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

[Alexitron]

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|

 

[Defennder]

What are x and L supposed to be here?

 

[Alexitron]

I guess real numbers.

 

[Alexitron]

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|