Download 4.4 Applications of determinants

Applications of Determinants:
Cramer’s rule:
For linear system
Ann x  b ,
if
det( A)  0 ,
then the system has the
unique solution,
x1 
det( An )
det( A1 )
det( A2 )
, x2 
,, x n 
,
det( A)
det( A)
det( A)
Ai , i  1,2,, n,
where
is the matrix obtained by replacing the i’th column of
A by b.
Example:
Please solve for the following system of linear equations by Cramer’s rule,
x1  3 x2  x3  2
2 x1  5 x2  x3  5
x1  2 x2  3 x3  6
[solution:]
The coefficient matrix A and the vector b are
1
A  2
1
3
5
2
1
1, b 
3
  2
  5

,
 6 
respectively. Then,
 2 3 1
A1    5 5 1, A2
 6 2 3
Thus,
x1 
1  2 1
1 3  2
 2  5 1, A3  2 5  5
1 6 3
1 2 6 
det( A3 )
det( A1 )  3
det( A2 ) 6

 1, x2 

 2, x3 
 3.
det( A)  3
det( A)  3
det( A)
1
Note:
If we are seeking numerical answers, then any method involving determinants
are much less efficient than Gauss-Jordan reduction.
Note:
The advantages of using determinants:
1. Sometimes, we do not need numerical answers but an expression for the
answer because we might need to further manipulate the answer.
2. Determinant plays a key role in the study of eigenvalues and eigenvectors
which will be introduced later.
Areas:
Consider the triangle with vertices ( x1 , y1 ), ( x2 , y2 ) and ( x3 , y3 ) .
( x2 , y 2 )
( x1 , y1 )
( x3 , y 3 )
( x1 ,0)
( x2 ,0)
( x3 ,0)
( x1 , y1 )
( x3 , y 3 )
 Area 
( x1 ,0)
( x3 ,0)
2
( y1  y3 )( x3  x1 )
2
( x2 , y 2 )
( x1 , y1 )
( x1 ,0)
( x2 ,0)
 Area 
( y1  y 2 )( x2  x1 )
2
( x2 , y 2 )
( x3 , y 3 )
 Area 
( x2 ,0)
( y 2  y3 )( x3  x2 )
2
( x3 ,0)
Therefore, the area of the triangle is
( y1  y 2 )( x2  x1 ) ( y 2  y3 )( x3  x2 ) ( y1  y3 )( x3  x1 )


2
2
2
x1 y1 1
1
1
 x2 y1  x1 y 2  x3 y 2  x2 y3  x3 y1  x1 y3   x2 y 2 1
2
2
x3 y 3 1 .
  x1

1
 det  x2
2
 x
 3
y1 1 

y 2 1 
y3 1 
3
Example:
Compute the area of the triangle with vertices (1,4), (3,1) and (2,6) .
[solution:]
The area is
  1 4 1 

 1
1
1
det   3 1 1    17   17  8.5 .
2
2
  2 6 1  2




Note: Suppose we have a parallelogram with vertices
( x1 , y1 ), ( x2 , y2 ) ,
( x3 , y3 ) and ( x4 , y4 ) ,
( x2 , y 2 )
( x1 , y1 )
( x3 , y 3 )
( x4 , y 4 )
Then, the area of the parallelogram is
  x1

1
2   det   x 2
2
 x
 3
y1
y2
y3
  x1
1 



1   det   x 2
 x
1 
 3
4
y1
y2
y3
1 


1 
1 