Let A be the real n x n matrix (aij) where aij = a for i < j, b (≠ a) for i > j, and ci for i = j. Show that det A = (b p(a) - a p(b) )/(b - a), where p(x) = ∏ (ci - x).
Solution
|c1 a a ... a | = |a a a ... a | + |c1-a 0 0 ... 0 |To evaluate the first determinant on the right, we subtract the first column from each of the others. Then expanding by the top row we get a D, where D has zeros below the main diagonal and hence is just the product of the elements on its diagonal. In other words, the first determinant is just a ∏2n (ci - b).
|b c2 a ... a | |b c2 a ... a | |b c2 a ... a |
|b b c3 ... a | |b b c3 ... a | |b b c3 ... a |
| ... | | ... | | ... |
|b b b ... cn| |b b b ... cn| |b b b ... cn|
If we expand the second determinant by the top row we get (c1 - a) det A', where A' is the (n-1) x (n-1) matrix formed by deleting the first row and column of A. So we can use induction. The result is trivial for n = 1. So assume it is true for n - 1. Then for n we have a ∏2n (ci - b) + b/(b - a) ∏1n (ci - a) - a/(b - a) (c1 - a) ∏2n (ci - b). Adding the first and third terms we get: a (1 - (c1 - a)/(b - a) ) ∏2n (ci - b) = - a/(b - a) ∏1n (ci - b). So the result is true for n.
© John Scholes
jscholes@kalva.demon.co.uk
30 Nov 1999