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Putnam
1978 Putnam
A2
Putnam 1978 A2
Putnam 1978 A2
Source: Putnam 1978
May 2, 2022
Putnam
matrix
determinant
Problem Statement
Let
a
,
b
,
p
1
,
p
2
,
…
,
p
n
a,b, p_1 ,p_2, \ldots, p_n
a
,
b
,
p
1
,
p
2
,
…
,
p
n
be real numbers with
a
≠
b
a \ne b
a
=
b
. Define
f
(
x
)
=
(
p
1
−
x
)
(
p
2
−
x
)
⋯
(
p
n
−
x
)
f(x)= (p_1 -x) (p_2 -x) \cdots (p_n -x)
f
(
x
)
=
(
p
1
−
x
)
(
p
2
−
x
)
⋯
(
p
n
−
x
)
. Show that
det
(
p
1
a
a
⋯
a
b
p
2
a
⋯
a
b
b
p
3
⋯
a
⋮
⋮
⋮
⋱
⋮
b
b
b
⋯
p
n
)
=
b
f
(
a
)
−
a
f
(
b
)
b
−
a
.
\text{det} \begin{pmatrix} p_1 & a& a & \cdots & a \\ b & p_2 & a & \cdots & a\\ b & b & p_3 & \cdots & a\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ b & b& b &\cdots &p_n \end{pmatrix}= \frac{bf(a) -af(b)}{b-a}.
det
p
1
b
b
⋮
b
a
p
2
b
⋮
b
a
a
p
3
⋮
b
⋯
⋯
⋯
⋱
⋯
a
a
a
⋮
p
n
=
b
−
a
b
f
(
a
)
−
a
f
(
b
)
.
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