Determinants
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Notation
The determinant of a square matrixn×n
A is denoted det(A) or |A|.
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Definition of Determinant
The determinant ofn×n
A = [aij ] is defined as
n!
n
X
Y
m(pk )
|n×n
A|=
(−1)
aipk (i) ,
k=1
i=1
where p1 , . . . , pn! are the n! distinct bijections (one-to-one and onto
functions) from {1, . . . , n} to {1, . . . , n} and
m(pk ) =
n−1 X
n
X
11[pk (i) > pk (i∗ )].
i=1 i∗ =i+1
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Example: n = 2
Suppose
"
A=
a11 a12
#
.
a21 a22
Then n! = 2,
p1 (1) = 1 p1 (2) = 2
m(p1 ) = 0
p2 (1) = 2 p2 (2) = 1 m(p2 ) = 1,
and
|A| =
2
2
X
Y
(−1)m(pk )
aipk (i)
k=1
i=1
= a11 a22 − a12 a21 .
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Example: n = 3
Find an expression for the determinant of


a11 a12 a13


.
A=
a
a
a
21
22
23


a31 a32 a33
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Example: n = 1
Suppose A = [a11 ]. Then n! = 1,
p1 (1) = 1 m(p1 ) = 0,
and
|A| =
1
X
(−1)m(pk )
k=1
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1
Y
aipk (i) = a11 .
i=1
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Verbal Description of Determinant Computation
1
Form all possible products of n matrix elements that contain
exactly one element from each row and exactly one element from
each column.
2
For each product, count the number of pairs of elements for which
the “lower” element is to the “left” of the “higher” element. (For
example, a31 is “lower” than a12 because a31 is in the third row
while a12 is in the first row. Also, a31 is to the “left” of a12 because
a31 is in the first column while a12 is in the second column.) If the
number of such pairs is odd, multiply the product by −1. If the
number of pairs is even, multiply the product by 1.
3
Add the signed products determined in steps 1 and 2.
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Prove that if A is an n × n matrix and c ∈ R, then
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|cA| = cn |A|.
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Prove that ifn×n
A is lower or upper triangular, then
|A| =
n
Y
aii .
i=1
(Note that this result implies |A| =
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Qn
i=1 aii
for a diagonal matrix.)
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Suppose A is an n × n matrix. Prove that |A| = |A0 |.
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Proof:
Let qk = p−1
k ; i.e., qk (j) = i ⇐⇒ pk (i) = j.
Note that
i < i∗ , pk (i) = j > pk (i∗ ) = j∗ ⇐⇒ j∗ < j, qk (j∗ ) = i∗ > qk (j) = i.
Thus, m(pk ) = m(qk ).
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Proof:
It follows that
n!
n
X
Y
m(pk )
|A| =
(−1)
aipk (i)
=
=
k=1
n!
X
i=1
n
Y
k=1
j=1
(−1)m(pk )
n!
X
(−1)
k=1
0
m(qk )
n
Y
aqk (j)j
aqk (j)j
j=1
= |A |.
2
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Find an expression for

 I 0 I 0
n×n
n×m

 = n×n n×m
B m×m
C
B m×m
C m×n
m×n
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.
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Result A.18 (b):
If A and B are each n × n matrices, then
|AB| = |A||B|.
(We will not go through a proof here. Most graduate-level linear
algebra textbooks contain a proof of this result.)
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Use Result A.18 (b) to prove
A nonsingular ⇐⇒ |A| =
6 0.
n×n
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Determinant of a Partitioned Matrix
If A is nonsingular,
A B
C D
= |A||D − CA−1 B|.
A B
C D
= |D||A − BD−1 C|.
If D is nonsingular,
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Proof:
A B
C D
A B A B A B = |A||A−1 | = |AA−1 | = |I| C D C D C D A B A−1 −A−1 B A B −1
= |A| |A | = |A| C D 0
C D I
I
0
= |A| = |A||D − CA−1 B|.
CA−1 D − CA−1 B 2
A similar argument proves the second result.
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