Post on 02-Jan-2016
description
transcript
3.5 Solution by Determinants
The Determinant of a Matrix
The determinant of a matrix A is denoted by |A|.
Determinants exist only for square matrices.
The Determinant for a 2x2 matrix
If A =
Then
This one is easy
A ad bc
a b
c d
Coefficient Matrix You can use determinants to solve a
system of linear equations You use the coefficient matrix of the
linear system Linear System Coeff Matrix
ax+by = ecx+dy = f
dc
ba
Cramer’s Rule Linear System Coeff Matrix
ax+by = e cx+dy = f
Let D be the coefficient matrix If det D ≠ 0, then the system has exactly one solution:
x
e b
Df dx
a b D
c d
andy
a eDc f
ya b D
c d
dc
ba
Example 1- Cramer’s Rule (2x2)
Solve the system:8x + 5y = 22x ─ 4y = −10
42
58The coefficient matrix is:42)10()32(
42
58
and
So:
42
410
52
xand
42
102
28
y
142
42
42
)50(8
42
410
52
x
242
84
42
480
42
102
28
y
Solution: (-1,2)
Example 1 (continued)
Value of 3 x 3 (4 x 4, 5 x 5, etc.) determinants can be found using so called expansion by minors.1 1 1
2 2 2 2 2 22 2 2 1 1 1
3 3 3 3 3 33 3 3
a b cb c a c a b
a b c a b cb c a c a b
a b c
The Determinant for a 3x3 matrix
Example 2 - Cramer’s Rule (3x3)
Solve the system:x + 3y – z = 1–2x – 6y + z = –33x + 5y – 2z = 4
14
4
253
162
131
453
362
131
z
Let’s solve for Z
The answer is: (2,0,1)!!!
Inverse Matrix
1
1 1
Matrix is an inverse of matrix if
A A
A A A A I
1000
0100
0010
0001
Using Matrix-Matrix Multiplication:
z
y
x
675
324
232
2x + 3y – 2z
–4x + 2y + 3z
5x + 7y + 6z
This gives us a simple way to write a system of linear equations.
675
324
232
A
z
y
x
X
2
1
28
B
Then the system
2x + 3y – 2z = –2
–4x + 2y + 3z = 1
5x + 7y + 6z = 28
can be written as:AX B
Solving Equations Using Inverse Matrices
If A is the matrix of coefficients, X is the matrix of variables and B is the matrix of constants, then a system of equations can be presented as a matrix equation…
A X B
…and we can solve it for X by multiplying both sides of the equation by A-1 from the left:
1 1
1
A A X A B
so
X A B
A X B
How to find the Inverse Matrix
For a 2x2 matrix:
a b
c dA =
If ad – bc ≠ 0 then:
d -b
-c aA-1 =
1
ad – bc
d
ad-bc
-b
ad-bc
-c
ad-bc
a
ad-bc
=
3 5
1 2B =A-1 =
2 -5
-1 3A =Is the inverse of
2 -5
-1 3AB =
3 5
1 2=
1 0
0 1= I
2 -5
-1 3BA =
3 5
1 2=
1 0
0 1= I
How to find the Inverse Matrix (cont’d)
Find the inverse of
1 =A
1 2
1 3A =
d -b
-c a
1
ad-bc
d
ad-bc
-b
ad-bc
-c
ad-bc
a
ad-bc
=
Using the formula:
a=1; b=2;
c=1; d=3
Since ad – bc = 3–2=1:
d -b
-c a =
3 -2
-1 1
1 =A
PropertiesReal-number multiplication is commutative:
Is matrix multiplication commutative?
baab No! BAAB
Real-number multiplication is associative:
Is matrix multiplication associative?
cabbca )()(
Yes! CABBCA )()(
Real-number multiplication has an identity:
Does matrix multiplication have an identity?
aaa 11
Yes! AAIIA
(but you must use an identity matrix of the proper size for A)
Real-number multiplication has inverses:
Does matrix multiplication have an identity?
111 aaaaUnless a = 0.
Yes! IAAAA 11
Unless det(A) = 0.