Advanced book on Mathematics Olympiad

Algebra 423

223.We know thatAA∗=A∗A=(detA)I 3 ,soifAis invertible then so isA∗, and
A=detA(A∗)−^1. Also, detAdetA∗=(detA)^3 ; hence detA∗=(detA)^2. Therefore,
A=±

√

detA∗(A∗)−^1.

Because

A∗=( 1 −m)

⎛

⎝

−m− 11 1

1 −m− 11

11 −m− 1

⎞

⎠,

we have

detA∗=( 1 −m)^3 [−(m+ 1 )^3 + 2 + 3 (m+ 1 )]=( 1 −m)^4 (m+ 2 )^2.

Using the formula with minors, we compute the inverse of the matrix

⎛

⎝

−m− 11 1

1 −m− 11

11 −m− 1

⎞

⎠

to be

1

( 1 −m)(m+ 2 )^2

⎛

⎝

−m^2 −m− 2 m+ 2 m+ 2

m+ 2 −m^2 −m− 2 m+ 2

m+ 2 m+ 2 −m^2 −m− 2

⎞

⎠.

Then(A∗)−^1 is equal to this matrix divided by( 1 −m)^3. Consequently, the matrix we
are looking for is

A=±

√

detA∗(A∗)−^1

=±

1

( 1 −m)^2 (m+ 2 )

⎛

⎝

−m^2 −m− 2 m+ 2 m+ 2

m+ 2 −m^2 −m− 2 m+ 2

m+ 2 m+ 2 −m^2 −m− 2

⎞

⎠.

(Romanian mathematics competition)

224.The series expansion

1

1 −x

= 1 +x+x^2 +x^3 +···

suggests that

(In−A)−^1 =In+A+A^2 +A^3 +···.

But does the series on the right converge?