This document provides information about determinants and their application to matrices and geometric shapes. It begins with definitions and examples of determinants of 2x2 and 3x3 matrices. The determinant of a matrix represents the area or volume of the geometric shape formed by the matrix's rows or columns. The document then provides an example of calculating the determinant of a 4x4 matrix using cofactor expansion. It includes two practice problems to calculate determinants. The final problems involve applying concepts like Kirchhoff's laws to solve for unknown currents in an electric circuit and traffic flows at intersections.

1. Additional examples &
problems
GE 111; Topics 11-12
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2. Determinant of 2-by-2 Matrix
a b 
A 

c d 
(a+c,b+d)
(a,b)
det(A) = ad – bc
if these are real numbers,
then A can be mapped into
area of the parallelogram.
(c,d)
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
2
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3. Determinant of 3-by-3 Matrix
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.
300
020
0 0 1 |det(A)| = |6|
320
121
0 2 1 |det(A)| = |-2| = 2
For more explanation and visualization see: MIT OCW
http://ocw.mit.edu/ans7870/18/18.013a/textbook/HTML/chapter04/section01.html
3
Diagram from C. Rocchini. Permission for reuse by GNU Free Documentation License
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http://en.wikipedia.org/wiki/Determinant

4. Example1: determinant of a 4x4
3 0 2 -1
-
+
-
+
-
+
1 2 0 -2
-
+
-
+
-
+
-
4 0 6 -3
+
-
+
-
+
-
+
5 0 2
A=
+
-
+
- +
0
First do cofactor expansion along 1 st row developing 3x3 matrices
2 0 -2
1 0 -2
1 2 -2
1 2 0
det (A) = 3 0 6 -3 - 0 4 6 -3 + 2 4 0 -3 - (-1) 4 0 6
0 2 0
5 2 0
5 0 0
5 0 2
Second do cofactor expansion along 1 st row of each 3x3 developing 2x2 determinants)
=3 2
6 -3
2
0
-0
0 -3
0
0
+ (-2)
0
6
0
2
- 0 +2 1
0 -3
0
0
...
det (A) = 20
4
Example from http://nebula.deanza.edu/~bloom/math43/Determinant4x4Matrix.pdf
GE 111 Engineering Problem Solving
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5. Example1: determinant of a 4x4
3 0 2 -1
-
+
-
+
-
+
1 2 0 -2
-
+
-
+
-
+
-
4 0 6 -3
+
-
+
-
+
-
+
5 0 2
A=
+
-
+
- +
0
Do cofactor expansion along column 2 developing 3x3 matrices
1 0 -2
3 2 -1
3 2 -1
3 2 -1
det (A) = -0 4 6 -3 + 2 4 6 -3 - 0 1 0 -2
5 2 0
5 2 0
5 2 0
+ 0 1 0 -2
4 0
6
Do cofactor expansion along row 3 of remaining 3x3 developing 2x2 determinants)
=2 5
2 -1
6 -3
-2
3 -1
4 -3
+0
3
2
4
6
det (A) = 20
5
Example from http://nebula.deanza.edu/~bloom/math43/Determinant4x4Matrix.pdf
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6. Example2: determinant of a 4x4
2
-2
A= 1
-1
5
-3
3
-6
-3
2
-2
4
-2
-5
2
3
0
0
1
0
-7
3
3
-3
5
-2
-2
2
4
9
2
5
2R4+R1
2R3+R2
R3+R4
Pick a pivot point wish to do cofactor
expansion along that can create ‘0’s
along a row or column; e.g. a31
-7 5 4
+1 3 -2 9
R2+R3 -3 2 5
-7 5 4
+1 3 -2 9
0 0 14
-7 5
det (A) = 1 (14
) = 1(14 (14-15) = -14
3 -2
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7. Problems 1&2 : Find determinants
6
2
A= 1
3
-1
2
1
1
0
-1
1
1
2
2
-3
6
0
B= 0
0
0
2
1
0
0
0
1
1
2
0
0
7
0
-2
-2
3
4
0
-2
-2
3
0
5
1
3
-1
2
5
1
3
-1
2
det (A) = 34
det (B) = 72
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8. Problem 3: A Traffic Light Assembly
C
Z
6m
D
A
3m
X
h
4m
B
4m
4m
6m
3m
4m
3m
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Y
Nov 2013

9. 3: A Traffic Light Assembly


9
The mass of the traffic light = 20 kg
h = 3.5 m. For equilibrium:
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10. 3: A Traffic Light Assembly
 0.597  0.8381 0.796 
A   0.796  0.4191  0.597


0.0995 0.3492 0.0995 


 0 
B 0 


196.2


det( A)  0.4702
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11. 3: A Traffic Light Assembly
 0.597  0.8381 0.796 
A   0.796  0.4191  0.597


0.0995 0.3492 0.0995 


FAB
 0 
B 0 


196.2


0
 0.8381 0.796
0
 0.4191  0.597
196.2 0.3492 0.0995 163.62


det( A)
0.4702
FAC
det( A)  0.4702
= 348.0
= 413.1
FAD = 174.0
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12. Problem 4: An Electric Circuit
Given the circuit in
the adjacent image,
what is the current
(ampere) for I1, I2,
and I3?
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)
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.
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13. Problem 4: An Electric Circuit
20I1  12I 3  80
15I 2  12 I 3  120
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.
I 3  I1  I 2
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14. Problem 4: An Electric Circuit
20I1  12I 3  80
20I1  12I 3  80
15I 2  12 I 3  120
15I 2  12 I 3  120
I 3  I1  I 2  0
I 3  I1  I 2
20 0 12
0 15 12
-1 -1 1
I1
I2
I3
80
= 120
0
I1 = 1, I2 = 4, I3 = 5
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15. Problem 5: Traffic Flow
15
800
600
Ruth Street
x2
x1
Tait Street
D
700
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x4
B
Preston Ave.
A
Clarence Ave.
Rush-hour traffic along one-way
Streets is shown in adjacent figure.
The numbers represent traffic flow
in vehicles per hour that enter and
400
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
400
allowed).
• All the cars that enter any one
intersection leave (no
breakdowns).
• Cars cannot go in reverse (all
flow must be nonnegative)
300
x3
C
700
500
Nov 2013

16. Problem 5: Traffic Flow
Flow through intersections and on streets between each
intersection (Vph)
Intersection
A
Flow
x1
B
x2
C
D
16
x2
1000
x3
x3
x1
1100
x4
1200
x4
1100
GE 111 Engineering Problem Solving
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17. Problem 5: Traffic Flow
Gauss-Jordan Elimination and create one column of x4’s
1
0
0
1
1
1
0
0
0
1
1
0
0
0
1
1
1
0
0
-R1+R4 0
1000
1100
1200
1100
0
1
0
0
1
-R3+R2 0
0
1
0 1 1100
0 -1 -100
1 1 1200
0 0
0
1
1
0
0
0 0
0 -1
1 1
0 1
1000
-100
1200
1100
x1 + x4 = 1100
x2 – x4 = -100
x3 + x4 = 1200
-R2+R1
1
0
0
1
0
1
0
0
0 1
0 -1
1 1
0 1
1100
-100
1200
1100
x1 = -x4 + 1100
x2 = x4 - 100
x3 = -x4 + 1200
As traffic flow must have nonnegative values:
100 ≤ x4 ≤ 1100
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18. Problem 5: Traffic Flow
800
600
Ruth Street
400
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?
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GE 111 Engineering Problem Solving
x2
B
x1
Tait Street
D
700
x4
Preston Ave.
A
400
Clarence Ave.
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.
300
x3
C
500
x1 = -x4 + 1100
x2 = x4 - 100
x3 = -x4 + 1200
Nov 2013
700