Determinant problem/Solution

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A solution: The approach used here is to use Gaussian Elimination. Divide the top row by m:

Δm=|1−1m−1m…−1m−1m−1…−1−1−1m…−1⋮⋮⋮⋱⋮−1−1−1…m|

Anticipate addition of rows along the second column:

m−1m=m2m−1m=m2−1m=(m+1)(m−1)m

This should become the leading element of the second row.

Anticipate addition of rows along the third and latter columns:

−1m−1=−1−mm=−(1+m)m

This should become the trailing elements of the second row.

Add the first row to the second row and the rows below it:

Δm=|1−1m−1m…−1m0(m+1)(m−1)m−(m+1)m…−(m+1)m0−(m+1)m(m+1)(m−1)m…−(m+1)m⋮⋮⋮⋱⋮0−(m+1)m−(m+1)m…(m+1)(m−1)m|

Divide the second row by m−1:

Δm(m−1)=|1−1m−1m…−1m0(m+1)m−(m+1)m(m−1)…−(m+1)m(m−1)0−(m+1)m(m+1)(m−1)m…−(m+1)m⋮⋮⋮⋱⋮0−(m+1)m−(m+1)m…(m+1)(m−1)m|

Anticipate addition along third column:

(m+1)(m−1)m−(m+1)m(m−1)
=(m+1)(m−1)2−(m+1)m(m−1)
=(m+1)[(m−1)2−1]m(m−1)
=(m+1)[(m−1+1)(m−1−1)]m(m−1)
=(m+1)(m−2)(m−1)

This should become the new leading element of the third row.

Anticipate addition along fourth and latter columns:

−(m+1)m−(m+1)m(m−1)
=−[(m+1)(m−1)+(m+1)m(m−1)]
=−(m+1)[m−1+1]m(m−1)
=−(m+1)(m−1)

This should become the trailing elements of the third row.

Add the second row to the rows below it:

Δm(m−1)=|1−1m−1m……−1m0(m+1)m−(m+1)m(m−1)……−(m+1)m(m−1)00(m+1)(m−2)(m−1)−(m+1)(m−1)…−(m+1)(m−1)00−(m+1)(m−1)(m+1)(m−2)(m−1)…−(m+1)(m−1)⋮⋮⋮⋮⋱⋮00−(m+1)(m−1)−(m+1)(m−1)…(m+1)(m−2)(m−1)|

Divide the third row by m−2:

Δm(m−1)(m−2)=|1−1m−1m……−1m0(m+1)m−(m+1)m(m−1)……−(m+1)m(m−1)00(m+1)(m−1)−(m+1)(m−1)(m−2)…−(m+1)(m−1)(m−2)00−(m+1)(m−1)(m+1)(m−2)(m−1)…−(m+1)(m−1)⋮⋮⋮⋮⋱⋮00−(m+1)(m−1)−(m+1)(m−1)…(m+1)(m−2)(m−1)|

Anticipate addition along the fourth column:

(m+1)(m−2)(m−1)−(m+1)(m−1)(m−2)
=(m+1)(m−2)2−(m+1)(m−1)(m−2)
=(m+1)[(m−2)2−1](m−1)(m−2)
=(m+1)[(m−1)(m−3)](m−1)(m−2)
=(m+1)(m−3)(m−2)

This should become the leading element of the fourth row.

Anticipate addition along the fifth and latter columns:

−(m+1)(m−1)−(m+1)(m−1)(m−2)
=−[(m+1)(m−2)+(m+1)](m−1)(m−2)
=−(m+1)[m−2+1](m−1)(m−2)
=−(m+1)(m−1)(m−1)(m−2)
=−(m+1)(m−2)

This should become the trailing elements of the fourth row.

Add the third row to the rows below it:

Δm(m−1)(m−2)=|1−1m−1m………−1m0(m+1)m−(1+m)m(m−1)………−(1+m)m(m−1)00(m+1)(m−1)−(m+1)(m−1)(m−2)……−(m+1)(m−1)(m−2)000(m+1)(m−3)(m−2)−(m+1)(m−2)…−(m+1)(m−2)000−(m+1)(m−2)(m+1)(m−3)(m−2)…−(m+1)(m−2)⋮⋮⋮⋮⋮⋱⋮000−(m+1)(m−2)−(m+1)(m−2)…(m+1)(m−3)(m−2)|

Divide the fourth row by m−3:

Δm(m−1)(m−2)(m−3)=|1−1m−1m………−1m0(m+1)m−(m+1)m(m−1)………−(m+1)m(m−1)00(m+1)(m−1)−(m+1)(m−1)(m−2)……−(m+1)(m−1)(m−2)000(m+1)(m−2)−(m+1)(m−2)(m−3)…−(m+1)(m−2)(m−3)000−(m+1)(m−2)(m+1)(m−3)(m−2)…−(m+1)(m−2)⋮⋮⋮⋮⋮⋱⋮000−(m+1)(m−2)−(m+1)(m−2)…(m+1)(m−3)(m−2)|

Add fourth row to rows below it:

Δm(m−1)(m−2)(m−3)=|1−1m−1m…………−1m0(m+1)m−(m+1)m(m−1)…………−(m+1)m(m−1)00(m+1)(m−1)−(m+1)(m−1)(m−2)………−(m+1)(m−1)(m−2)000(m+1)(m−2)−(m+1)(m−2)(m−3)……−(m+1)(m−2)(m−3)0000(m+1)(m−4)(m−3)−(m+1)(m−3)…−(m+1)(m−3)0000−(m+1)(m−3)(m+1)(m−4)(m−3)…−(m+1)(m−3)⋮⋮⋮⋮⋮⋮⋱⋮0000−(m+1)(m−3)−(m+1)(m−3)…(m+1)(m−4)(m−3)|

For purposes of calculating a determinant, the non-diagonal elements of an upper-triangular matrix do not matter. By induction the equation ends up looking thus:

Δm(m−1)(m−2)(m−3)...(2)(1)=|(m+1)(m+1)−1m−1m……………−1m0(m+1)m−(m+1)m(m−1)……………−(m+1)m(m−1)00(m+1)(m−1)−(m+1)(m−1)(m−2)…………−(m+1)(m−1)(m−2)000(m+1)(m−2)−(m+1)(m−2)(m−3)………−(m+1)(m−2)(m−3)0000(m+1)(m−3)−(m+1)(m−3)(m−4)……−(m+1)(m−3)(m−4)00000(m+1)(m−4)−(m+1)(m−4)(m−5)…−(m+1)(m−4)(m−5)000000(m+1)(m−5)…−(m+1)(m−5)(m−6)⋮⋮⋮⋮⋮⋮⋮⋱⋮000000…0(m+1)2|

The determinant of the upper triangular matrix equals the product of its diagonal elements:

Δm(m−1)(m−2)...(1)=(m+1)m(m+1)(m)(m−1)(m−2)...(3)(2)
Δ=(m+1)m(m+1)=(m+1)m−1.
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