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

AI Thread Summary
The discussion revolves around proving the symmetry of the expression [L^(n-1)]*(nx+L) based on the determinant of an nxn matrix M. Participants clarify that x and L are likely real numbers. The proof involves manipulating the matrix by abstracting columns and performing row additions to simplify the determinant. Ultimately, the transformation leads to the conclusion that the expression is indeed symmetric. The mathematical steps illustrate the relationship between the matrix elements and the determinant's properties.
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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