Linear algebra has applications in balancing chemical equations, cryptography, and geometry. Balancing chemical equations involves setting up a linear system to satisfy the law of conservation of mass such that the number of atoms is equal on both sides of the equation. Cryptography uses linear algebra operations like matrix multiplication for encryption and decryption of messages. In geometry, finding the equation of a circle through three points involves setting up a homogeneous linear system and calculating the determinant to obtain the equation of the circle.

1. APPLICATIONS OF
LINEAR ALGEBRA

2. APPLICATIONS
 Balancing Chemical Equation
 Cryptography
 In Geometry
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3. Balancing Chemical Equation

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Application of linear systems to chemistry is balancing a
chemical equation and also finding the volume of substance.
The rationale behind this is the Law of conservation of mass
which states the following:
“Mass is neither created nor destroyed in any chemical
reaction. Therefore balancing of equations requires the same
number of atoms on both sides of a chemical
reaction. The mass of all the reactants (the substances going
into a reaction) must equal the mass of the products (the
substances produced by the reaction).”

5. As an example consider the following chemical equation
C2H6 + O2 → CO2 + H2O.
Balancing this chemical reaction means finding values of x, y,
z and t so that
the number of atoms of each element is the same on both
sides of the
equation:
xC2H6 + yO2 → zCO2 + tH2O.
This gives the following linear system:
2x = z
6x = 2t
2y = 2z + t

6. The general solution of the above system is:
y = 7x/2
z = 2x
t = 3x
Since we are looking for whole values of the variables x, y z,
and t, choose x=2
and get y=7, z= 4 and t=6. The balanced equation is then:
2C2H 6 + 7O2 → 4CO2 + 6H2O.

7. Cryptography

8. Cryptography, to most people, is concerned with keeping
communications private. Indeed, the protection of sensitive
communications has been the emphasis of cryptography
throughout much of its history. Encryption is the
transformation of data into some unreadable form. Its
purpose is to ensure privacy by keeping the information
hidden from anyone for whom it is not intended, even
those who can see the encrypted data. Decryption is the
reverse of encryption; it is the transformation of encrypted
data back into some intelligible form.

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Example Let the message be:
GIT BELAGAVI
And the Encoding matrix be:
-3 -3 -4
0 1 1
4 3 4
We assign a number for each letter of the alphabet. For
simplicity, let us associate each letter with its position in the
alphabet: A is 1, B is 2, and so on. Also, we assign the number
27 (remember we have only 26 letters in the alphabet) to a
space between two words.

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G I T * B E L A G A V I
Since we are using a 3 by 3 matrix, we break the
enumerated message above into a sequence of 3 by 1
vectors:
7 27 12 1
9 2 1 22
20 5 7 9

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We now encode the message by multiplying each of the
above vectors by the encoding matrix. This can be done by
writing the above vectors as columns of a matrix and
perform the matrix multiplication of that matrix with the
encoding matrix as follows:
-3 -3 -4 7 27 12 1
0 1 1 9 2 1 22
4 3 4 20 5 7 9

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which gives the matrix
-128 -107 -67 -105
29 7 8 31
135 134 79 106
To decode the message, the receiver uses the technique of
inverse of encoding matrix. The inverse of this encoding
matrix, the decoding matrix is:
1 0 1
4 4 3
-4 -4 -3

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Thus, to decode the message, perform the matrix
multiplication:
1 0 1 -128 -107 -67 -105
4 4 3 29 7 8 31
-4 -4 -3 135 134 79 106
And get the matrix
7 27 12 1
9 2 1 22
20 5 7 9

14. The Columns of this matrix, written in linear form, give
the original message:
G I T * B E L A G A V I

15. In Geometry

16. Given three points A1 =(x1, y1), A2 =(x2, y2) and A3 =(x3, y3)
in the plan (and not on the same line), find the equation of
the circle going through these points.

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If M =(x, y) is an arbitrary point on the circle, then we can
write
where a, b, c and d are constants. Substituting the three
points in the above equation gives the following
homogeneous system in four equations and four variables
a, b, c and d:

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The system has infinitely many solutions. So, the determinant:
to find the equation of the circle going through the points A1
(1, 0), A2 (-1, 2) and A3 (3, 1), we write

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Which gives after simplification
this can be written as
The circle has (7/6, 13/6) as center and 37/18 as radius.

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MADE BY NAVEENCHANDRA HALEMANI
nhalemani@gmail.com