So, (nx+L)*[L^(n-1)] = [L^(n-1)]*(nx+L)Therefore, [L^(n-1)]*(nx+L) is symmetric.

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Homework Help Overview

The discussion revolves around proving a property related to an nxn matrix M, specifically involving the expression (nx+L)*[L^(n-1)] and its symmetry. The variables x and L are introduced, but their definitions are not fully clarified.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the structure of the matrix M and suggest methods for simplifying it through row operations. Questions are raised about the definitions of x and L, with some participants speculating they are real numbers.

Discussion Status

The discussion includes attempts to manipulate the matrix to demonstrate the symmetry property. While some progress is made in transforming the matrix, there is no explicit consensus on the definitions of the variables or the overall approach.

Contextual Notes

There is uncertainty regarding the definitions of x and L, which may impact the understanding of the problem. The original poster seeks assistance in proving the property without providing a complete solution.

Alexitron
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Can someone help me proove this one please?

If M is an nxn matrix

_____|x+L x x . . . x |
_____| x x+L x . . . x |
|M|= | x x x+L. . . x | = [L^(n-1)]*(nx+L)
_____| : : : : : : : : |
_____| x x x . . . x+L|
 
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What are x and L supposed to be here?
 
I guess real numbers.
 
Solved!

Abstract the n-1 column from the n column, then the n-2 from the n-1, then the n-3 from the n-2 etc.You get:

|x+L -L 0 0 . . .0 0 0 |
| x L -L 0 . . .0 0 0|
| x 0 L -L . . .0 0 0|
| x 0 0 L . . .0 0 0 |
| . . . . . . . . . . . |
| . . . . . . . . . . . |
| x . . . . . .L -L 0|
| x . . . . . .0 L -L|
| x . . . . . .0 0 L|

Then add The n row to the n-1, then the n-1 to n-2 etc and you get:

| nx+ L 0 0 0 . . . 0 0 0|
| (n-1)x L 0 0 . . .0 0 0|
| (n-2)x 0 L 0 . . .0 0 0|
| (n-3)x 0 0 L . . .0 0 0|
| . . . . . . . . . . . . . . .| = (nx+L)*L^(n-1)
| . . . . . . . . . . . . . . .|
| 3x . . . . . . . . . L 0 0|
| 2x . . . . . . . . . 0 L 0|
| x . . . . . . . . . . 0 0 L|
 

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