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Quiz 5 Solutions
April 24, 2005
Thursday
(1)
This was a homework problem (HW 9). Let
d
n
be the determinant of
the
n
×
n
matrix shown. Then by Laplace expanding along the ﬁrst row, we
see that
d
n
= (

1)
n
+1
d
n

1
for
n >
1 . Using this recursion, we ﬁnd that
d
n
= (

1)
∑
n
j
=2
j
+1
The summation in the exponent is equal to
n
(
n
+ 3)
/
2. This is an even
number when
n
(
n
+ 3)
/
2 is odd and an odd number otherwise. Elementary
arithmetic shows that
n
(
n
+ 3) is divisible by four whenever
n
or
n
+ 3 is,
and not divisible by four when
n
and
n
+ 3 are not. Therefore
d
n
is equal
to one whenever
n
is congruent to 1 or 4 (mod 4) and

1 otherwise.
(2)
(a) Recalling that det
AB
= (det
A
)(det
B
) for
n
×
n
matrices
A
and
B
, we see that
A
2
=
A
implies that (det
A
)
2
= (det
A
) Thus (det
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 Spring '05
 CONSANI
 Linear Algebra, Algebra, Determinant

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