1st Putnam 1938/B1 solution

1st Putnam 1938

Problem B1

Do either (1) or (2)

(1) Let A be matrix (a_ij), 1 ≤ i,j ≤ 4. Let d = det(A), and let A_ij be the cofactor of a_ij, that is, the determinant of the 3 x 3 matrix formed from A by deleting a_ij and other elements in the same row and column. Let B be the 4 x 4 matrix (A_ij) and let D be det B. Prove D = d³.

(2) Let P(x) be the quadratic Ax² + Bx + C. Suppose that P(x) = x has unequal real roots. Show that the roots are also roots of P(P(x)) = x. Find a quadratic equation for the other two roots of this equation. Hence solve (y² - 3y + 2)² - 3(y² - 3y + 2) + 2 - y = 0.

Solution

Answer: (2) The quadratic is A²x² + (AB + A)x + (AC + B + 1) = 0. The quartic in y has roots 0, 1, 2, 2.

(1) We have a_i1A_i1 + a_i2A_i2 + a_i3A_i3 + a_i4A_i4 = d. But a_i1A_j1 + a_i2A_j2 + a_i3A_j3 + a_i4A_j4 = 0 for i not equal to j (because it can be considered as an expansion of the determinant for the matrix derived from A by replacing row i by row j - the resulting matrix has two identical rows and hence zero determinant). So if we multiply the transpose of A by the matrix (A_ij) then we get d down the diagonal and zeros elsewhere. Hence d D = d⁴, so D = d³.

(2) It is obvious that if P(x) = x, then P(P(x)) = x.

P(P(x)) = x is A(Ax² + Bx + C)² + B(Ax² + Bx + C) + C = x. This is evidently a quartic and two of its roots are those of Ax² + (B - 1)x + C = 0. We could obtain the quadratic for the other two roots by multiplying out P(P(x)) - x and factorising it. But it is sufficient to obtain the coefficients of x⁴, x³ and x⁰. This gives us the sum of the four roots as -2B/A and their product as (AC + B + 1)C/A³. The sum and product of the two known roots are -B/A - 1/A and C/A. Hence the sum and product of the other two roots are -B/A + 1/A and (AC + B + 1)/A², so the roots are the roots of the quadratic A²x² + (AB + A)x + (AC + B + 1) = 0.

y² - 3y + 2 = 0 has roots 1 and 2. So these values are also roots of (y² - 3y + 2)² - 3(y² - 3y + 2) + 2 - y = 0. The other two roots are also the roots of x² + (-3 + 1)x + (2 - 3 + 1) = 0. These are obviously 0 and 2.

1st Putnam 1938