12b applications
TRANSCRIPT
Additional examples &
problems
1
GE 111; Topics 11-12
Nov 2013GE 111 Engineering Problem Solving
Determinant of 2-by-2 Matrix
dc
baA
(a,b) (a+c,b+d)
(c,d)
det(A) = ad – bcif these are real numbers,
then A can be mapped into
area of the parallelogram.
2 Nov 2013GE 111 Engineering Problem Solving
The area of the parallelogram is the absolute value determinant of the
matrix formed by the vectors representing the parallelogram's sides.
From ‘Determinant’ in Wikipedia
Determinant of 3-by-3 Matrix
3Diagram from C. Rocchini. Permission for reuse by GNU Free Documentation License
http://en.wikipedia.org/wiki/Determinant Nov 2013GE 111 Engineering Problem Solving
If values within matrix are
real then they can be
mapped to represent a
parallelepiped. The
volume of this
parallelepiped is the
absolute value of the
determinant of the matrix
formed by the rows r1, r2,
and r3.
For more explanation and visualization see: MIT OCW
http://ocw.mit.edu/ans7870/18/18.013a/textbook/HTML/chapter04/section01.html
3 0 0
0 2 0
0 0 1 |det(A)| = |6|
3 2 0
1 2 1
0 2 1 |det(A)| = |-2| = 2
Example1: determinant of a 4x4
Nov 2013GE 111 Engineering Problem Solving4
3 0 2 -1
1 2 0 -2
4 0 6 -3
5 0 2 0
A =
+ - + -
- + - +
+ - + -
- + - +
+ - +
- + -
+ - +
Example from http://nebula.deanza.edu/~bloom/math43/Determinant4x4Matrix.pdf
First do cofactor expansion along 1st row developing 3x3 matrices
2 0 -2
0 6 -3
0 2 0
1 0 -2
4 6 -3
5 2 0
1 2 -2
4 0 -3
5 0 0
1 2 0
4 0 6
5 0 2
det (A) = 3 - 0 + 2 - (-1)
6 -3
2 0
0 -3
0 0
0 6
0 2= 3 2 - 0 + (-2) - 0 +2 1
0 -3
0 0. . .
Second do cofactor expansion along 1st row of each 3x3 developing 2x2 determinants)
det (A) = 20
Example1: determinant of a 4x4
Nov 2013GE 111 Engineering Problem Solving5
3 0 2 -1
1 2 0 -2
4 0 6 -3
5 0 2 0
A =
+ - + -
- + - +
+ - + -
- + - +
+ - +
- + -
+ - +
Example from http://nebula.deanza.edu/~bloom/math43/Determinant4x4Matrix.pdf
Do cofactor expansion along column 2 developing 3x3 matrices
1 0 -2
4 6 -3
5 2 0
3 2 -1
4 6 -3
5 2 0
3 2 -1
1 0 -2
5 2 0
3 2 -1
1 0 -2
4 0 6
det (A) = -0 + 2 - 0 + 0
2 -1
6 -3
3 -1
4 -3
3 2
4 6= 2 5 - 2 + 0
Do cofactor expansion along row 3 of remaining 3x3 developing 2x2 determinants)
det (A) = 20
Example2: determinant of a 4x4
Nov 2013GE 111 Engineering Problem Solving6
2 5 -3 -2
-2 -3 2 -5
1 3 -2 2
-1 -6 4 3
A =
Pick a pivot point wish to do cofactor
expansion along that can create ‘0’s
along a row or column; e.g. a31
0 -7 5 4
0 3 -2 9
1 3 -2 2
0 -3 2 5
2R4+R1
2R3+R2
R3+R4
-7 5 4
3 -2 9
-3 2 5R2+R3
+1
-7 5 4
3 -2 9
0 0 14
+1
-7 5
3 -2det (A) = 1 (14 ) = 1(14 (14-15) = -14
Problems 1&2 : Find determinants
Nov 2013GE 111 Engineering Problem Solving7
6 2 1 0 5
2 1 1 -2 1
1 1 2 -2 3
3 0 2 3 -1
-1 -1 -3 4 2
A = det (A) = 34
6 2 1 0 5
0 1 1 -2 1
0 0 2 -2 3
0 0 0 3 -1
0 0 0 0 2
B = det (B) = 72
Problem 3: A Traffic Light Assembly
Nov 2013GE 111 Engineering Problem Solving8
AD
Z
X
Y
B
C
6 m
6 m4 m
3 m
4 m
4 m
3 m
3 m
h
4 m
3: A Traffic Light Assembly
Nov 2013GE 111 Engineering Problem Solving9
The mass of the traffic light = 20 kg
h = 3.5 m. For equilibrium:
3: A Traffic Light Assembly
Nov 2013GE 111 Engineering Problem Solving10
0995.03492.00995.0
597.04191.0796.0
796.08381.0597.0
A
2.196
0
0
B
4702.0)det( A
11
0995.03492.00995.0
597.04191.0796.0
796.08381.0597.0
A
2.196
0
0
B 4702.0)det( A
4702.0
62.163
)det(
0995.03492.02.196
597.04191.00
796.08381.00
A
FAB
Nov 2013GE 111 Engineering Problem Solving
3: A Traffic Light Assembly
= 348.0
= 413.1FAC
= 174.0FAD
Problem 4: An Electric Circuit
Nov 2013GE 111 Engineering Problem Solving12
Kirchoff’s voltage law: The sum of all voltage drops around a closed loop is zero.
Kirchoff’s current law: At any junction, current in must equal to current out.
Ohm’s law: current through a conductor between two points is directly proportional to the potential difference across two points. V (volts) = I (ampere) R(ohms)
Given the circuit in the adjacent image, what is the current (ampere) for I1, I2, and I3?
13
213 III
Nov 2013GE 111 Engineering Problem Solving
Kirchoff’s voltage law: The sum of all voltage drops around a closed loop is zero.
Kirchoff’s current law: At any junction, current in must equal to current out.
801220 31 II 1201215 32 II
Problem 4: An Electric Circuit
14
1201215 32 II
213 III
801220 31 II
1201215 32 II
0213 III
801220 31 II
20 0 12 0 15 12
-1 -1 1
I1I2I3
80120
0=
I1 = 1, I2 = 4, I3 = 5
Nov 2013GE 111 Engineering Problem Solving
Problem 4: An Electric Circuit
Problem 5: Traffic Flow
Nov 2013GE 111 Engineering Problem Solving15
Ruth Street
Tait Street Pre
sto
n A
ve
.
Cla
ren
ce
Ave
.
x1 x3
x2
x4
Rush-hour traffic along one-way
Streets is shown in adjacent figure.
The numbers represent traffic flow
in vehicles per hour that enter and
leave the streets. The variables;
x1, x2, x3, and x4 represent the
traffic flow on the streets between
the four intersections.
Assumptions:
• All the cars that enter the
system, leave (no parking is
allowed).
• All the cars that enter any one
intersection leave (no
breakdowns).
• Cars cannot go in reverse (all
flow must be nonnegative)
600
400
500
700
300
800
400
700
A B
D C
Problem 5: Traffic Flow
Nov 2013GE 111 Engineering Problem Solving16
Flow through intersections and on streets between each
intersection (Vph)
Intersection Flow
A x1 x2 1000
B x2 x3 1100
C x3 x4 1200
D x1 x4 1100
Problem 5: Traffic Flow
Nov 2013GE 111 Engineering Problem Solving17
Gauss-Jordan Elimination and create one column of x4’s
1 1 0 0 1000
0 1 1 0 1100
0 0 1 1 1200
1 0 0 1 1100
1 1 0 0 1000
0 1 0 -1 -100
0 0 1 1 1200
1 0 0 1 1100
-R3+R2
-R2+R1 1 0 0 1 1100
0 1 0 -1 -100
0 0 1 1 1200
1 0 0 1 1100
-R1+R4
1 0 0 1 1100
0 1 0 -1 -100
0 0 1 1 1200
0 0 0 0 0
x1 + x4 = 1100
x2 – x4 = -100
x3 + x4 = 1200
x1 = -x4 + 1100
x2 = x4 - 100
x3 = -x4 + 1200
As traffic flow must have nonnegative values:
100 ≤ x4 ≤ 1100
Problem 5: Traffic Flow
Nov 2013GE 111 Engineering Problem Solving18
There is a variety of solutions, however they must all
involve positive values of traffic flow. Placing
constraints on the system can provide more
information.
Example:
A water main breaks on Tait Street between
intersections C and D. They have to constrict traffic
during construction. The maximum flow capacity
along Preston (between intersections B and C) is 950
Vph, what is the maximum flow capacity that Tait
Street can handle during construction?
Ruth Street
Tait Street Pre
sto
n A
ve
.
Cla
ren
ce
Ave
.
x1 x3
x2
x4
600
400
500
700
300
800
400
700
A B
D C
x1 = -x4 + 1100
x2 = x4 - 100
x3 = -x4 + 1200