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Putnam
1951 Putnam
A1
A1
Part of
1951 Putnam
Problems
(1)
Putnam 1951 A1
Source:
5/25/2022
Show that the determinant:
∣
0
a
b
c
−
a
0
d
e
−
b
−
d
0
f
−
c
−
e
−
f
0
∣
\begin{vmatrix} 0 & a & b & c \\ -a & 0 & d & e \\ -b & -d & 0 & f \\ -c & -e & -f & 0 \end{vmatrix}
0
−
a
−
b
−
c
a
0
−
d
−
e
b
d
0
−
f
c
e
f
0
is non-negative, if its elements
a
,
b
,
c
,
a, b, c,
a
,
b
,
c
,
etc., are real.
Putnam