Transcript Thinking Mathematically by Robert Blitzer

Chapter 8
Matrices and Determinants
Matrix Solutions to Linear
Systems
Solving Linear Systems by Using
Matrices
A matrix gives us a shortened way of writing a system of equations. The first
step in solving a system of linear equations using matrices is to write the
augmented matrix. An augmented matrix has a vertical bar separating the
columns of the matrix into two groups. The coefficients of each variable are
placed to the left of the vertical line, and the constants are placed to the
right. If any variable is missing, its coefficient is 0. Here are two examples.
3x
x
x
+
+
+
y
y
3y
+
+
+
2z
2z
2z
=
=
=
31
19
25
3
1
1
1
1
3
2
2
2
31
19
25
x
+
2y
y
–
+
5z
3z
z
=
=
=
31
19
25
1
0
0
2
1
0
-5 -19
3 9
1 4
Example
Write the augmented matrix for the following
system:
2x-3y+z = -2
3x-5z = 6
4x-3y+z = 2
2  3 1  2

Solution:
3 0  5

4  3 1

6
2 
Text Example
Write the solution set for a system of equations represented by the matrix
1
0
0
2
1
0
-5 -19
3 9
1 4
Solution The system represented by the matrix is
1
0
0
2
1
0
-5 -19
3 9
1 4
1x + 2y – 5z = -19
0x + 1y + 3z = 9
0x + 0y + 1z = 4
This system can be simplified as follows.
x + 2y – 5z = -19
y + 3z = 9
z=4
Equation 1
Equation 2
Equation 3
Text Example cont.
Solution
The value of z is known. We can find y by back-substitution.
y + 3z = 9
y + 3(4) = 9
y + 12 = 9
y = -3
Equation 2
Substitute 4 for z.
Multiply.
Subtract 12 from both sides.
With values for y and z, we can now use back-substitution to find x.
x + 2y – 5z = -19
x + 2(-3) – 5(4) = -19
x – 6 – 20 = -19
x – 26 = -19
x =7
Equation 1
Substitute -3 for y and 4 for z.
Multiply.
Add.
Add 26 to both sides.
We see that x = 7, y = -3, and z = 4. The solution set for the system is
{(7, -3, 4)}.
Matrix Row Operations
•
1.
2.
3.
•
These row operations produce matrices that lead to
systems with the same solution set as the original system.
Two rows of a matrix may be interchanged. This is the
same as interchanging two equations in the linear system.
The elements in any row may be multiplied by a nonzero
number. This is the same as multiplying both sides of an
equation by a nonzero number.
The elements in any row may be multiplied by a nonzero
number, and these products may be added to the
corresponding elements in any other row. This is the
same as multiplying both sides of an equation by a
nonzero number and then adding equations to eliminate a
variable.
Two matrices are row equivalent if one can be obtained
from the other by a sequence of row operations.
Text Example
Use the matrix
3 18 -12 21
1 2 -3 5
-2 -3 4 -6
And perform each indicated row operation:
a. R1
R2
b. 3R1
c. 2R2 + R3
Solution
a. The notation R1
R2 means to interchange the elements in row 1 and
row 2. This results in the row-equivalent matrix.
1 2 -3 5
3 18 -12 21
-2 -3 4 -6
This was row 2; now it’s row 1.
This was row 1; now it’s row 2.
Text Example cont.
Solution
b. The notation 3R1 means to multiply each element in row 1 by 3. This
results in the row-equivalent matrix.
3•1 3•2 3•(-3) 3•5
3 18 -12 21
-2 -3
4
-6
=
1 2 -3 5
3 18 -12 21
-2 -3 4 -6
.
c. The notation 2R2 + R3 means to add 2 times the elements in row 2 to the
corresponding elements in row 3. Replace the elements in row 3 by these
sums. First, we find 2 times the elements in row 2:
2(1) or 2,
2(2) or 4,
2(-3) or –6,
2(5) or 10.
Now we add these products to the corresponding elements in row 3.
Although we use row 2 to find the products, row 2 does not change. It is
the elements in row 3 that change, resulting in the row-equivalent matrix
1
2
3 18
-2+2 -3+4
-3
5
-12
21
4-6 -6+10
=
3
1
0
18 -12 21
2 -3 5
1 -2 4
Solving Linear Systems Using
Gaussian Elimination
• Write the augmented matrix for the system
• Use matrix row operations to simplify the
matrix to one with 1s down the diagonal from
upper left to lower right, and 0s below the 1s
• Write the system of linear equations
corresponding to the matrix in step 2, and use
back-substitution to find the system’s solutions
Text Example
Use matrices to solve the system
3x + y + 2z = 31
x + y + 2z = 19
x + 3y + 2z = 25
Solution
Step 1 Write the augmented matrix for the system.
Linear System
3x + y + 2z = 31
x + y + 2z = 19
x + 3y + 2z = 25
Augmented Matrix
3
1
1
1
1
3
2
2
2
31
19
25
Text Example cont.
Solution
Step 2 Use matrix row operations to simplify the matrix to one with 1s
down the diagonal from upper left to lower right, and 0s below the 1s.
Our goal is to obtain a matrix of the form
1
0
0
a
1
0
b
d
1
c
e
f
.
Our first step in achieving this goal is to get 1 in the top position of the first
column.
We want 1 in
this position.
3
1
1
1
1
3
2
2
2
31
19
25
To get 1 in this position, we interchange rows 1 and 2. (We could also
interchange rows 1 and 3 to attain our goal.)
1
3
1
1
1
3
2
2
2
19
31
25
This was row 2; now it’s row 1.
This was row 1; now it’s row 2.
Text Example cont.
Solution
Now we want to get 0s below the 1 in the first column.
We want 0 in
these positions.
1
3
1
1
1
3
2
2
2
19
31
25
Let’s first get a 0 where there is now a 3. If we multiply the top row of
numbers by –3 and add these products to the second row of numbers, we
will get 0 in this position. The top row of numbers multiplied by –3 gives
-3(1) or –3, -3(1) or –3,
-3(2) or –6,
-3(19) or –57.
Now add these products to the corresponding numbers in row 2. Notice that
although we use row 1 to find the products, row 1 does not change.
1
1
2
19
3 + (-3) 1 + (-3) 2 + (-3) 31 + (-3)
1
3
2
25
=
1
0
1
We want 0 in
this position.
1
-2
3
2 19
-4 -26
2 25
Text Example cont.
Solution
We are not yet done with the first column. If we multiply the top row of
numbers by –1 and add these products to the third row of numbers, we will
get 0 in this position. The top row of numbers multiplied by –1 gives
-1(1) or –1, -1(1) or –1,
-1(2) or –2,
-1(19) or –19.
Now add these products to the corresponding numbers in row 3.
1
1
2
19
0
-2
-4
-26
1 + (-1) 3 + (-1) 2 + (-2) 25+(-19)
=
1
0
0
1
-2
2
2 19
-4 -26
0 6
We move on to the second column. We want 1 in the second row, second
column.
We want 1 in
this position.
1
0
0
1
-2
2
2 19
-4 -26
0 6
Text Example cont.
Solution
To get 1 in the desired position, we multiply –2 by its reciprocal, -1/2.
Therefore, we multiply all the numbers in the second row by –1/2 to get
1
› (0)
0
1
› (-2)
2
2
› (-4)
0
19
› (-26)
6
=
1
0
0
1
1
2
2
2
0
19
13
6
We want 0 in
this position.
We are not yet done with the second column. If we multiply the top row of
numbers by –2 and add these products to the third row of numbers, we will
get 0 in this position. The second row of numbers multiplied by –2 gives
-2(0) or 0, -2(1) or –2,
-2(2) or –4,
-2(13) or –26.
Now add these products to the corresponding numbers in row 3.
1
0
0+0
1
2
19
1
2
13
2 + (-2) 0 + (-4) 6+(-26)
=
1
0
0
1
1
0
2 19
2 13
-4 -20
Text Example cont.
Solution
We move on to the third column. We want 1 in the third row, third column.
1
0
0
We want 1 in
this position.
1
1
0
2 19
2 13
-4 -20
To get 1 in the desired position, we multiply –4 by its reciprocal, -1/4.
Therefore, we multiply all the numbers in the third row by –1/4 to get
1
1
2
19
0
1
2
13
-1/4(0) -1/4(0) -1/4(-4) -1/4(-20)
=
1
0
0
1
1
0
2
2
1
19
13
5
We now have the desired matrix with 1s down the diagonal and 0s below the 1s.
Step 3 Write the system of linear equations corresponding to the
matrix in step 2, and use back-substitution to find the system’s solution.
The system represented by the matrix in step 2 is
Text Example cont.
Solution
3
1
1
1
1
3
2
2
2
31
19
25
x + y + 2z = 19
y + 2z = 13
z = 5
We immediately see that the value for z is 5. To find y, we back-substitute 5
for z in the second equation.
y + 2z = 13
y + 2(5) = 13
y=3
Equation 2
Substitute 5 for x.
Solve for y.
Finally, back-substitute 3 for y and 5 for z in the first equation:
x + y + 2z = 19
Equation 1
x + 3 + 2(5) = 19
Substitute 3 for y and 5 for x.
x + 13 = 19
Multiply and add.
x=6
Subtract 13 from both sides.
The solution set for the original system is {(6, 3, 5)}.
Solving Linear Systems Using
Gauss-Jordan Elimination
1. Write the augmented matrix for the system.
2. Use matrix row operations to simplify the matrix
to one with 1s down the diagonal from upper left
to lower right, and 0s above and below the 1s.
3. Use the reduced row-echelon form of the matrix
in step 2 to write the system’s solutions set.
(Back-substitution is not necessary.)
Inconsistent and
Dependent Systems
and Their
Applications
Text Example
Use Gaussian elimination to solve the system
x – y – 2z = 2
2x – 3y + 6z = 5
3x – 4y + 7z = 12
Solution
Step 1 Write the augmented matrix for the system.
Linear System
Augmented Matrix
x – y – 2z = 2
2x – 3y + 6z = 5
3x – 4y + 4z = 12
1 -1 -2 2
2 -3 6 5
3 -4 4 12
Text Example cont.
Solution
Step 2 Attempt to simplify the matrix to one with 1s down the diagonal
and 0s below the 1s. Notice that the augmented matrix already has a 1in the
top position of the first column. Now we want 0s below the 1. To get the
first 0, multiply row 1 by 2 and add these products to row 2. To get the
second 0, multiply row 1 by 3 and add these products to row 3. Performing
these operations, we obtain the following matrix.
We want 1
in this
position.
1 -1 -2 2
0 -1 10 1
0 -1 10 6
Use the Previous matrix and:
Replace row 2 by -2R1 + R2 .
Replace row 3 by -3R1 + R3.
Text Example cont.
Solution
Step 2 Moving on to the second column, we obtain 1 in the desired
position by multiplying row 2 by 1.
1
1

1(0)  1(1)

0
1
2
 1
2

 
 1(10) 1(1) 0

 
10
6
 0
1
1
1
 2 2 
 
 101
 
10 6 
Text Example cont.
Solution
Now we want a 0 below the 1 in column 2. To get the 0, multiply row 2 by 1
and add these products to row 3. (Equivalently, add row 2 to row 3.) We
obtain the following matrix.
1

0

0
1
1
0
 2 2 
 
 101
 
0 5 
It is impossible to convert this last matrix to the desired form of 1s down the
main diagonal. If we translate the last row back into equation form, we get
0x + 0y + 0z  5,
which is false. Regardless of which values we select for x, y, and z, the last
equation can never be a true statement. Consequently, the system has no
solution. The solution set is , the empty set.
Nonsquare Systems
Up to this point, we have encountered only
square systems in which the number of
equations is equal to the number of variables. In
a nonsquare system, the number of variables
differs from the number of equations.
Example
Use Gaussian elimination to solve the following
system:
3x-y+z = 2
2x+3y-z = 7
z=t
Solution:
First, write the system in augmented matrix form:
3  1 1 2 
2 3  1 7 


0 0 1
t 
Example cont.
• (1/3)R1
3  1 1 2 
2 3  1 7 


0 0 1
t 
1  1 / 3 1 / 3 2 / 3 
2 3


1
7


0 0
1
t 
Example cont.
• -2R1 + R2
1  1 / 3 1 / 3 2 / 3 
2 3

1
7

0 0
1
t 
1/ 3
2 / 3
1  1 / 3
0 11/ 3  5 / 3 17 / 3 


0 0
1
t 
Example cont.
• 3/11 R2
1/ 3
2 / 3
1  1 / 3
0 11/ 3  5 / 3 17 / 3 


0 0
1
t 
1  1 / 3
0 1

0 0
2/3
 5 / 11 17 / 11
1
t 
1/ 3
Example cont.
x - (1/3)y + (1/3)z = 2/3
y - (5/11)z = 17/11
z=t
y - (5/11)t = 17/11
y = 17 + 5t
x - (1/3)(17+5t)/11 + (1/3)t = 2/3
33x - 17 - 5t + 11t = 22
33x + 6t = 39
33x = 39 - 6t
39  6t 13  2t
x

33
11
11
13  2t
x
11
17 + 5t
y
11
z t
Text Example
Use Gaussian elimination to solve the system
3x + 7y + 6z = 26
x + 2y + z = 8
Solution
3

1
We begin with the augmented matrix.
7
2
6 26 1
  
18  3
2
7
18  1
  
6 26 0
2
1
18 
 
3 2 
Because we now have 1s down the diagonal that begins with the upperleft
entry and a 0 below this 1, we translate the matrix back into equation form.
x + 2y + z  8
y + 3z  2
Equation 1
Equation 2
Text Example cont.
Solution
We can let z equal any real number and use backsubstitution to express x and
y in terms of z.
Equation 2
Equation 1
y + 3z  2
y  3z + 2
x + 2y + z  8
x + 2(3z + 2) + z  8
x  6z + 4 + z  8
x  5z + 4  8
x  5z + 4
Text Example cont.
Solution
With z = t, the ordered solution (x, y, z) enables us to express the system's
solution set as
{(5t + 4, 3t + 2, t)}
where t is any real number.
Matrix Operations
and Their
Applications
Notations for Matrices
We have seen that an array of numbers, arranged in rows and columns and
placed in brackets, is called a matrix. We can represent the matrix in two
different ways.
• A capital letter, such as A, B, or C, can denote a matrix.
• A lowercase letter enclosed in brackets, such as that shown below, can
denote a matrix.
A  [aij]
Matrix A with
elements ai j
A general element in matrix A is denoted by aij. This refers to the
element in the ith row and jth column. For example, a32 is the element
of A located in the third row, second column.
A matrix of order m X n has m rows and n columns. If m  n, a matrix
has the same number of rows as columns and is called a square matrix.
Text Example
Let
3
A  
4
2
5
0 

 1/ 5
a. What is the order of A?
b. If A  [aij], identify a23 and a12.
Solution
a. The matrix has 2 rows and 3 columns, so it is of order 2 X 3.
b. The element a23 is in the second row and third column. Thus, a23 
1 /5 .
The element a12 is in the first row and second column, and
consequently a12  2.
Definition of Equality of Matrices
• Two matrices A and B are equal if and only
if they have the same order m X n and aij =
bij for i = 1, 2, …, m and j = 1, 2, …, n
Properties of Matrix Addition
•
1.
2.
3.
4.
If A, B, and C are m X n matrices and 0 is
an m X n zero matrix, then the following
properties are true.
A+B = B+A Commutative Property
(A+B)+C = A+(B+C) Associative Property
A + 0 = 0 + A Additive Identity
A+(-A) = (-A) + A = 0 Additive Inverse
Example
• Add the following matrices
2  1 3   2 1 0 
4 0  5 + 1  3  1

 

Solution:
2  1 3   2 1 0 
4 0  5 + 1  3  1

 

2 + (2) (1) + 1 3 + 0



4
+
1
0
+
(

3
)
(

5
)
+

1
)


3
0 0


5

3

6


Definition of Scalar
Multiplication
If A  [aij] is a matrix of order m X n and c is
a scalar, then the matrix cA is the m X n
matrix given by
cA  [caij].
This matrix is obtained by multiplying each
element of A by the real number c. We call
cA a scalar multiple of A.
Example
If
1 3 
A
find  3 A

0  2
Example cont.
Solution:
1 3 
If A  
find  3 A

0  2 
3  3 *1  3 * 3 
1
 3



0  2   3 * 0  3 * 2
  3  9


6
0
Properties of Scalar
Multiplication
•
1.
2.
3.
4.
If A and B are m X n matrices, and c and d
scalars, then the following properties are
true.
(cd)A = c(dA) Associative Property
1A = A Scalar Identity Property
c(A+B) = cA + cB Distributive Property
(c+d)A = cA + dA Distributive Property
Definition of Matrix
Multiplication: 2X2 Matrices
 a b  e f 
AB  



c
d
g
h



ae + bg af + bh


ce
+
dg
cf
+
dh


Example
• Find the product:
3  2 1 
1 0  2 

3  1 7   2 0 4

 0  1 5 


Example cont.
Solution:
3  2 1 
1 0  2 


2
0
4
3  1 7  


 0  1 5 


1* 3 + 0 * 2 + 2 * 0 1* 2 + 0 * 0 + 2 * 1 1*1 + 0 * 4 + 2 * 5 
3 * 3 + 1* 2 + 7 * 0 3 * 2 + 1* 0 + 7 * 1 3 *1 + 1* 4 + 7 * 5


3

11
0
 13
9 

34
Definition of Matrix
Multiplication
The product of an m X n matrix, A, and an n
X p matrix, B, is an m X p matrix, AB,
whose elements are found as follows. The
element in the ith row and jth column of AB
is found by multiplying the each element in
the ith row of A by the corresponding
element in the jth column of B and adding
the products.
Properties of Matrix
Multiplication
•
•
•
•
(AB)C = A(BC) Associative Property
A(B+C) = AB+AC
(A+B)C = AC + BC Distributive Properties
c(AB) = (cA)B Associative Property of
Scalar Multiplication
Multiplicative
Inverses of Matrices
and Matrix
Equations
Definition of the Multiplicative
Inverse of a Square Matrix
Let A be an n X n matrix. If there exists an
n X n matrix A-1 such that
AA-1 = In and -1 A = In
then A -1 is the multiplicative inverse of A.
Text Example
Show that B is the multiplicative inverse of A, where
A=
-1 3
2 -5
and B =
5
2
3
1
Solution
To show that B is the multiplicative inverse of A, we
must find the products AB and BA. If B is the multiplicative inverse of A,
then AB will be the multiplicative identity matrix and BA will be the
multiplicative identity matrix. Because A and B are 2 X 2 matrices, n  2.
Thus, we denote the multiplicative identity matrix as I2; it is also a 2 X 2
matrix. We must show that
1 0 

AB  I2  
0 1
1 0 

BA  I 2  
0 1
Text Example cont.
Show that B is the multiplicative inverse of A, where
A=
-1
2
3
-5
and
B=
5
2
3
1
Solution
Let’s first show AB=I2
1 3 5 3 1(5) + 3(2)  1(3) + 3(1)  1 0 

 
 

AB  
2  52 1 2(5) + (5)(2) 1(3) + (5)(1) 0 1
Let’s now show BA=I2
5 31 3  5(1) + 3(2) 5(3) + 3(5)  1 0 

 
 

BA  
2 12  5 2(1) + 1(2) 2(3) + 1(5)  0 1
Both products give the multiplicative identity matrix. Thus, B is the multiplicative
5 3
inverse of A and we can designate B as A-1  


2 1
Example
• Find the multiplicative inverse of A.
3 2
A

1 4 
Solution
Let
x y 
A 

z
w


1
3 2  x y  1 0
AA  





1
4
z
w
0
1


 

1
3x + 2 z 3 y + 2w 1 0
1x + 4 z 1y + 4w   0 1

 

Example cont.
• Find the multiplicative inverse of A.
3 2
A

1 4 
Solution
3x + 2 z 3 y + 2w 1 0
1x + 4 z 1y + 4w   0 1

 

3x + 2 z  1
3 y + 2w  0
x + 4z  0
y + 4w  1
Example cont.
• Find the multiplicative inverse of A.
Solution
3 2
A

1
4


3x + 2 z  1
3 y + 2w  0
x + 4z  0
y + 4w  1
3x + 2 z  1
3 y + 2w  0
 3x  12z  0  3 y  12w  3
3x + 2 z  1
3 y + 2w  0
 3( x + 4 z  0)  3( y + 4w  1)
Example cont.
• Find the multiplicative inverse of A.
Solution
3 2
A

1
4


3x + 2 z  1
3 y + 2w  0
 3 x  12z  0  3 y  12w  3
 10z  1
1
z
10
 10w  3
3
w
10
Example cont.
• Find the multiplicative inverse of A.
Solution
3 2
A

1
4


1
z
10
x + 4z  0
3
w
10
y + 4w  1
4
x
10
2
y
10
Multiplicative Inverse of a 2x2
Matrix
a b 
If A  
, then

c d 
1 d  b 
1
A 


ad  bc  c a 
The matrix A is invertible if and only if ad-bc0. If
ad-bc=0, then A does not have a multiplicative inverse.
Procedure for Finding the Multiplicative
Inverse of an Invertible Matrix
•
1.
To find A-1 for any n X n matrix A for which A-1 exists:
Form the augmented matrix [A | I ], where I is the
multiplicative identity matrix of the same order as the
given matrix A.
2.
Perform row transformations on [A | I ] to obtain a
matrix of the form
[I | B ]. This is equivalent to
using Gauss-Jordan elimination to change A into the
identity matrix.
3.
Matrix B is A-1.
4.
Verify the result by showing that AA-1  I and A-1A  I.
Solving a System
-1
Using A
• If AX B has a unique solution, X  A-1B.
To solve a linear system of equations,
multiply A-1 and B to find X.
Text Example
Solve the system by using A-1, the inverse of the coefficient matrix
x – y + z
 2y + z
-2x – 3y
Solution
= 2
= 2
= 1/2
The linear system can be written as
1 -1 1
0 -2 1
-2 -3 0
x
y
z
2
2
=
1/2
The solution is given by X  A-1B. Consequently, we must find A-1. Using the inverse of
matrix A that we found previously,
3  3 1 2  1/ 2 

  

1
X  A B  0  2 1 2  1/ 2

  

2  3 01/ 2  1

Thus, x  1/2, y  -1/2, and z  1. The solution set is {(1/2, -1/2, 1)}.
Encoding a Word or Message
1. Express the word or message numerically.
2. List the numbers in step 1 by columns and form
a square matrix. If you do not have enough
numbers to form a square matrix, put zeros in
any remaining spaces in the last column.
3. Select any square invertible matrix, called the
coding matrix, the same size as the matrix in
step 2. Multiply the coding matrix by the square
matrix that expresses the message numerically.
The resulting matrix is the coded matrix.
4. Use the numbers, by columns, from the coded
matrix in step 3 to write the encoded message.
Decoding a Word or Message
That Was Encoded
1. Find the multiplicative inverse of the
coding matrix.
2. Multiply the multiplicative inverse of the
coding matrix and the coded matrix.
3. Express the numbers, by columns, from
the matrix in step 2 as letters.
Determinants and
Cramer’s Rule
Definition of the Determinant of
a 2 x 2 Matrix
a1 b1  is denoted by a1 b1
The determinant of the matrix



a2 b2
a2 b2 

And is defined by
a1 b1
a2
b2
 a1b2  a2b1
a
We also say that the value of the second-order determinant 1
a2
a1 b2 – a2b1.
b1
is
b2
Example
• Evaluate the determinant of:
Solution:
3  2
0 5 


3 2
 3 * 5  0 * 2
0 5
 15  0  15
Solving a Linear System in Two
Variables Using Determinants
• Cramer’s Rule a1 x + b1 y  c1
a2 x + b2 y  c2
then
c1 b1
a1 c1
c2 b2
a 2 c2
x
and y 
a1 b1
a1 b1
a2
b2
a2
b2
Example
• Use Cramer’s rule to solve the system:
3x  2 y  8
4 x + 3 y  5
Solution:
8
5
x
3
4
2
3
and y 
2
3
3
4
3
4
8
5
2
3
Example cont.
• Use Cramer’s rule to solve the system:
3x  2 y  8
4 x + 3 y  5
Solution:
 8 * 3  (2) * (5)
x
3 * 3  4 * (2)
 24  10  34


 2
9+8
17
3 * (5)  4 * (8)
y
3 * 3  4 * (2)
 15 + 32 17


1
9+8
17
Definition of a Third-Order
Determinant
a1 b1 c1
a2
b2
c2 
a3
b3
c3
a1b2 c3 + b1c2 a3 + c1a2b3  a3b2 c1  b3c2 a1  c3 a2b1
Definition of a 3 X 3 Matrix
• A third-order determinant is defined by
a1
b1
c1
b2
c2
b1 c1
a2 b2 c2  a1
 a2
b3 c3
b3
a3 b3 c3
c3
+ a3
b1
c1
b2
c2
Evaluating the Determinant of a 3 X 3 Matrix
1. Each of the three terms in the definition contains
two factors - a numerical factor and a secondorder determinant.
2. The numerical factor in each term is an element
from the first column of the third-order
determinant.
3. The minus sign precedes the second term.
4. The second-order determinant that appears in
each term is obtained by crossing out the row
and the column containing the numerical factor.
Text Example
Evaluate:
9 5
-2 -3
1 4
0
0
2
Solution
Note that the last column has two 0s. We will expand the determinant about
the elements in that column.
9
5 0
2  3
9 5
9
5
2  3 0  0
0
+2
1
4
1 4
2  3
1
4 2
 0  0 + 2[9(3)  (2)(5)]
 2(27+ 10)  2(17)  34
Determinants: Inconsistent and
Dependent-Systems
1. If D  0 and at least one of the
determinants in the numerator is not 0,
then the system is inconsistent. The
solution set is Ø.
2. If D  0 and all the determinants in the
numerators are 0, then the equations in the
system are dependent.