Advanced book on Mathematics Olympiad

426 Algebra

Second solution: Simply write

In=kAk+^1 −(k+ 1 )Ak+In=(A−In)(kAk−Ak−^1 −···−A−In),

which gives the inverse written in a different form.
(Mathematical Reflections, proposed by T. Andreescu)

228.Ifα =−1 then
(
A−^1 −

1

α+ 1

A−^1 BA−^1

)

(A+B)=In+A−^1 B−

1

α+ 1

A−^1 BA−^1 B

−

1

α+ 1

A−^1 B.

But(A−^1 B)^2 =A−^1 X(YA−^1 X)Y=αA−^1 XY=αA−^1 B. Hence in the above equality,
the right-hand side is equal to the identity matrix. This proves the claim.
Ifα=−1, then(A−^1 B)^2 +A−^1 B=0, that is,(In+A−^1 B)A−^1 B=0. This implies
thatIn+A−^1 Bis a zero divisor. Multiplying byAon the right we find thatA+Bis a
zero divisor itself. Hence in this caseA+Bis not invertible.
(C. Nast ̆ ̆asescu, C. Ni ̧ta, M. Brandiburu, D. Joi ̧ta, ̆ Exerci ̧tii ̧si Probleme de Algebra ̆
(Exercises and Problems in Algebra), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1983) ̆

229.The computation

(A−bIn)(B−aIn)=abIn

shows thatA−bInis invertible, and its inverse isab^1 (B−aIn). Then

(B−aIn)(A−bIn)=abIn,

which translates intoBA−aA−bB=On. Consequently,BA=aA+bB=AB,
proving that the matrices commute.

230.We have

(A+iB^2 )(B+iA^2 )=AB−B^2 A^2 +i(A^3 +B^3 )=In.

This implies thatA+iB^2 is invertible, and its inverse isB+iA^2. Then

In=(B+iA^2 )(A+iB^2 )=BA−A^2 B^2 +i(A^3 +B^3 )=BA−A^2 B^2 ,

as desired.
(Romanian Mathematical Olympiad, 1982, proposed by I.V. Maftei)

231.Of course, one can prove that the coefficient matrix is nonsingular. But there is a
slick solution. Add the equations and group the terms as