Do either (1) or (2)
(1) Let A be matrix (aij), 1 ≤ i,j ≤ 4. Let d = det(A), and let Aij be the cofactor of aij, that is, the determinant of the 3 x 3 matrix formed from A by deleting aij and other elements in the same row and column. Let B be the 4 x 4 matrix (Aij) 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.
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 ai1Ai1 + ai2Ai2 + ai3Ai3 + ai4Ai4 = d. But ai1Aj1 + ai2Aj2 + ai3Aj3 + ai4Aj4 = 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 (Aij) 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
© John Scholes
5 Mar 2002