Elementary Linear Algebra Anton & Rorres, 9 th Edition
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Transcript of Elementary Linear Algebra Anton & Rorres, 9 th Edition
Elementary Linear AlgebraAnton & Rorres, 9th Edition
Lecture Set – 02Chapter 2: Determinants
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Chapter Content
Determinants by Cofactor Expansion Evaluating Determinants by Row Reduction Properties of the Determinant Function A Combinatorial Approach to Determinants
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2-1 Minor and Cofactor
Definition Let A be mn
The (i,j)-minor of A, denoted Mij is the determinant of the (n-1) (n-1) matrix formed by deleting the ith row and jth column from A
The (i,j)-cofactor of A, denoted Cij, is (-1)i+j Mij
Remark Note that Cij = Mij and the signs (-1)i+j in the definition of
cofactor form a checkerboard pattern:
...
...
...
...
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2-1 Example 1
Let
The minor of entry a11 is
The cofactor of a11 is C11 = (-1)1+1M11 = M11 = 16
Similarly, the minor of entry a32 is
The cofactor of a32 is C32 = (-1)3+2M32 = -M32 = -26
8 4 1
6 5 2
4 1 3
A
168 4
6 5
8 4 1
6 5 2
4 1 3
11
M
266 2
4 3
8 4 1
6 5 2
4 1 3
32
M
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2-1 Cofactor Expansion The definition of a 3×3 determinant in terms of minors and
cofactors det(A) = a11M11 +a12(-M12)+a13M13
= a11C11 +a12C12+a13C13
this method is called cofactor expansion along the first row of A
Example 2
10)11)(1()4(3 4 5
4- 20
2 5
3 2 1
2 4
3 43
2 4 5
3 4 2
0 1 3
)det(
A
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2-1 Cofactor Expansion det(A) =a11C11 +a12C12+a13C13 = a11C11 +a21C21+a31C31
=a21C21 +a22C22+a23C23 = a12C12 +a22C22+a32C32
=a31C31 +a32C32+a33C33 = a13C13 +a23C23+a33C33
Theorem 2.1.1 (Expansions by Cofactors) The determinant of an nn matrix A can be computed by multiplying the
entries in any row (or column) by their cofactors and adding the resulting products; that is, for each 1 i, j n
det(A) = a1jC1j + a2jC2j +… + anjCnj
(cofactor expansion along the jth column)
and
det(A) = ai1Ci1 + ai2Ci2 +… + ainCin
(cofactor expansion along the ith row)
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2-1 Example 3 & 4 Example 3
cofactor expansion along the first column of A
Example 4 smart choice of row or column
det(A) = ?
1)3(5)2)(2()4(3 3 4
0 1 5
2 4
0 1 )2(
2 4
3 43
2 4 5
3 4 2
0 1 3
)det(
A
1002
12-01
2213
1-001
A
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2-1 Adjoint of a Matrix If A is any nn matrix and Cij is the cofactor of aij, then the
matrix
is called the matrix of cofactors from A. The transpose of this matrix is called the adjoint of A and is denoted by adj(A)
Remarks If one multiplies the entries in any row by the corresponding
cofactors from a different row, the sum of these products is always zero.
nnnn
n
n
C CC
C CC
C CC
21
22221
11211
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2-1 Example 5 Let
a11C31 + a12C32 + a13C33 = ?
Let
333231
232221
131211
aaa
aaa
aaa
A
131211
232221
131211
'
aaa
aaa
aaa
A
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2-1 Example 6 & 7
Let
The cofactors of A are: C11 = 12, C12 = 6, C13 = -16, C21 = 4, C22 = 2, C23 = 16, C31 = 12, C32 = -10, C33 = 16
The matrix of cofactor and adjoint of A are
The inverse (see below) is
042
361
123
A
16 10 12
16 2 4
16 6 12
16 16 16
10 2 6
12 4 12
)(adj A
16 16 16
10 2 6
12 4 12
64
1)(adj
)det(
11 AA
A
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Theorem 2.1.2 (Inverse of a Matrix using its Adjoint) If A is an invertible matrix, then
Show first that )(adj
)det(
11 AA
A
)det(2211 ACaCaCa ininiiii
IAAA )det()(adj
)(02211 jiCaCaCa jninjiji
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Theorem 2.1.3 If A is an n × n triangular matrix (upper triangular, lower
triangular, or diagonal), then det(A) is the product of the entries on the main diagonal of the matrix; det(A) = a11a22…ann
E.g.
44434241
333231
2221
11
0
00
000
aaaa
aaa
aa
a
A
?
40000
89000
67600
15730
38372
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2-1 Prove Theorem 1.7.1c A triangular matrix is invertible if and only if its diagonal
entries are all nonzero
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2-1 Prove Theorem 1.7.1d The inverse of an invertible lower triangular matrix is
lower triangular, and the inverse of an invertible upper triangular matrix is upper triangular
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Theorem 2.1.4 (Cramer’s Rule) If Ax = b is a system of n linear equations in n unknowns such
that det(I – A) 0 , then the system has a unique solution. This solution is
where Aj is the matrix obtained by replacing the entries in the column of A by the entries in the matrix b = [b1 b2 ··· bn]T
)det(
)det(,,
)det(
)det(,
)det(
)det( 22
11 A
Ax
A
Ax
A
Ax n
n
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2-1 Example 9 Use Cramer’s rule to solve
Since
Thus,
832
30643
62
321
321
31
xxx
xxx
x x
8 2 1
30 4 3
6 0 1
,
3 8 1
6 30 3
2 6 1
,
3 2 8
6 4 30
2 0 6
,
3 2 1
6 4 3
2 0 1
321 AAAA
11
38
44
152
)det(
)det(,
11
18
44
72
)det(
)det(,
11
10
44
40
)det(
)det( 33
22
11
A
Ax
A
Ax
A
Ax
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Exercise Set 2.1 Question 13
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Exercise Set 2.1 Question 23
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Exercise Set 2.1 Question 2 7