2.3 Linear Algebra 67213.LetAandBbe 3×3 matrices with real elements such that detA=detB =
det(A+B)=det(A−B)=0. Prove that det(xA+yB)=0 for any real numbers
xandy.
Sometimes it is more convenient to work with blocks instead of entries. For that
we recall the rule of Laplace, which is the direct generalization of the row or column
expansion. The determinant is computed by expanding over allk×kminors of some
krows or columns. Explicitly, givenA= (aij)ni,j= 1 , when expanding by the rows
i 1 ,i 2 ,...,ik, the determinant is given by
detA=∑
j 1 <j 2 <···<jk(− 1 )i^1 +···+ik+j^1 +···+jkMkNk,whereMkis the determinant of thek×kmatrix whose entries areaij, withi ∈
{i 1 ,i 2 ,...,ik}andj∈{j 1 ,j 2 ,...,jk}, whileNkis the determinant of the(n−k)×(n−k)
matrix whose entries areaijwithi/∈{i 1 ,i 2 ,...,ik}andj/∈{j 1 ,j 2 ,...,jk}. We exem-
plify this rule with a problem from the 4th International Competition in Mathematics for
University Students (1997).
Example.LetMbe an invertible 2n× 2 nmatrix, represented in block form as
M=
(
AB
CD
)
and M−^1 =(
EF
GH
)
.
Show that detM·detH=detA.
Solution.The idea of the solution is that the relation between determinants should come
from a relation between matrices. To this end, we would like to find three matrices
X, Y, Zsuch thatXY =Z, while detX=detM, detY=detH, and detZ=detA.
Since amongM,H, andA, the matrixMhas the largest dimension, we might try to set
X=Mand find 2n× 2 nmatricesYandZ. The equalityM·M−^1 =I 2 nyields two
relations involvingH, namelyAF+BH=0 andCF+DH=In. This suggests that
we should use bothFandHin the definition ofY. So we need an equality of the form
(
AB
CD)(
∗F
∗H
)
=
(
∗ 0
∗In)
.
We can try
Y=
(
InF
0 H)
.
The latter has determinant equal to detH, as desired. Also,