Calculating an n X n determinant

In summary, calculating an n x n determinant involves various methods including the cofactor expansion, row reduction to echelon form, or leveraging properties such as the triangular form. The cofactor expansion relies on breaking down the determinant into smaller (n-1) x (n-1) determinants, while row reduction simplifies the matrix to facilitate easier calculations. Determinants have specific properties, such as being zero if rows or columns are linearly dependent, and the determinant can be computed efficiently for larger matrices using algorithms like LU decomposition.
  • #1
TGV320
40
22
Homework Statement
Help in order to solve a determinant
Relevant Equations
Determinants
Hello,

I need some advice because I just can't figure out how to solve the problem. I could try to make the determinant triangular by adding all the b together, but that doen't seem a good way of solving the problem. Is there any direction I should be thinking of?

1699948294438.jpg
Thanks
 
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  • #2
Why not calculate the determinant for ##n = 2, 3, 4## and see whether a pattern emerges?
 
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  • #3
TGV320 said:
Homework Statement: Help in order to solve a determinant
Relevant Equations: Determinants

Hello,

I need some advice because I just can't figure out how to solve the problem. I could try to make the determinant triangular by adding all the b together, but that doen't seem a good way of solving the problem. Is there any direction I should be thinking of?

View attachment 335323Thanks
Hint: Follow PeroK's advice and find the determinant by expanding along the bottom row.

-Dan
 
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  • #4
By considering the Leibniz formula, one can figure out that only some terms survive, where the permutations do not contain zeroes.
 
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  • #5
Multiply the ##i^{th}## row by ##-a_i## and add it to the first. You just need to see what the top left element will be.
 
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  • #6
Hi,
Thanks for the advice.
I have figured it out,though I never thought about getting the answer by experimenting on it, always thought it to n. That way of doing it with n=2 then 3 is quite illuminating.

1700045361111.jpg
 
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