Chapter 3: Linear Systems and Matrices - Part 3/Slides

Determinants
p.202 to p.211
p.216 Problems 1 to 20
3.6

 det ( A ) is a real number (can be negative also)
 det ( A ) is also written as
 Only a square n x n matrix has a determinant
 det ( AB ) = det ( A ) det ( B )
 A matrix A is not invertible (singular matrix) if
and only if det ( A ) = 0
 A matrix A is invertible (nonsingular matrix) if
and only if det ( A ) ≠ 0
3.6 Determinants : Definition
A

 1x1 matrix : A = [ a11 ], then det (A) = a11
 2x2 matrix :
det (A) = a11 a22 - a12 a21
11 12
21 22
a a
a a
 
 
  
A
3.6 Determinants : Examples

 3x3 matrix
det (A) = a11 a22 a33 + a12 a23 a31 + a13 a21 a32
-a11 a23 a32 - a12 a21 a33 -a13 a22 a31
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
 
 
A
Note: This cannot be extended to higher order matrices.

 3x3 matrix
minus
det (A) =
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
 
 
A
11 12 13
22 23 21 23 21 22
32 33 31 33 31 32
det det deta a a
a a a a a a
a a a a a a
     
      
          
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
 
 
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
 
 
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
 
 

1 2 3
4 5 6
7 8 9
 
 
 
 
 
A det (A) = 0
1 2 3
0 0 0
7 2 5
 
 
 
 
 
B det (B) = 0
det (I) = 1
 3 3det 1
1 0 0
0 1 0
0 0 1
 
 
  
 
 

   

I I
 3 3det 0
0 0 0
0 0 0
0 0 0
 
 
 
 
 
0 0

Defn - Let A = [ aij ] be an n x n matrix.
Let Mpq be the (n – 1) x (n – 1) submatrix of A
obtained by deleting the pth row and qth column of
A. The determinant det(Mpq) is called the minor
of apq
Defn - Let A = [ aij ] be an n x n matrix.
The cofactor Apq of apq is defined as
Apq = (–1)p+q det(Mpq)
3.6 Determinants : Cofactor Expansion

1 2 3
4 5 6
7 8 9
 
 
 
 
 
A
   
   
   
2
11 11 11
3
12 12 12
4
31 31 31
5 6
1 det 3
8 9
4 6
1 det 6
7 9
2 3
1 det 3
5 6
A
A
A
 
 
 
 
 
 
 
 
 
    
   
    
M M
M M
M M
3.6 Determinants : Cofactor Expansion - Example

Comment
 Examine pattern of signs of term (–1) p+q
 
 
 
  
  
  
  
 
 
 
 
 
 
   
   
   
   
n = 3 n = 4

Theorem - Let A = [ aij ] be an n x n matrix. Then
det(A) = ai1Ai1+ ai2Ai2 + L + ain Ain
(expansion of det(A) with respect to row i )
det(A) = a1j A1j+ a2j A2j + L + anj Anj
(expansion of det(A) with respect to column j )
 det(A) can be calculated by expansion along the
row (or column) of our choice.

Expand the determinant of a 3 x 3 matrix with
respect to first row 11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
 
 
A
   
   
   
1 1 22 23
11 22 33 23 32
32 33
1 2 21 23
12 21 33 23 31
31 33
1 3 21 22
13 21 32 22 31
31 32
1
1
1
a a
A a a a a
a a
a a
A a a a a
a a
a a
A a a a a
a a



   
   
   
So det(A) = a11A11 + a12A12 + a13A13

Expand the determinant of a 3 x 3 matrix with
respect to first column of A
   
   
   
1 1 22 23
11 22 33 23 32
32 33
2 1 12 13
21 13 32 12 33
32 33
3 1 12 13
31 12 23 13 22
22 23
1
1
1
a a
A a a a a
a a
a a
A a a a a
a a
a a
A a a a a
a a



   
   
   
So det(A) = a11A11 + a21A21 + a31A31
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
 
 
A

 Evaluate
Pick row or column with large number of
zeros, such as column 2
 
1 2 3 4
4 2 1 3
det
3 0 0 3
2 0 2 3





A

Example (continued)
 
   
            
            
         
       
12 22 32 42
1 2 2 2
1 2 3 2
1 2 3 2
det 2 2 0 0
4 1 3 1 3 4
2 1 3 0 3 2 1 3 0 3
2 2 3 2 2 3
2 1 1 3 3 2 3 2 1 4 3 3 3
2 3 1 3 3 2 3 2 1 1 3 3 4
2 9 6 2 12 9 2 3 9 6 2 3 12
2 15 6 2 45 30 2 9 2 15 48
A A A A
 
 
 
   
 
     
 
            
           
          
          
A

3 0 0 0
0 4 0 0
0 0 8 0
0 0 0 5
 
 
 
 
  
3.6 Determinants: Triangular & Diagonal Matrices
9 0 0
0 0 0
0 0 2
 
 
 
  
0 0 0
0 0 0
0 0 0
 
 
 
  
1 0 0
0 1 0
0 0 1
 
 
 
  
8 0
0 3
 
 
 
2 0
0 0
 
 
 
0 0
0 1
 
 
 
0 0
0 0
 
 
 
3 10
0 1 5
0 0 2
e 
 
 
   4 0 0
0 9 0
3 2 1
 
 
 
  
3 0 0 0
4 0 0
7 8 0
5
e
e
e e


 
 
 
 
 
  

 Theorem - Let A be a square matrix.
Then det ( A ) = det ( AT )
 Since det ( A ) = det ( AT ), the properties of
determinants with respect to row manipulations are true
also for the corresponding column manipulations
 Theorem - If a row (column) of A consists entirely of
zeros, then det ( A ) = 0.
 Proof - Let the i th row of A be all zeros. Each term of
det ( A ) contains a factor from the i th row of A. So,
det ( A ) = 0.
3.6 Determinants : Properties

 Theorem – Interchange two rows (columns) of A
If matrix B is obtained from matrix A by interchanging
two rows (columns) of A,
then det (B) = - det (A)
 Theorem - If two rows (columns) of A are equal, then
det (A) = 0.
 Proof - Suppose rows r and s of A are equal.
Interchange rows r and s to obtain matrix B. Then
det (B) = - det (A). However, B = A so
det (B) = det (A). Thus det (A) = 0.

 Theorem – Multiply row (column) i of A by c ≠ 0.
If B is obtained from A by multiplying a row (column) of
A by a real number c, then
det ( B ) = c det ( A ).
 Proof - Suppose the ith row of A is multiplied by c to
get B. Each term of det ( B ) contains a single factor
from the ith row of B. Thus, each term of det ( B )
consists of a single factor of c times the corresponding
term of det ( A ). So, det ( B ) = c det ( A ).

 Theorem – Add c times row (column) r of A to row
(column) s of A, r ≠ s
If B = [ bij ] is obtained from A = [ aij ] by adding to each
element of rth row (column) of A, c times the
corresponding element of the sth row (column), r ≠ s, of
A, then
det(B) = det(A)

 Theorem - Let A = [ aij ] be an upper (lower) triangular
matrix, then
det(A) = a11a22 … ann.
The determinant of a triangular matrix is the product of the
elements on the main diagonal.

 Theorem - If A is an nxn matrix, then Ax = 0
has a nontrivial solution if and only if det(A) = 0.
 Proof - Ax = 0 has a nontrivial solution if and only if A
is singular. If A is singular, then det(A) = 0.
 Theorem - If A is an nxn matrix, then Ax = b
has a unique solution if and only if det(A) 0.
≠

 Theorem - If A and B are n x n matrices, then
det(AB) = det(A) det(B)
 Proof - If A or B is singular, then AB is singular and the result
follows immediately. So, suppose that A and B are both
nonsingular. Then A and B can be expressed as the product of
elementary matrices.
A = E1E2 L Ep and B = F1F2 L Fq
det( A ) = det( E1 ) det( E2 ) L det( Ep )
det( B ) = det( F1 ) det( F2 ) L det( Fq )
det( AB ) = det( E1E2 L Ep F1F2 L Fq ) =
det( E1 ) det( E2 ) L det( Ep ) det( F1 ) det( F2 ) L det( Fq ) =
det( A ) det( B )

 Theorem - If A is nonsingular, then
det(A–1) = 1/det(A)
 Proof - Let In be the nxn identity. Then AA–1 = In
and det(In) = 1.
By the preceding theorem, det(A) det(A–1) = 1,
so det(A–1) = 1/det(A)

AB BA but det(AB) = det(BA)
1 2 1 2 1 1
0 2 1 1 1 0
2 1 1 2 1 1
   
   
   
   
   
 
   A B
2 2 0 0 3 0
0 1 1 1 4 2
7 0 1 4 5 4
   
   
   
   
   

    
 
AB BA
det(AB) = det(BA) = det(A) det(B) = -12
3.6 Determinants: Matrix Multiplication

Problems 3.6
p. 216
Problems : 1 to 20

Chapter 3: Linear Systems and Matrices - Part 3/Slides

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