12b applications

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