Finding determinant of Matrix n*n

Lazar

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Jul 1, 2014
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Hi guys ,

I am truying to solve this example under b)
I solved a) it is n! I added 1st row to all oters row so that I would get a matrix that under it's main diagonal has all zeroes and than just
1*2*...*n and that is n! now underr b

matrices.JPG

but for b the same thing cant work , i tried to add columns instead of rows but not working ...
If possible could you give me a hint :) .
 
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Subtracting the first row from each succeeding row gives
\(\displaystyle \left|\begin{array}( 4 & 3 & 3 & ... & 3 \\ -1 & 1 & 0 & ... & 0 \\ -1 & 0 & 1 & ... & 0 \\ & & ... & & \\ -1 & 0 & 0 & ... & 1\end{array}\right|\)

Try to expand that on, say, the second row.
 
Is the answer
4+3(n-1)
I' ll post my work in notebook later just need to get a camera so that I could shoot a photo and upload it ?
Thank you really much
HallsofIvy :D
 
Since I am having problem with camera attaching to a computer I can't post the photo but
here is my way of thinking
I took the 2nd row multipled it with -3 and added it to the 1st row , next
took the 3rd row etc ...... , and repeated the process until the end for the n-th row
so I would get above the main diagonal all zeroes than I just multiplied the main diagonal
and it has all 1 except the 1st number witch is 4 + 3 + 3+ 3 ... +3 , since it has n-1 rows that were added
it is 4+ 3(n-1 ) :D .
 
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