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IDENTITY AND INVERSE MATRICES
Lesson 4.7
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Identity Matrix:
If you multiply a matrix by an identity matrix (I) the result is the same as the original matrix.
If Matrix A is a square matrix then it is also commutative. A I I A
A I A
_ _
_ _
a b a b
c d c d
1 0
0 1
a b a b
c d c d
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0 0 00 0
0 0 0 0 or 0
11
1 11
1 11
0 or 0 0 0 0
0
2x2 3x3
00 1
4x4
0 0
•Ones on the principal diagonal and all the rest are zeros.
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Inverse Matrix
If two inverse matrices are multiplied the resulting product is the Identity Matrix.
1A A I ? ? 1 0
? ? 0 1
a b
c d
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For a 2x2 inverse matrix
1If then 1
det
d bM
c a
a bM
c d M
1 9 41
5 218 2
2 4If , then
9 05MM
For example:
921
52
29 41 or
15 22M
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What about inverse matrices of other sizes?
Use calculator for now. If you go on in math, you will learn that
in Linear Algebra.
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Using Inverse Matrices for Encryption and Decryption
Basic phrase: Bulldogs Translate to a numerical string where A=1,
B=2, etc.
2 21 12 12 4 15 7 19 Put this string into a matrix,
maybe a 4x2 2 21
12 12
4 15
7 19
[ 0 N 14
A 1 O 15
B 2 P 16
C 3 Q 17
D 4 R 18
E 5 S 19
F 6 T 20
G 7 U 21
H 8 V 22
I 9 W 23
J 10 X 24
K 11 Y 25
L 12 Z 26
M 13
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Multiply this matrix by another random matrix, in this case, if we do right hand multiplication, a simple 2x2 would work.
For example. 2 1
1 3
2 21
12 12 2 1
4 15 1 3
7 19
17 65
12 48results in
7 49
5 64
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This matrix is converted back to a string of numbers.
-17 65 12 48 -7 49 5 34
17 65
12 48
7 49
5 64
To get back to the original message, the receiver would then use the inverse of our encryption matrix to get it back.
3 17 7
1 27 7
17 65
12 48
7 49
5 64
2 21
12 12 to get back to
4 15
7 19
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[ 0 N 14
A 1 O 15
B 2 P 16
C 3 Q 17
D 4 R 18
E 5 S 19
F 6 T 20
G 7 U 21
H 8 V 22
I 9 W 23
J 10 X 24
K 11 Y 25
L 12 Z 26
M 13
2 21
12 12
4 15
7 19
Which translates back to:
Bulldogs
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Work Time 10 minutes.
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Using Matrix Equations to solve systems of equations4 2 17
3 4 10
x y
x y
Can be converted to a matrix equation, where the variables are in their own matrix, like this:
4 2 17
3 4 10
x
y
xA B
y
Coefficient matrix ∙ variable matrix = constant matrix
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4 2 17
3 4 10
x
y
xA B
y
Use inverse matrices to solve:
Multiply both sides by A-1 because A-1A gives the identity.
1 1
I
xA B
yA A
12
4x
yI
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Try this one:
•First, turn it into a matrix equation
•Then label A and B
•Then multiply both sides by the inverse of A. (since it is a 3x3, use calculator to do calculation.)
•Make sure you check your solution.
5 7 3 2
2 5 5 21
4 2 13 5
x y z
x y z
x y z
5 7 3 2
2 5 5 21
4 2 13 5
x
y
z
1 1
x
A A y A B
z
A B
3
2
1
x
y
z
(3, 2,1)