Page 34 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Then aox = b and aox
1 2
aox = aox = b
1 2
x = x (by left cancellation law)
1 2
In a similar manner, we can prove that the equation

y o a = b
has the unique solution
–1
y = b o a .
Theorem 7: If corresponding to any element a G; there is an element O which satisfies one of
a
the conditions
a + O = a or O + a = a
a a
then it is necessary that O = o, where O is the identity element of the group.
a a
Proof: Since o is the identity element,
We have
a + o = a … (i)
also, it is given that

a + O = a … (ii)
a
Hence, from (i) and (ii)
a + O = a + O
a
or O = o (by left cancellation law)
a
Again, we have
o + a = a … (iii)
and O + a = a (given) … (iv)
a
Hence, from (iii) and (iv), we get

O + a = o + a
a
so that O = o (by right cancellation law.)
a
Modulo System

It is of common experience that railway time-table is fixed with the provision of 24 hours in a
day and night. When we say that a particular train is arriving at 15 hours, it implies that the train
will arrive at 3 p.m. according to our watch.
Thus all the timing starting from 12 to 23 hours correspond to one of 0, 1, 3… 11 o’clock as
indicated in watches. In other words all integers from 12 to 23 one equivalent to one or the other
of integers 0, 1, 2, 3, …, 11 with modulo 12. In saying like this the integers in question are divided
into 12 classes.
In the manner described above the integer could be divided into 2 classes, or 5 classes or m
(m being a positive integer) classes and then we would have written mod 2 or mod 5 or mod m.
This system of representing integers is called modulo system.

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