Example In an earlier example, we computed a 4 by 4 determinant by cofactor expansion only. The computation involes two 3 by 3 determinants. Now we mix two methods and get a more efficient computation.

det[	2	1	4	2	] = det[	0	1	3	2	]	= -2det[ 1 3 2 ] = -2det[ -5 3 -1 ] 2 1 2 0 1 1 2 1 1 0 1 0
	0	2	1	2		0	2	1	2			= 2det[	-5	-1	] = -10.
	0	2	1	1		0	2	1	1				0	1
	2	0	1	0		2	0	1	0

In the computation, the red equalities = mean row/columns operations. The first red equality means (-1)[row 4] + [row 1], and the second one means (-2)[col 2] + [col 1] and (-1)[col 2] + [col 3]. Moreover, the green equalities = mean cofactor expansions, along the darkened row or column.

Example By [col 3] + [col 2] and (-1)[row 2] + [row 3], we have

det[	1 - λ	3	-3	] = det[	1 - λ	0	-3	] = det[	1 - λ	0	-3	].
	-3	7 - λ	-3		-3	4 - λ	-3		-3	4 - λ	-3
	-6	6	-2 - λ		-6	4 - λ	-2 - λ		-3	0	1 - λ

Expanding along [col 2], the determinant is

(4 - λ)det[	1 - λ	-3	] = (4 - λ)[(1 - λ)2 - (-3)2] = (4 - λ)(1 - λ - (-3))(1 - λ + (-3)).
	-3	1 - λ

Thus we conclude

det[	1 - λ	3	-3	] = (4 - λ)2(- 2- λ).
	-3	7 - λ	-3
	-6	6	-2 - λ

Example The 4 by 4 Vandermonde matrix is

V4(a, b, c, d) = [	1	a	a2	a3	].
	1	b	b2	b3
	1	c	c2	c3
	1	d	d2	d3

To find its determinant, we apply (-a)[col 3] + [col 4], (-a)[col 2] + [col 3], (-a)[col 1] + [col 2], and then expand along the first row

detV4(a, b, c, d) = det[	1	0	0	0	] =	det[ b - a b(b - a) b2(b - a) ]. c - a c(c - a) c2(c - a) d - a d(d - a) d2(d - a)
	1	b - a	b(b - a)	b2(b - a)
	1	c - a	c(c - a)	c2(c - a)
	1	d - a	d(d - a)	d2(d - a)

Note that in the 3 by 3 matrix, [row 1] is (b - a) multiplied to (1, b, b²), and similarly for [row 2] and [row 3]. By the third operation, we get

detV4(a, b, c, d) = (b - a)(c - a)(d - a)det[	1	b	b2	] = (b - a)(c - a)(d - a)detV3(b, c, d).
	1	c	c2
	1	d	d2

This suggests a pattern for reducing the determinant of a Vandermonde matrix to another Vandermonde matrix of smaller size. Following the pattern, we get

detV₄(a, b, c, d)
= (b - a)(c - a)(d - a)detV₃(b, c, d)
= (b - a)(c - a)(d - a)(c - b)(d - b)detV₂(c, d)
= (b - a)(c - a)(d - a)(c - b)(d - b)(d - c)detV₁(d)
= (b - a)(c - a)(d - a)(c - b)(d - b)(d - c).

In general, the n by n Vandermonde matrix is

Vn(a1, a2, a3, ..., an) = [	1	a1	a12	. .	a1n-1	].
	1	a2	a22	. .	a2n-1
	1	a3	a32	. .	a3n-1
	:	:	:		:
	1	an	an2	. .	ann-1

From the relation

detV_n(a₁, a₂, a₃, ..., a_n) = (a_n - a₁)...(a₃ - a₁)(a₂ - a₁)detV_n-1(a₂, a₃, ..., a_n),

we get

detV_n(a₁, a₂, a₃, ..., a_n) = ∏_1≤i<j≤n(a_j - a_i).