224 SolutionsProblems of Chapter 4
4.1 In order to prove thatdet A = an An + a^A^ H h ainAin,(property 11) where Aij's are cofactors (Aij = (—l),+J'detMjj, where the
minor Mij is formed from A by deleting row i and column j);
without loss of generality, we may assume that i = 1.
Let us apply some row operations,On «22 «13
«21 ^22 «23
«31 a22 «33Onl «n2 On3Oln
«2n
«3n
->•Oil «22 «13 - - - 0,\n
0 022 «23 ' • • a 2 „
0 a 3 2 Q33 • • • Ctzn0 a„2 «n3 "• «rwhere atj = "aija;^1 +a^ai1, i,j = 2,..., n. In particular, a 22 = -a'*a»o»+^0 «°n.
Furthermore,>!->•an ^22 ^13 '
0 Q22 C*23 •
0 «22 «33 •0 an2 an3 •- 0,in
- a2n
- a3n
Otnn _—•an «22 «13 • •
0 C*22 Q23 • '
0 0 /? 33 • •din
OLln
fan0 0 /3„ 3 • • • Pnwhere /3y = -"'J^0 "^"", M = 2,..., n. In particular,-0:230:32 + «33Q22
/3 33 =
«22
(ai2«3i ~ fl32Qii)(a23aii - Q13Q21) + (033011 ~ Qi303i)(o220ii - 012O21)
«ii(022011 - ai 2 a 2 i)If we open up the parentheses in the numerator, the terms without an cancel
each other, and if we factor an out and cancel with the same term in the
denominator, we will have
A»3 =
Q12Q23Q31 + Q13Q32Q21 — OllQ23032 ~ 013031022 ~ 012021033 + Q11Q22Q33
~a 12 a1l + 022&11If we further continue the row operations to reach the upper triangular form,
we will have
A-+ •an
0
00022 «13 "
Q22 a23 •
0 /3 33 •0 0 •- oi„
- ain
Snn