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1. Deduce from the theorems on determinants the following:
(1) If a column is zero, the determinant is zero.
(2) det(
A
) = det(
A
t
), where
A
t
is the transposed matrix.
(3) If a row is zero, the determinant is zero.
(4) Let
A
be a matrix in “block form”:
A
=
A
1
?
?
0
A
2
.
.
.
0
0
A
k
.
Here each
A
i
is a square matrix say of size
r
i
, and
A
2
starts at the
r
1
+ 1 column and
r
1
+ 1 row,
etc. Prove that
det(
A
) = det(
A
1
) det(
A
2
)
···
det(
A
k
)
.
Conclude that the determinant of a triangular matrix is given by
det
a
11
?
0
a
22
.
.
.
0
0
a
kk
=
a
11
a
22
···
a
kk
.
(Here each
a
ii
is a scalar).
Proof.
(1) Let
A
be a matrix such that its
i
th column
v
i
is zero. Then
det(
A
) = det(
v
1

. . .

v
i

. . .

v
n
)
= det(
v
1

. . .

v
i
+
v
i

. . .

v
n
)
= det(
v
1

. . .

v
i

. . .

v
n
) + det(
v
1

. . .

v
i

. . .

v
n
) = 2 det(
A
)
.
The claim follows.
(2) Let
A
= (
a
ij
) be a square
n
×
n
matrix. Note that for every two permutations
σ, τ
we have
a
σ
(1)1
···
a
σ
(
n
)
n
=
a
σ
(
τ
(1))
τ
(1)
···
a
σ
(
τ
(
n
))
τ
(
n
)
.
We apply that for
τ
=
σ

1
and obtain
det(
A
) =
X
σ
∈
S
n
sgn
(
σ
)
a
σ
(1)1
···
a
σ
(
n
)
n
=
X
σ
∈
S
n
sgn
(
σ
)
a
1
σ

1
(1)
···
a
nσ

1
(
n
)
=
X
σ
∈
S
n
sgn
(
σ

1
)
a
1
σ

1
(1)
···
a
nσ

1
(
n
)
=
X
σ
∈
S
n
sgn
(
σ
)
a
1
σ
(1)
···
a
nσ
(
n
)
.
(We used
sgn
(
σ
) =
sgn
(
σ

1
) and that summing over all
σ
∈
S
n
is the same as summing over all
σ

1
∈
S
n
). It remains to note that if we let
A
t
= (
b
ij
) (with
b
ij
=
a
ji
) then
X
σ
∈
S
n
sgn
(
σ
)
a
1
σ
(1)
···
a
nσ
(
n
)
=
X
σ
∈
S
n
sgn
(
σ
)
b
σ
(1)1
···
b
σ
(
n
)
n
= det(
A
t
)
.
(3) Apply (1) and (2)!
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 Spring '08
 Goren
 Determinant

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