Algebra 419214.We reduce the problem to a computation with 4×4 determinants. Expanding
according to the rule of Laplace, we see that
x^2 =∣∣
∣
∣∣
∣∣
∣
a 0 b 0
c 0 d 0
0 b 0 a
0 d 0 c∣∣
∣
∣∣
∣∣
∣
and x′^2 =∣∣
∣
∣∣
∣∣
∣
b′a′ 00
d′c′ 00
00 b′a′
00 d′c′∣∣
∣
∣∣
∣∣
∣
.
Multiplying these determinants, we obtain(xx′)^2.
(C. Co ̧sni ̧t ̆a, F. Turtoiu,Probleme de Algebr ̆a(Problems in Algebra), Editura Tehnic ̆a,
Bucharest, 1972)
215.First, suppose thatAis invertible. Then we can write
(
AB
CD
)
=
(
A 0
CIn)(
In A−^1 B
0 D−CA−^1 B)
.
The matrices on the right-hand side are of block-triangular type, so their determinants
are the products of the determinants of the blocks on the diagonal (as can be seen on
expanding the determinants using the rule of Laplace). Therefore,
det(
AB
CD
)
=(detA)(det(D−CA−^1 B))=det(AD−ACA−^1 B).The equality from the statement now follows form that fact thatAandCcommute.
IfAis not invertible, then since the polynomial det(A+In)has finitely many zeros,
A+Inis invertible for any sufficiently small>0. This matrix still commutes with
C, so we can apply the above argument toAreplaced byA+In. The identity from the
statement follows by letting→0.
216.Applying the previous problem, we can write
det(In−XY )=det(
InX
Y In)
=(− 1 )ndet(
Y In
InX)
=(− 1 )^2 ndet(
In Y
XIn)
=det(In−YX).Note that we performed some row and column permutations in the process, while keeping
track of the sign of the determinant.
217.Forkeven, that is,k= 2 m, the inequality holds even without the assumption from
the statement. Indeed, there existsarbitrarily small such that the matrixB 0 =B+In
is invertible. Then
det(A^2 m+B 02 m)=detB 02 mdet(
(AmB 0 −m)^2 +In