Page 471 - Matrix Analysis & Applied Linear Algebra

6.1 Determinants 467

are any 3 × 3 nonzero minors. There are exactly four 3 × 3 minors, and they
are

123 121 131 231

456 =0, 451 =0, 461 =0, 561 =0.

789 781 791 891

Since all 3 × 3 minors are 0, we conclude that rank (A)=2. You should be
able to see from this example that using determinants is generally not a good
way to compute the rankof a matrix.
In (6.1.12) we observed that the determinant of a product of elementary
matrices is the product of their respective determinants. We are now in a position
to extend this observation.

Product Rules

• det (AB) = det (A)det (B) for all n × n matrices. (6.1.15)

A B
• det = det (A)det (D)if A and D are square. (6.1.16)
0 D
Proof of (6.1.15). If A is singular, then AB is also singular because (4.5.2)
says that rank (AB) ≤ rank (A). Consequently, (6.1.14) implies that

det (AB)=0=det (A)det (B),
so (6.1.15) is trivially true when A is singular. If A is nonsingular, then A
can be written as a product of elementary matrices A = P 1 P 2 ··· P k that are
of Type I, II, or III—recall (3.9.3). Therefore, (6.1.12) can be applied to produce

det (AB) = det (P 1 P 2 ··· P k B) = det (P 1 )det (P 2 ) ··· det (P k )det (B)
= det (P 1 P 2 ··· P k ) det (B) = det (A)det (B).
A r×r 0
Proof of (6.1.16). First consider the special case X = , and use the
0 I
definition to write det (X)= x x . But
σ(p) x 1j 1 2j 2 ··· x rj r r+1,j r+1 ··· x n,j n


1 r r +1 ··· n
 ···
1 when p = ,
x =
x rj r r+1,j r+1 ··· x n,j n j 1 ··· j r r +1 ··· n
0 for all other permutations,

so, if p r denotes permutations of only the first r positive integers, then

x
x
det (X)= x 1j 1 2j 2 ··· x rj r r+1,j r+1 ··· x n,j n = x 1j 1 2j 2 ··· x rj r = det (A).
x
σ(p) σ(p r )