x_{n} \\ x_{1} & x_{2}-a_{2} & x_{3} & \cdots & x_{n} \\ x_{1} & x_{2} & x_{3}-a_{3} & \cdots & x_{n} \\ \vdots & \vdots & \vdots & & \vdots \\ x_{1} & x_{2} & x_{3} & \cdots & x_{n}-a_{n}\end{array}\right|$

$a_{i} \neq 0, i=1,2, \cdots, n$

The result should be

$(-1)^{n-1} a_{1} a_{2} a_{3} \cdots a_{n}\left(\sum_{i=1}^{n} \frac{x_{i}}{a_{i}}-1\right)$

I want to use the FindSequenceFunction to get the result. Here is my code:

Clear["Global`*"];
Format[a[n_]] := Subscript[a, n];
Format[x[n_]] := Subscript[x, n];

NewMatrix[n_Integer?Positive] := 
 Module[{i = 1, j = 1, M = Array[m, {n, n}]}, 
  For[i = 1, i <= n, i++, 
   For[j = 1, j <= n, j++, 
    If[j == i, m[i, j] = x[j] - a[j], m[i, j] = x[j]]]]; M]

tab = Table[Det[NewMatrix[i]], {i, 2, 10}]

$\left\{a_{1} a_{2}-a_{2} x_{1}-a_{1} x_{2},-a_{1} a_{2} a_{3}+a_{2} a_{3} x_{1}+a_{1} a_{3} x_{2}+a_{1} a_{2} x_{3}, a_{1} a_{2} a_{3} a_{4}-a_{2} a_{3} a_{4} x_{1}-a_{1} a_{3} a_{4} x_{2}-a_{1} a_{2} a_{4} x_{3}-a_{1} a_{2} a_{3} x_{4},\right.$, $-a_{1} a_{2} a_{3} a_{4} a_{5}+a_{2} a_{3} a_{4} a_{5} x_{1}+a_{1} a_{3} a_{4} a_{5} x_{2}+a_{1} a_{2} a_{4} a_{5} x_{3}+a_{1} a_{2} a_{3} a_{5} x_{4}+a_{1} a_{2} a_{3} a_{4} x_{5},$, $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}-a_{2} a_{3} a_{4} a_{5} a_{6} x_{1}-a_{1} a_{3} a_{4} a_{5} a_{6} x_{2}-a_{1} a_{2} a_{4} a_{5} a_{6} x_{3}-a_{1} a_{2} a_{3} a_{5} a_{6} x_{4}-a_{1} a_{2} a_{3} a_{4} a_{6} x_{5}-a_{1} a_{2} a_{3} a_{4} a_{5} x_{6},-a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7}+$ $\left.\quad a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} x_{1}+a_{1} a_{3} a_{4} a_{5} a_{6} a_{7} x_{2}+a_{1} a_{2} a_{4} a_{5} a_{6} a_{7} x_{3}+a_{1} a_{2} a_{3} a_{5} a_{6} a_{7} x_{4}+a_{1} a_{2} a_{3} a_{4} a_{6} a_{7} x_{5}+a_{1} a_{2} a_{3} a_{4} a_{5} a_{7} x_{6}+a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} x_{7}\right\}$

Then,

FindSequenceFunction[tab, n]

FindSequenceFunction $[\left\{a_{1} a_{2}-a_{2} x_{1}-a_{1} x_{2},-a_{1} a_{2} a_{3}+a_{2} a_{3} x_{1}+a_{1} a_{3} x_{2}+a_{1} a_{2} x_{3}, a_{1} a_{2} a_{3} a_{4}-a_{2} a_{3} a_{4} x_{1}-a_{1} a_{3} a_{4} x_{2}-a_{1} a_{2} a_{4} x_{3}-a_{1} a_{2} a_{3} x_{4},\right.$, $-a_{1} a_{2} a_{3} a_{4} a_{5}+a_{2} a_{3} a_{4} a_{5} x_{1}+a_{1} a_{3} a_{4} a_{5} x_{2}+a_{1} a_{2} a_{4} a_{5} x_{3}+a_{1} a_{2} a_{3} a_{5} x_{4}+a_{1} a_{2} a_{3} a_{4} x_{5},$, $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}-a_{2} a_{3} a_{4} a_{5} a_{6} x_{1}-a_{1} a_{3} a_{4} a_{5} a_{6} x_{2}-a_{1} a_{2} a_{4} a_{5} a_{6} x_{3}-a_{1} a_{2} a_{3} a_{5} a_{6} x_{4}-a_{1} a_{2} a_{3} a_{4} a_{6} x_{5}-a_{1} a_{2} a_{3} a_{4} a_{5} x_{6},-a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7}+$ $\left.\quad a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} x_{1}+a_{1} a_{3} a_{4} a_{5} a_{6} a_{7} x_{2}+a_{1} a_{2} a_{4} a_{5} a_{6} a_{7} x_{3}+a_{1} a_{2} a_{3} a_{5} a_{6} a_{7} x_{4}+a_{1} a_{2} a_{3} a_{4} a_{6} a_{7} x_{5}+a_{1} a_{2} a_{3} a_{4} a_{5} a_{7} x_{6}+a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} x_{7}\right\},n]$

It doesn't work.

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  • $\begingroup$ I've noticed very few functions that return answers in terms of Sum/Inactive[Sum]. For instance D[], maybe DSolve[], but I don't recall FindSequenceFunction doing it. Is there an example? It might help with finding an approach to the problem. $\endgroup$
    – Michael E2
    Apr 18 at 15:02
  • $\begingroup$ Sorry, I haven't found such an example so far. I don't know if others can do it. @Michael E2 $\endgroup$
    – lotus2019
    Apr 19 at 2:18

1 Answer 1

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Not an answer,only a comment about the simple version of how to calculate such determinant.

The matrix can be writed as

$$\begin{pmatrix}1\\1\\ \vdots \\ 1\end{pmatrix}(x_1,x_2,\cdots,x_n)-\mathrm{diagonal}(a_1,a_2,\cdots,a_n)$$

Clear[det];
det[n_Integer?Positive] := 
 Det[Transpose[{ConstantArray[1, n]}] . {Array[Indexed[x, #] &, n]} - 
   DiagonalMatrix[Array[Indexed[a, #] &, n]]];
det[5]

enter image description here

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  • $\begingroup$ I'll only note that using the formula here (basically Sherman-Morrison-Woodbury), we have $\det(\mathbf e\mathbf x^\top-\operatorname{diag}(\mathbf a))=\det(-\operatorname{diag}(\mathbf a))(1-\mathbf x^\top(\operatorname{diag}(\mathbf a))^{-1}\mathbf e)=(-1)^n (1-\sum_{j=1}^n\frac{x_j}{a_j})\prod_{j=1}^n a_j$ $\endgroup$ Jun 4 at 22:42
  • $\begingroup$ To check: Table[Simplify[Det[Transpose[{ConstantArray[1, n]}] . {Array[Indexed[x, #] &, n]} - DiagonalMatrix[Array[Indexed[a, #] &, n]]] == (-1)^n (1 - Sum[Indexed[x, j]/Indexed[a, j], {j, 1, n}]) Product[Indexed[a, j], {j, 1, n}]], {n, 2, 11}]. $\endgroup$ Jun 4 at 22:42

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