Determinants (Part - 1) - Mathematics, Engineering Video Lecture - Engineering Mathematics

good morning viewers today's topic is
determinants I will start this topic by
giving you first the basic concepts
regarding determinants I will introduce
the concept and then we will talk about
the properties of determinants to start
with we'll consider a system of two
equations in two variables x and y the
two equations are a 1 X plus B 1 Y is
equal to C 1 a 2 X plus B 2 y is equal
to C 2 these two equations can be
written in the determinant form as we
have discussed in my earlier lectures as
X is equal to B where a is a 2 by 2
matrix and XY is a column vector while
on the right hand side we have the
column matrix C 1 C 2 we know how to
solve these two equations from our
school days Algebra one can very easily
obtain X is equal to C 1 B 2 minus C 2 B
1 divided by a 1 B 2 minus a 2 B 1 while
Y can be simplified as a 1 C 2 minus a 2
B 1 divided by a 1 B 2 minus a 2 B 1 now
we try to represent the expression A 1 B
2 minus a 2 B 1 in this form and on the
right hand side what I have written is
called a determinant consisting of the
matrix a 1 B 1 in the first row and a 2
B 2 in the second row this is enclosed
by pair of vertical lines because this
is a determinant this determinant is
equal to the left hand side because one
can very easily check that this left
hand side is equal to a 1 B 2 this
product minus the product B 1 a 2
similarly one can write down the
expression C 1 B 2 minus B 1 C 2 is
equal to C 1 B 2 minus B 1 C 2
then the expression X which we have
obtained in my earlier slide the
expression X is equal to C 1 B 2 minus C
2 B 1 divided by a 1 B 2 minus a 2 B 1
as obtained in my earlier slide is this
can be written as the matrix C 1 B 1 C 2
B 2 then we have encode enclosed in
vertical lines divided by the matrix a 1
B 1 a 2 B 2 vertical lines we call this
as a determinant the expression X is
equal to C 1 B 2 minus C 2 B 1 divided
by a 1 B 2 minus a 2 B 1 can be written
as X is equal to C 1 B 1 divided C 2 B 2
divided by a 1 B 1 a 2 B 2 now here we
have used the notation which we have
used here we call this as a determinant
of the matrix a 1 B 1 a 2 B 2 using this
notation we write down why in this
particular form we have already written
this expression as we have already eaten
this expression as this here similarly
we can obtain a 1 C 2 C 1 minus C 1 a 2
as this determinant now given this
matrix a as a 1 1 a 1 2 in the first row
a 2 1 a 2 2 in the second row one can
define the determinant of the matrix as
or we denote it by determinant a we can
define it as a 1 1 a 2 2 minus a 1 2 A 2
1 this is a 1 1 a 2 2 the first term and
a 1 2 A 2 1 the second term with a minus
sign in between now the example is if I
have given a matrix 1 7 2 8 then its
determinant can be written as this which
can be computed as 1 into 8 is 8 minus 7
2 7 into 2 is 14 and the result is minus
6
now this is the way we define matrices
for this the way we define determinants
for two by two matrices but to define
determinants in more general form I will
now give you some basic concepts first
thing is to define a permutation a
permutation is 1 but 1 to 1 on 2 mapping
from the set 1 to 3 n onto itself for
example if I have been given a set of
first n natural numbers then a Sigma is
denoted as a permutation can be defined
as this here the number 1 is nothing to
J 1 2 is mapping to J 2 and n is mapping
to Jane what are J 1 J 2 J they are
simply the rearrangement of numbers 1 2
2 in this is one one and onto mapping so
more member will be repeated by writing
down J 1 J 2 J and no number will be
dropped off we can write down the
determinant we can write down the
permutation in a form like Sigma I is
equal to J I where ji lies between 1 to
N and I lies between 1 and N or the
other way of writing this permutation as
Sigma is equal to J 1 J 2 2 J n so we
have understood that permutation is
simply a rearrangement of numbers for
example if we consider the set of first
T natural numbers 1 2 3 then 1 3 2 is a
permutation on this set
1 2 3 we will denote this permutation as
Sigma is equal to 1 2 3 this is the set
and 1 is mapping 2 1 2 is mapping 2 3 3
is mapping to 2 so this way for a given
set of 3 natural numbers there will be 6
permutations possible and they are 1 2 3
the first combinate for permutation then
the next may be 1 3 2 next 2 3 1 2 1 3 3
1 2 & 3 2 1 so in all there are 6
permutation possible so if SN is a set
of all permutations then we say
that there are factorial n permutations
possible on the given set of n numbers
we further define an ordered pair PQ of
distinct integers P and Q then Q is said
to be an inversion if P is greater than
Q so if we consider the permutation
Sigma is equal to J 1 J 2 J n then for
given Sigma we consider the set Phi
Sigma which is an or set of ordered
pairs J 1 then we can say j2 j1 j1 next
number J 3 next is next ordered pair is
j1 J n then we start with J 2 J 2 J 3 J
3 J 2 J 4 if you go up to J 2 J N and
finally we'll have J n minus 1 and Jane
so we call this set as Pi Sigma now in
this set we consider how many inversions
are there let us say we consider this
ordered pair 1 greater than 2 not true
so it is not an inversion 1 greater than
3 is not an inversion the number of
inversions in the set is called the
number of inversions of the permutations
I think I will extend this with an
example
next slide a permutation is Sigma is
even if it has even number of inversions
and we call the permutation is odd if it
has odd number of inversions so for this
let us consider the permutation 3 2 4 1
on the set of first four natural numbers
on this permutation we define the set
Phi Sigma as 3 2 3 4 3 1 these are the
first three then we start with 2 2 4 2 1
these are the ordered pairs then we
start with 4 4 and 1 so in all we are
having 6 ordered pairs in Phi Sigma now
if we consider the number of inversions
in this then 3 2 is an inversion because
P is greater than Q 3 4 is not an
inversion 3 1 is an inversion because 3
is greater than 1 similarly we will have
2
two one and four one are inversions so
in this set we are in all having four
inversions and consequently the above
permutation is an even permutation now
we consider a square matrix of order N
and we try to form the product of M
numbers then these numbers this pie is
denoting the product PI a IJ i we write
it as a 1 a 1 a 2 J 2 J 3 J 3 and a and
J n now if we consider for example Sigma
as this permutation then this product I
is equal to 1 to 4 a IJ I is will be a 1
1 a 2 4 8 3 2 and a 4 3 so if we
consider this product and consider the
square matrix then this product
basically means that we are considering
this term multiplied by this term this
term and this term now one can very
easily observe that in by formulating
this product we are considering only one
term from a corresponding row and column
so in the first row and column will have
only this term from the second row we
are having this term from the third row
we are cancelling this term and the
fourth row I am casting distant so time
conceived this term or other column is
selected
Moke other column element is selected if
I am consider stem no other column
element is selected so I am selecting
one down from each column and row and
forming a product so this is one product
but actually there will be factorial
three pictorial for such products
possible so if you are having M Wells
square matrix then they will be exactly
factorial n such products now to each
such product if we assign plus or minus
sign according as the permutation is
even or odd then the algebraic sum of
all these product is defined as a
determinant of the matrix now this way
we define determinant a as summation
Sigma over the set of
permutations signal Sigma determines the
sign and this is the product
this signal Sigma is defined as minus
one if Sigma is odd and it is one if
Sigma is even now let us see how what is
the meaning of this in context with a
two by two matrix for which we have
already introduced the determinant so if
we give this matrix 2 by 2 matrix then
its determinant is a11 a.22 minus a12
a21 now one can see that if we consider
this a 1 1 then we cannot test of this
element and this element only this
product is possible so we have a11 a.22
minus the second level I am consider
this element and this element if I am
something then these two elements cannot
be considered so we have only option
left with a 1 1 so we'll have a 1 to a 2
1 now this is coming with negative sign
because this combination this 1 2 to 1
this permutation is an odd permutation
so will have a negative sign so we'll be
applying this definition we are forming
product of two terms and then this B
uses concept and what we have is the
determinant of this 2 by 2 matrix so
this is what we have already defined you
are the similar concept is used for
defining determinant of third order
matrix so if I have a third order matrix
as a11 a12 a13 a21 a.22 a23 a31 a32 a33
then we can denote it by this symbol
where a is the 3 by 3 matrix or I can
write it over Tremblant a or I can write
it like this
now this is a 3 by 3 matrix so there
will be factorial 3 terms possible so
total are 6 terms which are possible and
each term is actually a product of three
terms a 1 1 a 2 2 a 3 and all these are
different every time I am searching only
one term from a particular row and
column this one can check in each and
every term now but if we consider these
two terms then one can very easily see
that a1 1
can is common from these two terms and
what remains is a 2 to a 3 3 minus a 2 3
a 3 2 and these two can be written down
in the form of a determinant this is a 2
2 a 3 3 minus a 2 3 a 3 2 similarly one
can write down in a 1 2 can be taken out
from these two terms and a 1 3 can be
taken out from these two terms and then
determinant a can be written in this
particular form
I have taken the terms a 1 1 a 1 2 A 1 3
outside and then I am multiplying by a 2
by 2 determinant now this a 1 1 a 1 2 A
1 3 happens to be the elements in the
first row so that means I am defining
the determinant in terms of first row
but this is not all we can define
determinant in terms of any rows or
columns like what we can see in the next
slide so if we start with determinant a
is equal to this then we can rearrange
the terms in a different manner yes so
that we can now have a 1 1 a 2 1 and a 3
1 and they are multiplied by
corresponding 2 by 2 determinants so
this is not all one can take out one can
expand in terms of any rows or any
columns this we have already discussed
now we will take the example so if we
have to obtain the determinant of this 3
by 3 matrix then I am expanding about
the first row then 1 is taken out then
we talk about we evaluate this
determinant 1 1 - 3 2 - this is the
negative sign then we will talk about 2
1 product minus 3 1 so the next term is
4 multiplied by 4 minus 1 and the final
result is 11
what we have done for two-by-two matrix
can we do it for a 3x3 matrix for this
let us consider the matrix 3x3 matrix
a11 a12 a13 a21 a.22 a23 a31 a32 a33 to
evaluate this determinant what one can
do is can repeat this first column here
and can repeat this second column here
then form the product a 1 1 a 2 2 a 3 3
a 1 2 a 2 3 a 3 1 and a 1 3 a 2 1 a 3 2
these products are written down here and
then we take the product of these terms
like a 1 3 a 2 3 a 3 1 a 1 1 a 2 3 a 3 2
and a 1 2 a 2 1 a 3 3 and these terms
are written with a negative sign that is
how you evaluate a 3 by 3 determinant so
if we illustrate this with this example
then I have copied down 1 2 3 here and
this column is copied down here then I
obtain this product this product and
this product minus this product this
product and this product and finally
what I get is this 236 4 6 4 and 12 this
comes out to be 20 now this is the way
we are if we have evaluated third order
determinant but if you keep on applying
this method for evaluating determinants
then things will become very tedious so
we can evaluate higher other
determinants in terms of low order
determinates this also I have explained
with 3 by 3 determinant but now I will
give you more basic concepts related to
this
so for determinants of higher order
matrices we have a general formula
involving determinants of lower order
matrices this is called - now Muller is
defined as m IJ it is denoted as
my J it is a sub is a sub matrix of
order n minus 1 it is a square sub
matrix of order n minus 1 which is
obtained from a by deleting the eyath
row that is this is row and jth column
the determinant of the sub-matrix m IJ
is called the minor of the element a IJ
of E for example if I have this 3 by 3
matrix then the determinant of a 1 1 if
we denoted by M 1 1 then it is obtained
by the determinant of this 2x2 square
sub-matrix while M 2 2 m 2 2 is this
element is the determine is the minor of
this element and this - cymatics is
obtained by deleting this row and this
column and what we have is this minor
now ml 1 this is some matrix M 1 1 and M
2 2 our sub matrices and then one is a
minor of M 1 1 which is the determinant
of this while M 2 2 is the determinant
of this sub matrix a 1 1 a 1 3 a 3 1 and
a 3 3 so for given matrix a the
determinant a is this and it is computed
as a 1 1 multiplied by a.22 a23 a32 a33
that is this minus a 2 1 this multiplied
by a second order determinate and then a
3 1 multiplied by this actually this
formula has already been obtained in my
earlier slides using this concept one
can write down determinant a as a 1 1
miler m 1 1 minus a 1 2 multiplied by m
1 2 plus a 1 3 multiplied by m 1 3 this
minus n plus this minus m plus is coming
because we have to have alternate size
and this formula can be used for these
signs now we say a 1 J M 1 J the sign
which is to be taken is minus 1 raised
to power J plus I and this is when I
expand this around the first row but as
I told you this can be done for each and
every row or column so if I take any row
from 1 2 3 then this formula can be used
for evaluating determinants and this
formula can be used for evaluating
determinants when I am considering
columns now I have introduced a concept
of fineness but the problem with minuses
one has to remember the sign minus plus
which term is to be taken with minus
sign which term is to be taken by with
the sign so what I do is what we do is
we introduce another term called as
cofactor associated with the element a
IJ we denoted by a IJ it is defined as
AI J is equal to minus 1 raised to power
I plus J determinant m IJ so that minus
1 I plus J is taken with the cofactor
and in fact we do not have to remember
this formula one can use this jacquard
pattern to associate sign with
appropriate m IJ the first M start with
plus the next term is minus then plus
minus and so on so this checkered
pattern can be used for evaluating
determinants using this one can write
down the determinant a as a 1 1 a 1 1
plus a 1 2 A 1 2 A 1 3 a 1 3 here a 1 1
is the cofactor of a 1 1 a 1 2 is the
cofactor of a 1 2 A 1 3 is a cofactor of
a 1 3 so in short we can write down this
expression as summation J is equal to 1
2 3 a 1 J and J this is cofactor then
determinant a he'll also be written in
this I can take any value from 1 2 3
that means it can be expanded about any
row and this means J is taking values
from 1 2 3 that means it can be expanded
about any column for general square
matrix of order n the formula can be
written in this particular form where I
will take values from 1 to N and J will
take values from 1 to n this is what we
have obtained earlier and determinant a
is equal to Sigma belonging to s n so
that is how this definition is being
derived from the definition which I have
introduced in the beginning now
illustrate this with the help of an
example let us say a be a matrix such
that it is each row and each column is
exactly one nonzero element then its
determinant is nonzero see if I am
taking here a 4 by 4 determinant then
every row or column has only one entry
nonzero so when you form this
determinant and form a factorial four
terms then you will notice that only one
term will remain rest of the term will
be this other terms will be 0 and what
is that term that term is a 1 3 a 1 3 a
2 2 this term a 3 1 this term and a 4 4
this term Rezo terms will be 0 and we
are we are putting this with negative
sign because this is of order this is of
inversion 1 so it is an odd inversion
that is why we are having a negative
sign so what is the determinant of this
determine this matrix that is 1 into 2
into 3 into 1 and the final result is 6
now after defining the determinants I
will discuss the properties of
determinants these properties are
important because they can be used to
evaluate this this can be used to
compute the determinant in a convenient
manner let us see the properties one by
one if all the elements in a row of the
matrix a
zero then determinant a is zero now
first let me give you the proof of this
determinant a as I told you can be
written as J is equal to 1 to N a KJ a
KJ
this a KJ is a cofactor of a KJ now if
all the elements in a row of the matrix
a is 0 then you can use that particular
row for expanding the determinant so let
us say let us say K is the row in which
will have all elements 0 then we can use
this definition since a KJ is 0 so for
all values of J 2 J 1 to n so this
everything will be 0 for example if we
have this determinant in which the third
row has all elements 0 then the final
result will be 0 one can very easily
check we start with 0 whatever be the
determinant this term will not
contribute same thing will happen with
this this multiplied by its cofactor
will not contribute anything this
multiplied by its cofactor will also not
contribute anything and that is what we
are having 0 now this if I have given
then the row of a matrix has all zero
elements but it may happen that a column
of a matrix has all zero elements the
proof will be on the same line and the
result will be that if all the elements
in a column of a matrix a is 0 then
determinant a is 0 now the next property
says that if the given matrix a has two
or more identical rows then still the
determinant a will be 0 now to prove
this result I will start with a 2 by 2
matrix and in this case a to 2 and the
first row is identical with the second
row so a 2 2 a 2 3 minus a 2 3 a 2 2
this is 0 now this can this concept can
be generalized for Emeth order
determinate because m8 order determinant
can be retained terms of n minus 1 1 is
order determinate
this n minus one eighth order
determinant will be written in terms of
n minus two with order determine it
ultimately everything will be written in
terms of second order determinate and
since all the secondary determinants are
zero here so the final result will be
zero so all second order - are zero in
this case and all higher order - are
also zero because of this and that is
how the result is proved for example in
this particular matrix the second row
and third row are identical and that is
why if you evaluate it the final result
will be zero the mixed property says
that if the given matrix a has two or
more identical columns then determinant
a is zero so earlier we have proved with
rows and this we have proved for columns
the proof will go exactly on the same
lines as we have done earlier now this
property says that matrix a is
multiplied by a scalar K to get the
matrix B then determinant B is equal to
K determinant a for example let us say
we have a matrix a with elements a IJ if
in the Emmet row we have K times a MJ
for all values of 1 to n that means that
if a row of a matrix a is multiplied by
a scalar K so let us say I am
multiplying emeth row of the matrix by K
and I am getting a new matrix b MJ then
determinant B will be K times
determinant a the proof of this is
simple I am adding determinant B by
definition it is J is equal to 1 to n
summation BM j and its cofactor BM j is
equal to now for BM j I have K times a
mg so this k AMJ into am j now k can be
taken outside so what remains is am j
into am j and that is the determinant a
so determining B is K times
dominant II so this is the result is
fruit and we Lestrade which deserve by
this example where we are having a
matrix 1 2 3 3 2 1 4 2 6 3 by 3 matrix
now this particular row is 2 times this
particular row so if I take 2 outside
then this row becomes 2 1 3 and this is
the same as 2 this is the same as if
this Row is multiplied by 2 so it is 2 4
6 3 2 1 2 1 3 so this determinant is
equal to this determinant now in the
next property we discuss what happens
when the two rows of a matrix are
identical what happens to is determinant
so if two rows of a matrix are
proportional then determinant a is equal
to 0 now to prove this we assume that ma
throw an eye at row of the given matrix
a are proportional that means a and J is
equal to K IJ for all values of J
between 1 to n then we one can find
admit one can determine a matrix B which
is same as B IJ is equal to AI J for all
values of I am not equal to M but it
means all the rows of B and a are
identical except the M F through in the
a Metro of the matrix B is given by BM j
is equal to AI J that means it is
identical with the IJ through now that
means one can take K outside from the B
matrix and what we have is determinant a
is equal to K times determinant B this
is a result which we have proved in my
earlier property now since determinant B
has two identical rows namely a metro
and IA throw that means determinant B is
0 so IMM rows are identical in B so
determinant B is 0 and that means
determinant a is
also zero now look at this example here
we are having one two three three two
one two four six that is the first row
and the third row they are proportional
so one can write down it is two times
this so this is my B matrix now in this
D matrix one two three and one two three
in the first row and the third row are
identical so this determinant is zero
and that proves that this 3 by 3
determinant has value 0 now if a matrix
C is multiplied by a constant K that
means each and every row of the matrix a
is multiplied by K then the terminal C
is equal to K raised to power n
determinant a where n is the order of
the matrix the idea is each row is from
each row one can take K outside and that
way there we in such case taken out and
the final result is K determinant C is
equal to KN determinant e now if a row
of a matrix a is multiplied by a scalar
K to get the matrix B then determinant B
is equal to K determinant a this is what
we have proved in my earlier result and
if you apply it for each and every row
then we will be having this now the
properties which we have discussed so
far they are applicable to rows
we are proved in four rows but they are
applicable to columns also so all the
properties are applicable to rows as
well as columns now if the matrix B is
obtained by interchanging two rows
namely R in s of the matrix a then
determining B is negative of determinant
a that means when you interchange two
rows of a matrix then this minus sign
will come in between now the proof of
this back
the proof of this thing can be proved
very easily from this because if you are
changing two rows then actually the sire
signal Sigma will change the inversion
will change and one minus sign will be
coming because of this that is why
determinant B is equal to minus
determinant a now in the next set of
properties we will see the effect of
elementary transformational Amenti
operations on matrices and we will see
what is their effect on determinants so
we have first elementary operation where
we interchange you throw with jethro of
the matrix now if we change our throw
and jethro we denoted by determinant a
RI and RJ are interchange this is used
as a suffix then this is equal to minus
times a this I have proved in in my
earlier slide now if you replace I a
throw by K times I a throw then ki will
go to RI and in that case if we use our
earlier results then determinant of a by
interchanging k RI to RI is equal to
determinant a divided by K I will
explain this with examples the third
elementary operation is when we replace
row J by a nonzero K times Rho plus Rho
J so k RI plus RJ will be replaced by RJ
that means determinant of a after
replacing this by RJ is equal to a so
let me explain this with example so let
us say I have a matrix a as ABC in the
first row rst in the second row u VW in
the third row and its value is 2 let us
say this the ABC R such that this
determinant is equal to 2 then ABC UVW
rst this
is actually obtained from this
determinant by interchanging the second
row and third row so because we are
changing two rows so the minus sign will
come since this cell if this determinant
is two so this determinant will be minus
two
now all these elementary row operations
are all they all the results related
with rows are applicable to columns also
so if I write down this determinant as
three ABC 3 R st 3 u VW that simply
means that I am multiplying this column
by 3 and I am getting this this 3 times
a 3 times this is this actually related
with this not with this so I have to
have to change it it is not with this I
have to show this reference this
determinant is a pair this determined is
obtained from this determinant by
multiplying this column by 3 so this
column is 3 a 3 R 3 you wrestle things
remain the same be s v c t w since this
is this is multiplied by 3 so the effect
of this is that it is 3 times 2 and the
final result is 6 now this matrix is
obtained by replacing the second column
as B plus 2 C s plus 2 T V Plus 2 W this
is actually obtained by adding this plus
2 times this column so it is the second
column C 2 multiplied by 2 times third
column so each element of second row
built second column will become B plus 2
C s plus 2 T V Plus 2 W and the result
will not be change it will remain as 2
only so this determinant and this
determinant both have value 2
now the value of a determine remain
unaltered if roles are written as
columns or vice versa because all the
properties which we have discussed so
far they are applicable to rows as well
as to columns and with this we have
another property that is determinant a
is equal to determinant of a transpose
because the transpose a is actually
obtained from the matrix a by writing
rows as columns so that is why
determinant a is equal to determinant of
a transpose the things we will progress
simplified if we talk about some special
matrices like diagonal matrix now if we
have a diagonal matrix a having elements
as c1 c2 CN a diagonal matrix a means it
has only the diagonal elements nonzero
rest of the elements are zero so the
determinant of a in that case will be c1
c2 CN because this is actually true from
the result which we have discussed in my
first property so if you have this
diagonal element diagonal matrix then
its determinant will be 1 2 & 3 that is
6 so we have to simply multiply the
diagonal elements and that will be the
value of determinant
now this result can be generalized to a
triangular matrix so if we have a
triangular matrix with main diagonal
elements as C 1 C 2 C n then the
determinant is obtained as the product
of diagonal elements let us see that is
proof for this so let us consider this
triangular matrix this is actually an
upper triangular matrix in which all the
elements shown on the upper side of the
diagonal they are nonzero while all the
elements on the lower side of the
diagonal are 0 so if this determinant is
to be evaluated this this determinant
comes out to be the product of die
elements the proof is simple what we
have to do is we'll consider this Ava
will we can expand this determinant
along this but first column so we'll be
having a 1 1 multiplied by this minor
plus 0 multiplied by its corresponding
minor will not contribute anything 0
multiplied by its minor will not
contribute anything 0 multiplied by this
minor will not contribute anything so we
will have only a 1 1 multiplied by this
minor now for this minor we evaluate
along this particular column so a 2 2
multiplied by this minor will contribute
but rest of the terms will not
contribute similarly we'll go to the
next lower lower order minor and this
way will continue and finally we will
have only this term left with us so what
we have is the product a 1 1 a 2 2 NL is
the value of this lower triangular
matrix now this is the case when we are
having lower triangular matrix if we
will have an upper triangular matrix in
which only the these terms in the lower
diagonal they are nonzero and these
terms are 0 so the result can be proved
on the same very lines only thing is
instead of taking this particular Row
one has to take this particular column
in that case and determinant a will be
simply the product of diagonal terms so
whether we have a triangular matrix or
we have a diagonal matrix its
determinant will be suffer the top
diagonal terms now we have seen number
of properties for evaluating
determinants if you have to apply these
properties then things will be very
complicated especially when we have
higher order determinants because the
computational effort required will be
much more so what we do is we we can
apply basic definition but the lot of
computation effort will be required for
that purpose but if you use element of
elementary transformations to a given
matrix and simplify
to some suitable form and then one can
very easily evaluate the determinant
with least efforts so what we can do is
we can try to transform our matrix maybe
to our tried I try and triangular matrix
so that some effort is required to could
reduce it into triangular matrix but
then evaluation of determinant will be
simple because in that case you have to
simply get the product of diagonal terms
so that is the way to evaluate the
diagonals let me explain this with an
example
so let us take a given matrix is a if
you have to evaluate this 3x3 matrix
then what you have to do is one way may
be you just evaluate this multiplied by
this tag minor minus this multiplied by
its corresponding minor Plus this
multiplied by its corresponding minor
but that will require a lot of
computational effort now we will try to
make this matrix as a triangular matrix
so that we can get the determinant by
simply evaluating the product now for
this if we apply the we multiply the
first row by 3 that is this row is
multiplied by 3 so 3 into 6 into 12
since I am changing the row by a
multiple so these two will be equal if
will be having 1 by 3 on this side so
the after applying this operation simply
means 1 by 3 of this now since these two
are identical so what I can do is r2
minus r1 is the place by r2 so say
second row is the place by r2 minus r1
so this is my new second row this is
this minus this so it is 0 1 minus 6 is
minus 5 to minus 12 is minus 10 and
third row remains the same now on this
determinant I apply this transformation
this elementary
operation that is second row is minus 1
by 5 times r2 so if I apply this row
then since I am multiplying by minus 1
by 5 so minus 5 will appear here and
this second row will become 0 1 because
minus 5 has been taken out n minus 2 n
will have 2 here now for the third row
this row is changed here how I have
obtained this row I have multiplied this
row by 3 so we'll have 3 here 9 here and
12 here and since I am multiplying it by
3 so I divided by 3 here so the factor
is minus 5 by 9
so this matrix is obtained now again I
can apply this operation third row is
changed by R 3 minus R 1 because these
two rows are identical now so you can be
this element is identical so if I
subtract I will get 0 then 1 2 0 3 and 0
is close to an triangular matrix but
still there are certain terms which will
be not making it to be a diagonal
element so I write down as R 3 I am
interchanging 2 rows now so 1 & 2 will
be here and it is 0 3 0 3 0 is appearing
here since I am interchanging 2 rows so
this minus sign will be absorbed here
now it is 3 R 3 is replaced by R 3 that
means third row is replaced by 3 times
this so 3 times this means 3 means here
and 2 into 3 is 6 and since I am
multiplying by 3 so we'll be having PI
by 27 here so now these two becomes
identical soil again apply this
elementary transformation and what I
have is simply this this and this will
get cancelled so 0 3 0 will remain the
same but this 0 3 minus 3 0 6 minus 0 is
6 now this is now a triangular matrix
and its determinant will simply be the
product of these this
this term and this term and what I have
is this term now in the beginning it may
appear that we are doing lot of effort
in reducing a given matrix into a
triangular matrix ticket because lot of
elementary operations have to be done
but if for three mathematics it may be
true but when the number of as the size
of matrix is increasing then the effort
involved in reducing the triangular
matrix is much small as compared to
evaluating factorial n products and
simplifying the determinant so this
effort is much less as can be seen from
this 3 by 3 matrix now we can see that
if we evaluate the determinant by normal
definition which I have given you
earlier then this comes out to be 10 so
that result is the same so the
determinant evaluated by the usual
definition also comes out to be the same
as we have obtained in the case of
triangular reduction finally the
determinant of 3a is 3 into 3 because a
is a 3 by 3 matrix so the determinant of
a is known and since I am multiplying 3
Emmys I am multiplying each and every
row by 3 so this C raised to power 3
that is 27 into 10 so determinant of 3a
comes out to be 270 now in this example
we will try to evaluate this determinant
by using properties of the determinants
which we have discussed so far in this
example we are trying to show that the
value of this determinant is equal to B
minus a into C minus a into B minus C
now this determinant is evaluated using
elementary row operations will try to
reduce this determinant into a
triangular matrix now to reduce it into
a triangular matrix we apply a series of
elementary transformations the
transformation is that the second row is
the place by r2 minus r1 so here I am
writing the new second row it is r2
minus r1 of this so it is b square minus
a square B minus a and 1 minus 1 is 0
simultaneously I apply r3 minus r1 to
get new r3 so that means C square minus
a square C minus a and 0 so this matrix
is obtained after applying these two
elementary operations now on this I
apply these elementary operations the
idea is one can very easily see that B
minus a is can be taken outside from
this row and C minus ay can be taken
outside from this row so I say r2
divided by B minus a is r2 and r3
divided by C minus a is ir3 so the row
will not be if I apply these operations
on this matrix then what I get is B
minus a into C minus a is taken outside
what we have is B plus a from here 1 and
0 in the second row C plus a will remain
from the by c minus c plus a will be the
remainder of b square minus c a square
when we take C minus a outside then 1
and 0 now in this if you expand around
this column then it will be plus minus
and plus so will be having 1 into
determinant into minus this and what is
the minor it is B plus a multiplied by C
plus a so these two remains as such but
when you simplify it is B plus a minus C
plus a so will have B minus C so this is
evaluated as B minus a into C minus a
into B minus C
in the next example we try to verify
that this data this determinant is equal
to this determinant now for this we
again apply a series of operations C 3
plus C 1 is equal to C 1 so if I
multiply if I add the third column and
the first column then this may be out a
will be cancelled out B minus C plus B
will have only minus si si minus a plus
C will be having minus a now if you
compare this matrix with our final
objective one can note down that the
columns are actually interchanged so to
get the final result we can interchange
the column C 3 and C 1 so we apply this
transformation so C 3 becomes C 1 and C
1 becomes C 3 this way and when we apply
we get this result and since we are
interchanging two rows two columns here
so this minus sign and minus sign due to
this interchange will be absorbed and
what we have is the final result a 1 b b
1 c c 1 a which is the same as this
viewers with this we have arrived at the
end of this lecture on determinants in
this lecture I have defined determinants
I have started with the 2x2 map 2 by 2
determinant then I have introduced how
to evaluate a 3 by 3 determinant then we
have given a general definition for an
authority terminate we have discussed
some properties of the determinants and
using those properties we see how we can
apply elementary transformation and the
determinant will reduce to determinant
of a triangular matrix and then the
triangular matrix determinant can easily
be computed so that is all for today's
lecture q