Advanced book on Mathematics Olympiad

Algebra 427

3 (x 1 +x 2 +x 3 )+ 3 (x 4 +x 5 +x 6 )+···+ 3 (x 97 +x 98 +x 99 )+ 3 x 100 = 0.

The terms in the parentheses are all zero; hencex 100 =0. Taking cyclic permutations

yieldsx 1 =x 2 = ··· =x 100 =0.

232.Ifyis not an eigenvalue of the matrix

⎛

⎜

⎜

⎜⎜

⎝

01001

10100

01010

00101

10010

⎞

⎟

⎟

⎟⎟

⎠

,

then the system has the unique solutionx 1 =x 2 =x 3 =x 4 =x 5 =0. Otherwise, the

eigenvectors give rise to nontrivial solutions. Thus, we have to compute the determinant

∣∣

∣∣

∣

∣∣

∣∣

∣

−y 1001

1 −y 100

01 −y 10

001 −y 1

1001 −y

∣∣

∣∣

∣

∣∣

∣∣

∣

.

Adding all rows to the first and factoring 2−y, we obtain

( 2 −y)

∣

∣∣

∣∣

∣∣

∣

∣∣

11111

1 −y 100

01 −y 10

00 1−y 1

1001 −y

∣

∣∣

∣∣

∣∣

∣

∣∣

.

The determinant from this expression is computed using row–column operations as fol-

lows:

∣∣

∣∣

∣

∣∣

∣∣

∣

11111

1 −y 100

01 −y 10

00 1−y 1

1001 −y

∣∣

∣∣

∣

∣∣

∣∣

∣

=

∣∣

∣∣

∣

∣∣

∣∣

∣

10 00 0

1 −y− 10 − 1 − 1

01 −y 10

00 1−y 1

1 − 1 − 10 −y− 1

∣∣

∣∣

∣

∣∣

∣∣

∣

=

∣

∣∣

∣∣

∣

∣∣

−y− 10 − 1 − 1

1 −y 10

01 −y 1

− 1 − 10 −y− 1

∣

∣∣

∣∣

∣

∣∣

=

∣

∣∣

∣∣

∣

∣∣

−y− 10 − 1 − 1

−y −y 0 − 1

01 −y− 1

− 10 −y−y

∣

∣∣

∣∣

∣

∣∣