25231 19211 31213 25=2z2z+3y3y+x 19=2z2z+y+x 31=2z2z+y+3x3x 4100 9310 -19-521 4=z 9=3z3z+y...
TRANSCRIPT
252311921131213
25=2z+3y+x19=2z+y+x31=2z+y+3x
41009310
-19-521
4=z9=3z+y
-19=5z–2y+x
8.1 Matrices & Systems of Equations
• Augmented Matrix
• Row Operations
• Row Echelon Form –> Gaussian Elimination
• Reduced Row Echelon Form -> Gauss-Jordan Elimination
Augmented Matrix Example
Augmented Matrix in Row Echelon Form (Different Example)
* Reduced Row Echelon Form has 1’s along the diagonal and 0’s everywhere else
To solve a system of equations1.Write system of equations as an augmented matrix2.Use row operations to convert to row echelon form3.Find variables using back-substitution
WE’LL DO EXAMPLES IN CLASS!
Matrix Row Operations1. Two rows of a matrix may be interchanged.
2. The elements in any row may be multiplied by a nonzero number. (3R2)
3. Multiply a row by a non-zero number and add to another row
(Example: 2R1 + R3 -> R3)
-64-3-25-321
21-12183
Perform row operations:
Interchange: R1 R2 3R1 2R2 + R3R3
-64-3-221-121835-321
-64-3-25-321
63-36549
4-2105-321
21-12183
8.2 Matrix Algebra
A matrix of order m x n has m rows and n columns
A [aij] (Matrix a with elements aij)
Order: 2 x 3a23 = -1/5
a12. = 2
131
012
504
312
Matrix Addition
=
635
300
Scalar Multiplication
3 3 =
3 6 0
12 15 3 / 5
**Note: A matrix where all elements are zero is known as the zero matrix : 0
Equality of MatricesTwo matrices A and B are equal (A = B) if they have the sameOrder (number of rows and columns are equal) and each (i,j) entryOf A is equal to the corresponding (i,j) entry of B.
X – 8 6Y + 7 5 =
10 V15 R
x – 8 = 10 x = 18Y + 7 = 15 y = 8V = 6 and R = 5
X + Y 2 3 x - y =
4 23 1
x + y = 4X – y = 1So, x = 5/2 and y = 3/2
Matrix Multiplication
Note: AB is often not equal to BATry A = 2 3 X B = 3 4 -1 0 5 1
8.3 The Matrix InverseFor the Real Number System: A x 1 = A and 1 x A = ASo, 1 is the multiplicative Identity
For Matrices: A x I = A and I x A = ASo, I is the multiplicative Identity Matrix. I is a square matrix (2x2, 3x3, etc.) with 1’s in the Diagonal and 0’s elsewhere
5 7 1 0 5 7 1 0 5 76 8 0 1 6 8 0 1 6 8X = = X
For the Real Number System (A) (1/A) = 1 and (1/A)(A) = 1 (Identity)So, 1/A is the multiplicative inverse of A and A is the multiplicative inverse of 1/A
For Matrices: AA-1 = I and A-1A = I (If AB = BA = I then A & B are inverses)
-1 3 5 3 1 0 5 3 -1 3 2 -5 2 1 0 1 2 1 2 -5= =
Finding a Matrix Inverse or Proving None
To prove a matrix has no inverse, suppose has an inverse
Then,
Since the two matricies are equal y ouMust have 0 = 1, but this is False,Thus, No Inverse.
Find the multiplicative inverse of 2 1 5 3 2 1 w x 1 0 53 y z 0 1
2w + y = 1 2x + z = 05w + 3y = 0 5x + 3z = 1
Solve the system of equationsX = -1 w = 3 So, A-1 = 3 -1Y = -5 z = 2 -5 2
=
Special rule for inverse of a 2 x 2 matrixLet A= a b If ad – bc =0 the no inverse c d Otherwise, A-1 =
Another Method for Finding an Inverse
1 1 0 1 0 0
0 3 1 0 1 0
2 3 3 0 0 1
A I
Start With:
Perform Row Operations untilThe left is the Identity Matrix
2 1 1
6 3 11 1 0 0
8 8 82 3 1
0 1 08 8 86 1 3
0 0 18 8 8
R R R
��������������
1
6 3 1
8 8 8 6 3 12 3 1 1
2 3 18 8 8 8
6 1 36 1 3
8 8 8
A
Thus, the inverse is:
Solving Systems of Equations using InverseTo solve the system, solveThe matrix equation : AX = B X = A-1B
4
3 7
2 3 3 21
x y
y z
x y z
A X = B
Write the linear system in matrix form, then find the inverse. We know the inverse of this matrix since we found it in the previous example.
1
6 3 11
2 3 18
6 1 3
7
4
21
X A B
21
21
6 3 11
2 3 18
6 1
7
7
7
4
4
3 214
So, the solution set is {(3, 1, 4)}
8.4 Determinants and Cramer’s Rule
A a b
c d
The determinant of a 2 x 2 matrix, A is denoted det(A), |A| or
|A| = ad - bc
a b
c d
Example:
The determinant of a square n x n matrix, A, (n ≥ 3).is the sum of the entries in any row of A (or column of A), multiplied by their respective cofactors.
The minor Mij of the element aij is the determinant of the (n–1) × (n–1) matrix obtained by deleting the ith row and the jth column of A.
The cofactor Aij of the entry aij is given by: 1i j
ij ijA M
11
Finding Minors and Cofactors
a. Find minors M11 and M32
b. Find cofactors A11 and A32
c. Find |A|
To find M11 delete the first row and first columnThen, find the determinant: (-3)(7) – (-6)(1) = -21 - -6 = -21 + 6 = -15
To find M32 delete the third row and second columnThen, find the determinant: (-6)(2) – (4)(5) = -12 - 20 = -32
To find cofactors A11 & A32
A11 = (-1)1+1 (-15) = (-1)2 (-15) = 1(15) = 15A32 = (-1)3+2 (-32) = (-1)5 (-32) = -1(-32) = -32
1i j
ij ijA M
12
Example: Find the Determinant
Recall: The determinant of a square n x n matrix, is the sum of the entries in any row (or column), multiplied by their respective cofactors.
1i j
ij ijA M
13
© 2010 Pearson Education, Inc. All rights reserved
CRAMER’S RULE FOR SOLVINGTWO EQUATIONS IN TWO VARIABLES
a1x b1y c1
a2 x b2 y c2
The system
D a1 b1
a2 b2
,
x Dx
D and y
Dy
D
provided that D ≠ 0, where
Dx c1 b1
c2 b2
, and Dy a1 c1
a2 c2
.
two variables has a unique solution (x, y) given by
of two equations in
14© 2010 Pearson Education, Inc. All rights reserved
Using Cramer’s Rule to Solve Systems of Equations
Solve the system (2 x 2):
5 1
6
4
2 3
x y
x y
D 5 4
2 315 8 7
43 24 21.
3
1
6xD
530 2 28.
1
62yD
x Dx
D
21
7 3 and y
Dy
D
28
74.
1
4
7
0
5 2 3
4 3
x y z
x y z
x y z
Solve the system(3 x 3)
7 1 1
5 2 3
4 3 1
2 3 1 1 1 17 5 4
3 1 3 1 2 3
7 2 9 5 1 3 4 3 2
7 7 5 4 4 5 9
D
1i j
ij ijA M Recall:
1 1
2 3
3 1
2 3 1 1 1 1
3 1 3 1
1
2 3
1 2 9 4 1 3
1 7 4 4
0
1 4 0
9
4xD
1
4
0
1 4 0
7 1
5 3
4 1
5 3 7 1 7 1
4 1 4 1 5 3
1 5 12 4 7 4
17 44 27
yD
1
4
0
1 4 0
7 1
5 2
4 3
5 2 7 1 7 1
4 3 4 3 5 2
15 8 4 21 4
23 4 17 45
zD
x Dx
D
9
9 1
y Dy
D
27
93
z Dz
D
45
95
|A| = a11A11+a21A21+a31A31