#### Transcript Cramer Rule

```Determinants
Every n  n matrix A is associated with
a real number called the determinant
of A, written  A. The determinant of a
2  2 matrix is defined as follows.
5.3 - 1
Determinant of a 2  2 Matrix
 a1 1
If A = 
 a21
A 
a1 1
a1 2
a21
a22
a1 2 
 , th e n
a22 
 a1 1a 2 2  a 2 1a1 2 .
5.3 - 2
Note Matrices are enclosed with
square brackets, while determinants
are denoted with vertical bars. A matrix is
an array of numbers, but its determinant
is a single number.
5.3 - 3
Determinants
The arrows in the following diagram will
remind you which products to find when
evaluating a 2  2 determinant.
5.3 - 4
Example 1
Let A
3
=
 6
EVALUATING A 2  2
DETERMINANT
4
.

8
Find A.
Solution
Use the definition with
a1 1   3, a1 2  4, a 2 1  6, a 2 2  8 .
A  3
8

a11 a22
6
4
a21 a12
  24  24
  48
5.3 - 5
Determinant of a 3  3 Matrix
If
 a1 1

A =  a21
 a 3 1
a1 2
a22
a32
a1 3 

a 2 3 , th e n

a 3 3 
a1 1
a1 2
a1 3
A  a21
a22
a 2 3  ( a 1 1a 2 2 a 3 3  a 1 2 a 2 3 a 3 1  a 1 3 a 2 1a 3 2 )
a31
a32
a 3 3  ( a 3 1a 2 2 a1 3  a 3 2 a 2 3 a1 1  a 3 3 a 2 1a1 2 ).
5.3 - 6
Evaluating
The terms on the right side of the equation in the
definition of A can be rearranged to get
a1 1
a1 2
a1 3
A  a21
a22
a 2 3  a1 1 ( a 2 2 a 3 3  a 3 2 a 2 3 )  a 2 1 ( a1 2 a 3 3  a 3 2 a1 3 )
a31
a32
a 3 3  a 3 1 ( a1 2 a 2 3  a 2 2 a 1 3 ).
Each quantity in parentheses represents the
determinant of a 2  2 matrix that is the part of the
matrix remaining when the row and column of the
multiplier are eliminated, as shown in the next slide.
5.3 - 7
Evaluating
a1 1 ( a 2 2 a 3 3  a 3 2 a 2 3 )
a 2 1 ( a1 2 a 3 3  a 3 2 a1 3 )
a 3 1 ( a1 2 a 2 3  a 2 2 a1 3 )
 a1 1

a21

 a 3 1
a1 2
 a1 1

a21

 a 3 1
a1 2
 a1 1

a21

 a 3 1
a1 2
a22
a32
a22
a32
a22
a32
a1 3 

a23

a 3 3 
a1 3 

a23

a 3 3 
a1 3 

a23

a 3 3 
5.3 - 8
Cramer’s Rule for Two
Equations in Two Variables
Given the system a1 x  b1 y  c 1
a 2 x  b2 y  c 2
if then the system has the unique solution
x 
where
D 
a1
b1
a2
b2
,
Dx
D
Dx 
and y 
c1
b1
c2
b2
Dy
D
,
, and Dy 
a1
c1
a2
c2
.
5.3 - 9
Caution As indicated in the
preceding box, Cramer’s rule does not
apply if D = 0. When D = 0 the system is
inconsistent or has infinitely many
solutions. For this reason, evaluate D first.
5.3 - 10
APPLYING CRAMER’S RULE
TO A 2  2 SYSTEM
Use Cramer’s rule to solve the system
Example 4
5 x  7 y  1
6x  8y  1
Solution
Dy
Dx
By Cramer’s rule,
and y  D . Find D first,
D
since if D = 0, Cramer’s rule does not apply. If
D ≠ 0, then find Dx and Dy.
x 
D 
5
7
6
8
 5(8 )  6(7 )   2
Dy 
5
1
6
1
Dx 
1
7
1
8
  1(8 )  1(7 )   1 5
 5(1)  6(  1)  1 1
5.3 - 11
APPLYING CRAMER’S RULE
TO A 2  2 SYSTEM
By Cramer’s rule,
Example 4
Dy
15 15
11
11
x 


and y 

  .
D
2
2
D
2
2
Dx
11
 15
,



2 
 2
The solution set is
as can be verified
by substituting in the given system.
5.3 - 12
General form of Cramer’s
Rule
Let an n  n system have linear equations of the
form a1 x 1  a 2 x 2  a 3 x 3   a n x n  b .
Define D as the determinant of the n  n matrix of all
coefficients of the variables. Define Dx1 as the determinant
obtained from D by replacing the entries in column 1 of D
with the constants of the system. Define Dxi as the
determinant obtained from D by replacing the entries in
column i with the constants of the system. If D  0, the
unique solution of the system is
x1 
D x1
D
,
x2 
Dx2
D
,
x3 
Dx3
D
,
, xn 
D xn
D
.
5.3 - 13
APPLYING CRAMER’S RULE
TO A 3  3 SYSTEM
Use Cramer’s rule to solve the system.
x  y  z20
2x  y  z  5  0
x  2y  3z  4  0
Example 5
Solution
x 
2x 
y  z  2
y  z  5
x  2y  3z  4
Rewrite each
equation in the form
ax + by + cz + = k.
5.3 - 14
APPLYING CRAMER’S RULE
TO A 3  3 SYSTEM
Verify that the required determinants are
Example 5
1
2
1 1
1
1
D  2
1
1   3, D x   5
1
1  7,
1
2
3
2
3
1
2
Dy  2
5
1
4
1
4
1
1
1   2 2, D z  2
1
3
2
1
2
 5  21.
4
5.3 - 15
APPLYING CRAMER’S RULE
TO A 3  3 SYSTEM
Example 5
Thus,
x 
Dx
D

7
3
 
7
3
and z 
,
Dz
D
y 

Dy
D
21
3
so the solution set is

22
3

22
3
,
 7,
 7 22

, 7 .
  ,

 3 3
5.3 - 16
Note When D = 0, the system is
either inconsistent or has infinitely
many solutions. Use the elimination
method to tell which is the case. Verify that
the system in Example 6 is inconsistent, so
the solution set is ø.
5.3 - 17
```