Evaluate 111abca2b2c2 (the Vandermonde determinant).
Solution — Step by Step
R2→R2−R1 and R3→R3−R1:
100ab−ac−aa2b2−a2c2−a2
b2−a2=(b−a)(b+a) and c2−a2=(c−a)(c+a).
Factor (b−a) from R2 and (c−a) from R3:
(b−a)(c−a)100a11a2b+ac+a
Only the (1,1) entry contributes:
(b−a)(c−a)⋅1⋅11b+ac+a
=(b−a)(c−a)⋅(c+a−b−a)
=(b−a)(c−a)(c−b).
(b−a)(c−a)(c−b)=(a−b)(b−c)(c−a)⋅(−1)?.
Working out the signs: (b−a)=−(a−b), (c−a)=−(a−c), (c−b)=−(b−c). Three negatives → overall sign (−1)3=−1.
Wait — let me recheck. (b−a)(c−a)(c−b): factor (b−a)=−(a−b), (c−b)=−(b−c). Two negatives so (b−a)(c−b)=(a−b)(b−c). Combined with (c−a): (a−b)(b−c)(c−a). No extra sign.
Final answer:(a−b)(b−c)(c−a).
Why This Works
The Vandermonde structure is everywhere — interpolation, polynomial roots, signal processing. The determinant vanishes whenever two of a,b,c are equal, because that forces two rows to be identical. So the determinant must be divisible by (a−b), (b−c), and (c−a).
The result is a polynomial in a,b,c of degree 0+1+2=3 (one entry from each row), and (a−b)(b−c)(c−a) is exactly degree 3. Identical degrees + same vanishing → equal up to a sign, which we fix by checking one specific case.
Alternative Method
Direct cofactor expansion along the first row gives 1⋅(bc2−cb2)−a(c2−b2)+a2(c−b).
Factor the bracket: bc−ab−ac+a2=a(a−b)−c(a−b)=(a−b)(a−c).
So determinant =(c−b)(a−b)(a−c)=(a−b)(b−c)(c−a). Same result.
Common Mistake
Sign errors in tracking (a−b) vs (b−a) are the main trap. The “standard” form (a−b)(b−c)(c−a) is cyclic — many textbooks use this to avoid sign confusion. Always verify with a specific test: try a=1,b=2,c=3 → determinant =(1)(1)(−2)⋅1=−2 (computed directly), and (a−b)(b−c)(c−a)=(−1)(−1)(2)=2. Hmm, sign mismatch — let me redo: original det with a=1,b=2,c=3: 111123149=1(18−12)−1(9−4)+1(3−2)=6−5+1=2. And (a−b)(b−c)(c−a)=(−1)(−1)(2)=2. Match! Always sanity-check with numbers.
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