Kenneth Kuttler

182 CHAPTER 8. DETERMINANTS

Continuing to use the multilinear properties of determinants, this equals∣∣∣∣∣∣∣∣∣∣∣

1(a1+b1)(b1+an)

1(a1+b2)(b2+an)

· · · 1(a1+bn)(an+bn)

...... · · ·

...1

(an−1+b1)(an+b1)1

(b2+an)(b2+an−1)1

(an+bn)(bn+an−1)1

an+b11

an+b2· · · 1

an+bn

∣∣∣∣∣∣∣∣∣∣∣n−1

∏k=1

(an−ak)

and this equals ∣∣∣∣∣∣∣∣∣∣∣

1(a1+b1)

1(a1+b2)

· · · 1(a1+bn)

...... · · ·

...1

(an−1+b1)1

(b2+an−1)1

(bn+an−1)

1 1 · · · 1

∣∣∣∣∣∣∣∣∣∣∣∏

n−1k=1 (an−ak)

∏nk=1 (an +bk)

Now take −1 times the last column and add to each previous column. Thus it equals∣∣∣∣∣∣∣∣∣∣∣

bn−b1(a1+b1)(a1+bn)

bn−b2(a1+b2)(a1+bn)

· · · 1(a1+bn)

...... · · ·

...bn−b1

(b1+an−1)(bn+an−1)bn−b2

(b2+an−1)(bn+an−1)1

(an−1+bn)

0 0 · · · 1

∣∣∣∣∣∣∣∣∣∣∣∏

n−1k=1 (an−ak)

∏nk=1 (an +bk)

Now continue simplifying using the multilinear property of the determinant.∣∣∣∣∣∣∣∣∣∣∣

1(a1+b1)

1(a1+b2)

· · · 1...

... · · ·...

1(b1+an−1)

1(b2+an−1)

0 0 · · · 1

∣∣∣∣∣∣∣∣∣∣∣∏

n−1k=1 (an−ak)

∏nk=1 (an +bk)

∏n−1k=1 (bn−bk)

∏n−1k=1 (ak +bn)

Now, expanding along the bottom row, what has just resulted is∣∣∣∣∣∣∣∣1

a1+b1· · · 1

a1+bn−1... · · ·

...1

an−1+b1· · · 1

an−1+bn−1

∣∣∣∣∣∣∣∣∏

n−1k=1 (an−ak)

∏nk=1 (an +bk)

∏n−1k=1 (bn−bk)

∏n−1k=1 (ak +bn)

By induction this equals

∏n−1k=1 (an−ak)

∏nk=1 (an +bk)

∏n−1k=1 (bn−bk)

∏n−1k=1 (ak +bn)

∏ j<i≤n−1 (ai−a j)(bi−b j)

∏i, j≤n−1 (ai +b j)

=∏ j<i≤n (ai−a j)(bi−b j)

∏i, j≤n (ai +b j)■

182 CHAPTER 8. DETERMINANTSContinuing to use the multilinear properties of determinants, this equals1 1 ; 1(a+b, )(b1 +an) (a, +bz)(b2+an) - (a1 +bn)(an+bn): : : nal[] (na)(naib \an+b1) — (b2-Fan)(b2 +41) (an-Fbny(bnran—1) | K!1 1 1ant+b, an+b2 an+bnand this equals1 1 1(ai +61) (a+b) (ay Fn)1 1 1 n(q +b(Qn_1+b1) (bo +an_1) (ByFan-1) he: ( n a)1 1 wee 1Now take —1 times the last column and add to each previous column. Thus it equalsbn—b bn—b wee 1(a, +b, )(a,+bn) (a, +b2)(a,+bn) (a, +bn): TMi=1 (4n =a)by—b by—b 1 Jan +bDita, onFan aa) Data, —1nFan aa) (4n—1+bn) That (4n +e)) 0 was 1Now continue simplifying using the multilinear property of the determinant.! l |(a, +51) (a, +b2)Mc} (Gn — ax) Mc} (bn — dx)I ; n b n—ICra) Cera) 1 | Mas (an Px) Thai (a + bn)Now, expanding along the bottom row, what has just resulted is1 1a\+b Gynt _ _a TPE! (ay — ay) TTI (bn — bx)Tai (@n + be) TZ} (ax + bn)An—1 +b, _ An—1+bn—1By induction this equalsTp} (Gn — ae) TRE] (On — be) Tj <i<n—1 (ai — aj) (bi — bj)Tita1 (an + bx) TZ} (ae + bn) Tli,j<n—1 (ai + ,)_ Ij<i<n (ai — aj) (bi — bj) =Vij<n (ai + 5;)