Putnam 47/A6 solution

7th Putnam 1947

Problem A6

A is the matrix

a  b  c

d  e  f

g  h  i

det A = 0 and the cofactor of each element is its square (for example the cofactor of b is fg - di = b²). Show that all elements of A are zero.

Solution

a²e² - b²d² = (ei - fh)(ai - cg) - (fg - di)(ch - bi) = (ae - bd) i² + (cd - af) hi + (bf - ce) gi = (g³ + h³ + i³) i = 0, since 0 = det A = g³ + h³ + i³. Hence ae = ±bd. Similarly cd = ±af, bf = ±ce. Multiplying the three equations together we get abcdef = - abcdef unless at least one of the equations has a plus sign. In the first case, at least one of a, b, c, d, e, f is zero. In the second case, the element corresponding to the cofactor is zero - for example ae = bd implies i² = 0 and hence i = 0. So either a member of the first two rows is zero, or a member of the last row is zero.

wlog we may assume a = 0. That implies b or d = 0 also. [Note that if, for example, i was the zero element, then we would have ei = ±fh, by an argument similar to that above and hence f or h = 0). If b = 0, then since a³ + b³ + c³ = 0, we have also c = 0. Similarly, if d = 0, then g = 0. So we now have a complete row or column zero. But now the square of any other element is a linear combination of elements in the that row or column and hence zero. Suppose, for example, a = b = c = 0. Then g² = bf - ce = 0, and similarly for the other five elements.

Comment. This is surprisingly hard.

7th Putnam 1947