Advanced book on Mathematics Olympiad

72 2 Algebra

Solution.If we addInto the left-hand side of the identity from the statement, we recognize
this expression to be the polynomialP(X)=(X+A)(X+B)(X+C)evaluated at the
identity matrix. This means that

(In+A)(In+B)(In+C)=In.

This shows thatIn+Ais invertible, and its inverse is(In+B)(In+C). It follows that

(In+B)(In+C)(In+A)=In,

or

BCA+BC+BA+CA+A+B+C=On.

Subtracting this relation from the one in the statement and grouping the terms appropri-
ately, we obtain

ABC−BCA=(B+C)A−A(B+C).

The conclusion follows. 

Here are other examples.

227.LetAbe ann×nmatrix such that there exists a positive integerkfor which

kAk+^1 =(k+ 1 )Ak.

Prove that the matrixA−Inis invertible and find its inverse.

228.LetAbe an invertiblen×nmatrix, and letB=XY, whereXandYare 1×n,
respectively,n×1 matrices. Prove that the matrixA+Bis invertible if and only
ifα=YA−^1 X =−1, and in this case its inverse is given by

(A+B)−^1 =A−^1 −

1

α+ 1

A−^1 BA−^1.

229.Given twon×nmatricesAandBfor which there exist nonzero real numbersa
andbsuch thatAB=aA+bB, prove thatAandBcommute.

230.LetAandBben×nmatrices,n≥1, satisfyingAB−B^2 A^2 =InandA^3 +B^3 =
On. Prove thatBA−A^2 B^2 =In.