|A|∙|B|=|AB| Matrix Component Proof?

Question

Hello all,

I am preparing for an exam tomorrow and we have been asked to prove that the determinant of square matrix 'A' multiplied by the determinant of square matrix 'B' is equal to the determinant of the product of matrix 'A' and matrix 'B'. The proof must be in terms of components (i.e. in terms of a_ij and b_ij).

Any help would be greatly appreciated!!

Grateful Student

Answer 1

Favorite Answer

Andrew, are you assuming A and B are general n x n matrices, or are you proving this for a special case? Just in case, I'll work with n x n matrices. Let A = (a_ij) and B = (b_ij). Then AB = (c_ij), where c_ij = ∑(k = 1, n) a_ik b_kj. Hence

|AB| = ∑(p ∈ Sn) sign(p) c_1p(1) ••• c_np(n)

= ∑(p ∈ Sn) sign(p) (∑(k(1) = 1, n) a_1k(1) b_k(1)p(1)) ••• (∑(k(n) = 1, n) a_nk(n) b_k(n)p(n))

= ∑(k(1) = 1,n) ••• ∑(k(n) = 1, n) (a_1k(1)••• a_nk(n)) ∑(p ∈ S(n) sign(p) (b_k(1)p(1) ••• b_k(n)p(n)).

= ∑(k(1),..., k(n) = 1, n) (a_1k(1) ••• a_nk(n)) |b_k(i)j| (*)

Now if there exist l and m such that k(l) = k(m), then the matrix (b_k(i)j) has two equal rows and |(b_k(i)j)|= 0. So the sum in (*) is equal to the sum over all n-tuples (k(1),...,k(n)) where each of k(1),...,k(n) are distinct. Given such an n-tuple, there exists a permutation q ∈ Sn such that q(i) = k(i) for all 1 ≤ i ≤ n. Conversely, given a permutation q ∈ Sn, the elements q(1),...,q(n) are distinct. Hence the sum in (*) is equal to

∑(q ∈ Sn) (a_1q(1) ••• a_nq(n)) |(b_q(i)j)|

= ∑(q ∈ Sn) (a_1q(1) ••• a_nq(n)) • sign(q) |(b_ij)|

= ∑(q ∈ Sn) sign(q) a_1q(1) ••• a_nq(n) • |B|

= |A| |B|.

Answer 2

Hint: Resort to the definition of determinant

Answer 3

A=___and B=

(a c)______(e g)

(b d)______(f h)

so

AB=

(ae+cf ag+ch)

(be+df bg+dh)

then

det(AB)= (ae+cf)(bg+dh) - (be+df)(ag+ch)

= (abeg + adeh + bcfg + cdfh)

_______ - abeg - bceh - adfg - cdfh

= abeg - abeg + cdfh - cdfh +

________+ (ad - bc)eh - (ad-bc)fg

so

det(AB) = (ad-bc)(eh-fg)

finally

det(AB) = det(A) . det(B)

et voilà !!

hope it' ll help !!

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