Determinants

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MAC 2103
Module 3
Determinants
1
Learning Objectives
Upon completing this module, you should be able to:
1.
Determine the minor, cofactor, and adjoint of a matrix.
2.
Evaluate the determinant of a matrix by cofactor
expansion.
3.
Determine the inverse of a matrix using the adjoint.
4.
Solve a linear system using Cramer’s Rule.
5.
Use row reduction to evaluate a determinant.
6.
Use determinants to test for invertibility.
7.
Find the eigenvalues and eigenvectors of a matrix.
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2
Determinants
There are three major topics in this module:
Determinants by Cofactor Expansion
Evaluating Determinants by Row Reduction
Properties of the Determinant
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3
What is a Determinant?
A determinant is a real number associated with a square matrix.

a b
c d
Determinants are commonly used to test if a matrix is
invertible and to find the area of certain geometric figures.
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How to Determine if a Matrix is Invertible?
The following is often used to determine if a square matrix is invertible.
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5
Example
Determine if A-1 exists by computing the determinant of
the matrix A.
a)
b)
Solution
5 9
det(A)

 (5)(1)  (4)(9)  31
a)
4 1
A-1 does exist
b)
det(A) 
9 3
 (9)(1)  (3)(3)  0
3 1
A-1 does not exist
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6
What are Minors and Cofactors?
We know we can find the determinants of 2 x 2 matrices; but can we find the
determinants of 3 x 3 matrices, 4 x 4 matrices, 5 x 5 matrices, ...?
In order to find the determinants of larger square matrices, we need to
understand the concept of minors and cofactors.
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7
Example of Finding Minors and Cofactors
Find the minor M11 and cofactor A11
for matrix A.
Solution
To obtain M11 begin by crossing out the first row and column
of A.
The minor is equal to
det B = 6(5) ( 3)(7)
= 9
Since A11 = ( 1)1+1M11, A11 can
be computed as follows:
A11 = ( 1)2( 9) = 9
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8
How to Find the Determinant of Any
Square Matrix?
Once we know how to obtain a cofactor, we can find the determinant of any
square matrix. You may pick any row or column, but the calculation is easier if
some elements in the selected row or column equal 0.
n
n
a A
ij
ij
i 1
for any column j
Rev.F09
or
a A
ij
ij
j 1
for any row i
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9
Example of Finding the Determinant
by Cofactor Expansion
Find det A, if
Solution To find the determinant of A, we can select any
row or column. If we begin expanding about the first
column of A, then
det A = a11A11 + a21A21 + a31A31.
A11 = 9 from the previous example
A21 = 12 and A31 = 24
det A = a11A11 + a21A21 + a31A31
= ( 8)( 9) + (4)( 12) + (2)(24)
Now, try to
find the
determinant
of A by
expanding
the first row
of A.
= 72
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10
How to Find the Adjoint of a Matrix?
The adjoint of a matrix can be found by taking the
transpose of the matrix of cofactors from A.
In our previous example, we have found the cofactors
A11, A21, A31. If we continue to solve for the rest of the
cofactors for matrix A, namely A12, A22, A32 , A13, A23, and
A33 , then we will have a 3 x 3 matrix of cofactors from A
as follows:
 A11

 A21
 A
 31
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A12
A22
A32
A13 

A23 
A33 

11
How to Find the Adjoint of a Matrix? (Cont.)
The transpose of this 3 x 3 matrix of cofactors from A is
called the adjoint of A, and it is denoted by Adj(A).
 A11

Adj(A)   A12
 A
 13
A21
A22
A23
A31 

A32 
A33 

What are we going to do with this Adj(A)? We can use it
to help us find the A-1 if A is an invertible matrix.
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12
How to Find A-1 Using the Adjoint of a Matrix?
Theorem 2.1.2: If A is an invertible matrix, then
1
A 
Adj(A)
det(A)
1
Note:
1. The square matrix A is invertible if and only if det(A) is not
zero.
2. If A is an n x n triangular matrix, then det(A) is the product of
the entries on the main diagonal of the matrix (Theorem
2.1.3.)
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13
What is Cramer’s Rule?
Cramer’s Rule is a method that utilizes determinants to solve systems of linear
equations. This rule can be extended to a system of n linear equations in n
unknowns as long as the determinant of the matrix is non-zero.
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14
Example of Using Cramer’s Rule to
Solve the Linear System
Use Cramer’s rule to solve
the linear system.
Solution In this system a1 = 1, b1 = 4, c1 = 3, a2 = 2, b2
= 9 and c2 = 5
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15
Example of Using Cramer’s Rule to
Solve the Linear System (cont.)
E = 7, F = 1 and D = 1
The solution is
Note that Gaussian elimination with backward substitution is usually
more efficient than Cramer’s Rule.
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16
What Are the Limitations on the Method of
Cofactors and Cramer’s Rule?
The main limitations are as follow:
1.
2.
3.
A substantial number of arithmetic operations are needed to
compute determinants of large matrices.
The cofactor method of calculating the determinant of an n x
n matrix, n > 2, generally involves more than n! multiplication
operations.
Time and cost required to solve linear systems that involve
thousands of equations in real-life applications.
Next, we are going to look at a more efficient method to find the
determinant of a general square matrix.
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17
Evaluating Determinants by Reducing the
Matrix to Row-Echelon Form
Let A be a square matrix. (See Theorem 2.2.3)
(a) If B is the matrix that results from scaling
by a scalar k, then
det(B) = k det(A).
(b) If B is the matrix that results from either
rows interchange or columns interchange,
then
det(B) = - det(A).
(c) If B is the matrix that results from row
replacement, then
det(B) = det(A).
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Just keep these in
mind when A is a
square matrix:
1. det(A)=det(AT).
2. If A has a row of
zeros or a column
of zeros, then
det(A)=0.
3. If A has two
proportional rows
or two proportional
columns, then
det(A)=0.
18
How to Evaluate the Determinant by Row
Reduction?
Let’s look at a square matrix A.
 0 3 1 


A 1 1 2 
 3 2 4 
We can find the determinant by reducing it
into row-echelon form.
Step 1: We want a leading 1 in row 1. We
can interchange row 1 and row 2 to
accomplish this.
1 1 2
1 1 2
det(A)  0 3 1   0 3 1
3 2 4
3 2 4
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How to Evaluate the Determinant by Row
Reduction? (Cont.)
Step 2: We want a leading 1 in row 2. We
can take a common factor of 3 from row 2 to
accomplish this (scaling).
1 1 2
det(A)  3 0 1 13
From Step 1:
1 1 2
det(A)   0 3 1
3 2 4
3 2 4
Step 3: We want a zero at both row 2 and
row 3 below the leading 1 in row 1. We can
add -3 times row 1 to row 3 to accomplish
this (row replacement).
1
det(A)  3 0
1
1
2
1
3
0 1 2
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How to Evaluate the Determinant by Row
Reduction? (Cont.)
Step 4: We want a zero below the leading 1
in row 2. We can add row 2 to row 3 to
accomplish this (row replacement).
1 1
det(A)  3 0 1
0 0
2
1
3
From Step 3:
1
det(A)  3 0
5
3
Step 5: We want a leading 1 in row 3. We
take a common factor of -5/3 from row 3 to
accomplish this (scaling).
1
1
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1
3
0 1 2
Remember: If A is an n x
n triangular matrix, then
det(A) is the product of
the entries on the main
diagonal of the matrix.
1 1 2
 5 
 5 
1
det(A)  (3)   0 1 3  (3)   (1)  5
 3
 3
0 0 1
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2
21
Let’s Look at Some Useful
Basic Properties of Determinants
• Let A and B be n x n matrices and k is any
scalar. Then,
det(kA)  k det(A)
n
Question:
Is det(A+B) =
det(A) + det(B) ?
det(AB)  det(A)det(B)
• If A is invertible, then
1
det(A ) 
det(A)
1
This is because A-1A=I, det(A-1A) =det(I) =1;
det(A-1) det(A) = 1, and so
1
det(A ) 
, det(A)  0.
det(A)http://faculty.valenciacc.edu/ashaw/
Remember: If A is
an n x n triangular
matrix, then
det(A) is the
product of the
entries on the
main diagonal of
the matrix.
1
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What are Eigenvalues and EigenVectors?
An eigenvector of an n x n matrix Aris a nontrivial
r
(nonzero) vector x such that Ax   x , where
a scalar called an eigenvalue.
 is
Linear systems of this
r form can be rewritten as follows:
r
r
 x  Ax  0
r
r r
( I  A)x  Bx  0
The system has a nontrivial solution x if and only if
det( I  A)  det(B)  0.
This is the so called characteristic equation of A and r r
therefore B has no inverse, and the linear system Bx  0
has infinitely many solutions.
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Example
r r
Express the following linear system in the form ( I  A) x  0.
x1  2x2   x1 Find the characteristic equation, eigenvalues
2x1  x2   x2 and eigenvectors corresponding to each of the
eigenvalues.
The linear system can be written in matrix form as
 1 2 


 2 1  
 x1  


 x2  
 x1 
x1 
  
 with A   1 2  ,


x2 
x2 

2
1





1 2   x1   0 



2 1   x2   0 


 x1 

x
 x2 
 1 0   x1   1 2   x1   0 






0
1
x
2
1
x
0

  2  
  2  

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24
Example (Cont.)
   0   1 2   x1   0 
  0     2 1   x    0 

  2  

 
   1 2   x1   0 




2


1
x
0


  2  
r r
which is of the form ( I  A) x  0.
Thus,
   1 2
I  A  
 2   1

.

Can you tell what is the characteristic equation for A?
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Example (Cont.)
The characteristic equation for A is
det( I  A)  0
 1
2
or
2
0
 1
(  1)(  1)  (2)(2)  0
(  1)2  4  0
 2  2  1  4  0
 2  2  3  0
(  3)(  1)  0
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Example (Cont.)
Thus, the eigenvalues of A are: 1  3, 2  1
By definition, x is an eigenvector of Arif and only if x
r
is a nontrivial solution of ( I  A) x  0.
   1 2   x1   0 




2


1
x
0


  2  
If   3 , then we have
 2 2   x1   0 




2
2
x
0

  2  

that is
Thus, we can form the augmented matrix and solve
by Gauss Jordan Elimination.
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27
Example (Cont.)
Let’s form the augmented matrix and solve by
Gauss Jordan Elimination.
r1  2 2

r2  2 2
1
2
0 

0 
r1  r1  1 1

r2
 2 2
0 

0 
 1 1
r1

2r1  r2  r2  0 0
Thus,
x1  x2  0
x1  x2  t
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0 

0 
a free variable, t (,)
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Example (Cont.)
Solving this system yields: x1  t
x2  t
So the eigenvectors corresponding to 1  3
are the nontrivial solutions of the form 
x1   t 
 1 
  t 
x1  
 x2   t 
 1 
Similarly, if   1, then we have
 2 2   x1   0 




2
2
x
0

  2  

 2x1  2x2

 2x1  2x2
Rev.F09
  0 


0
 

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Example (Cont.)
Let’s form the augmented matrix and solve by
Gauss Jordan Elimination.
r1  2 2

r2  2 2
0 

0 
 12 r1  r1  1 1

r2
 2 2
0 

0 
 1 1 0 
r1


2r1  r2  r2  0 0 0 
x1  x2  0
Thus,
x1  x2  t
x1  t, x2  t,t (, )
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Example (Cont.)
Solving this system yields: x1  t
x2  t
So the eigenvectors corresponding to 2  1
are the nontrivial solutions of the form
 x1   t 
 1 

x2  
t

x
t
1
 2  



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What have we learned?
We have learned to:
1.
Determine the minor, cofactor, and adjoint of a matrix.
2.
Evaluate the determinant of a matrix by cofactor expansion.
3.
Determine the inverse of a matrix using the adjoint.
4.
Solve a linear system using Cramer’s Rule.
5.
Use row reduction to evaluate a determinant.
6.
Use determinants to test for invertibility.
7.
Find the eigenvalues and eigenvectors of a matrix.
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32
Credit
Some of these slides have been adapted/modified in part/whole from the
text or slides of the following textbooks:
• Anton, Howard: Elementary Linear Algebra with Applications, 9th Edition
• Rockswold, Gary: Precalculus with Modeling and Visualization, 3th Edition
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