Determinants (Part - 2) - Mathematics, Engineering Video Lecture - Engineering Mathematics

good morning viewers I will be
continuing with the determinants as a
recap to my last lecture we define the
determinants first determinant of a
square matrix a is a number denoted by a
determinant a actually we associate a
number with a given matrix especially a
square matrix then we say the
determinant is uniquely associate some
number with a square matrix we have
discussed some schemes for computing
determinants of square matrices of order
two or three like if they have a second
order matrix then its determinant is
computed as a11 a.22 minus a12 a21 while
for a third order determinant is
calculated as for the top a 1 1 a 2 2 a
3 3 a 1 2 a 2 a 3 3 a 3 1 a 1 3 a 2 1 a
3 2 these are summed up and then we
calculate the sum of terms a 1 3 a 3 2 a
3 1 a 1 1 a 2 3 a 3 2 a 1 2 a 2 1 a 3 3
and take the difference but this type of
definition cannot be continued for high
order determinants so we define
determinants for NH order square matrix
as determinant a is equal to Signum of
the permutation Sigma and the product of
I is equal to 1 to N of a IJ I now this
is explained with a 4x4 matrix and we
say that we can calculate the product of
terms like a 1 1 from the first row
first column a 2 for from the second row
4th column a 3 2 from the third row
second column a for 3 from the fourth
row third column the idea is we'll take
only one term from each row and column
this way factorial n permutations are
possible we
calculate this product of all such terms
and then take the sum but then we define
determinant a as summation of minus 1
raised to power J + i AI J minor m IJ m
IJ this actually denotes the minor this
we have defined in the last term and
this definition can be applied to any
row or any column this way we define a
determinant in terms of lower order
metric determinants this is true for our
you throw this definition can be applied
to any row while this definition can be
applied to any column
the result is determinant a now we have
further defined cofactors a IJ for the
idea is here we are having minus 1 get
raised to Phi J plus I am I J so we
absorb this term with this minor and we
define a IJ is the cofactor of a IJ
element of the matrix a so we define
determinant a in this order in this in
this particular manner where I varies
from 1 to n and if we consider column
wise expansion then we can take this
definition and J varies from 1 to n
after that we have discussed some
properties and we have proved these
properties these properties actually
help us in evaluating determinants more
conveniently like if the elements of a
row are 0 then determinant a is 0 we
need not to compute all the terms which
we have discussed we simply say
determinant a 0 if any row or column it
becomes 0 similarly if the two rows are
identical then determinant a is 0 you
don't have to calculate all the - if two
rows are proportional then determinant a
is 0 and if two rows are multiplied by K
then if a row is multiplied by
then k times determinant a will be the
value of new determinant if in the
interchange of two rows occurs then
determinant is multiplied by minus 1 so
determinant a becomes minus determinant
a after interchanging of two rows
similarly we had applied elementary
transformations to corresponding
matrices and then we see what is the
relationship between the determinant of
the matrix before applying the operation
and after applying the operation so
before applying the operation that is
determinant a but when we change it the
determinant remains the same but if you
apply this elementary transformation
then determinant obtained after applying
this transformation is / K but this
elementary transformation will not
change the value of the determinant now
these transformations actually help us
in compute in evaluating the
determinants more easily and with less
computational a computational effort now
let us try to have some idea about the
computational effort involved in the
matrices because matrices are used
extensively in engineering and
scientific applications so let us have
some idea about the computational effort
involved in it
cofactor it if we use the cofactor
expansion and to evaluate the
determinant then the computational
effort is of the order of factorial in
this vector n is coming because we have
seen that for a determinant there are
factorial n product type of products
possible they have to be multiplied they
have to be added and subtracted so the
computational effort may be
computational effort employed will be of
order of factorial n now you will be
having a better idea if you try to see
it on a system when n is equal to 1 2
and so on so if n is equal to 5 and we
are evaluating a determinant on a
machine which can compute 10s for three
operations per second
then for n is equal to five the
computational time will be of order Oh
point zero zero zero three seconds okay
if we increase the matrix size to be 10
by 10 then the computational effort
required by the Machine will be 10
seconds and if you further increase the
size of the matrix to 15 then the
determinant will require 4 into 10 this
with 6 seconds of computer time and this
amounts to be 40 days if you further
increase the matrix size to 20 then the
huge effort is required and the computer
time will be 7 depends for 6 seconds
which comes out to be these many years
and if you further increase it say to 25
then we require approximately 10 lists
about 12 years to compute a matrix to
compute the determinant of a matrix of
size 25 now engineering applications
require large matrices to be handled and
we have to require we have to we have to
be care and computing determinants of
higher orders so if this is the scenario
then it will be difficult to use
determinants in practice so we will not
be using cofactor expansion rather then
we'll be using other method which is
called the triangular ization method for
computing determines let us first
compare let us see how the computation
effort in this case will be so if we in
this particular case the computation
effort is of order 2 raise to n cube
divided by 3 so this is the order of the
computational effort requiring computing
determinants using this method so this
is a table for giving the values for 20
because if you are using regularization
method then for M is equal to 20 the
computational effort will reduce to
point 0 0 5 seconds if you come
it with the earlier case then this is
very small and when we increase n to be
25 then 0.01 second will be required to
compute the determinant by this
particular method so although the method
is the same definition perfected
definition remains the same but triangle
ization method reduces the computation
effort tremendously for four or five
order matrices therefore it may not be
very may not be compared or if you are
doing hand computation then maybe we
find that regulation method is not so if
there is mercy but for higher order
matrices things are really more
comfortable when we use strangulation
method now after this we'll be
discussing some more properties of
determinants they will be helpful in
applying determinants and matrices to
engineering problems now for this let us
consider a matrix n by n square matrix
consisting of a 1 1 a 1 2 A 1 J a 1 n
elements in the first row a 2 1 a 2 2 a
2 J a 2 n in the second row this is the
eyath row of the matrix while this is
the last row this indicates that there
may be number of rows in between and
this also indicates there may be number
of columns in between here or here now
the matrix the determinant B is the
determinant of this matrix this matrix
is the same as this matrix the only
difference is in the jth column so a11
in the B matrix is the same as here so
always and columns are same except this
particular column so the determinant C
is the same as determinant a but the
difference is in the J through the JH
Row is having a 1j and v 1j the sum of
the two
in fact the jet column of a and J column
of B are added to give the jet column of
C matrix so if we define C matrix in
this particular manner then CI K is
equal to AI K is equal to BI K for all K
not equal to J and C IJ is equal to AI J
plus B IJ for jet column then we have a
result which says that determinant of C
is equal to determinant of a plus
determinant of B let us prove this
result we can expand the matrix about G
H column to get the determinant a so
this way we write down determinant AS
summation i is equal to 1 to n AI J
multiplied by a IJ e IJ being the
cofactor of a IJ that is this is the
matrix a so we have obtained determinant
a about this column same thing you can
be done for B now one can observe that
al J is equal to B IJ by because AI is a
the cofactor of a IJ element is obtained
by deleting this column and this row B
IJ is the cofactor in the B matrix by
eliminating this column and this row but
this is the only difference in a and B
so the determine this cofactor a IJ will
be equal to B IJ and this means a plus B
determined a plus determinant B is equal
to AI J plus B IJ this is taken and a IJ
is common so this is equal to AI j plus
b IJ multiplied by a IJ and this simply
means that determinant a by
we'll be is equal to this determinant
where the Jets element is the sum of the
two now this is nothing but what we have
that is the C matrix this simply means
the determinant of a plus determinant of
B is equal to determinant of C now it
gives me a feeling the determinant of F
is determinant of B will be equal to
determined of a plus B but actually it
is not because if I add two determinants
a and B the result will be 2 times a 1 1
2 times a 1 2 but this will be a 1 J
plus B 1 J and this will also be 2 a 1 n
so that means a plus B is not equal to
determinant of a plus B that means this
is this what do we mean by this this
simply means that we first add the
matrices and then take the determinant
so if we first add the matrices so this
is a resultant matrix and taking the
determinant will be getting this since
that elements are different than this
that means a plus B is not equal to
determinant of a plus B so this is what
we have determinant of a plus
determinant of B is not equal to
determinant of a plus B now this I can
explain this I can verify with an
example also so if I have a determinant
a as 1 2 2 1 4 2 0 1 3 and determinant V
as 1 1 0 1 2 0 2 2 3 then the
determinant C will be first column is
copied as such third column is copied as
such but the second column is 2 plus 1 3
4 plus 2 6 1 plus 0 1 so determinant C
is this let us calculate the determinant
the determinant determinant a is first
column first element first 1 into 12
minus 2 minus I am expanding about this
6 minus 2 and
I actually I can expend about any row or
any column but since this particular
matrix has one element 0 in the first
column so I will prefer to compute
determinant by expanding about the first
column the idea is this way I will be
performing less computation as compared
to this column this row or any other
column so I will be computing only these
two terms not the third term so this
gives me the result as 6 X pi the
determinant of B is concerned I'll up
use the same determine same column for
evaluating determinant and it comes out
to be 1 into 6 minus 0 minus 1 and it is
3 minus 0 so the final result is 3
similarly the third determinant 1 into
18 minus 2 minus 1 into 9 minus 2 and
that gives me 9 and one can check that
determinant C which is 9 is equal to
determined a plus determinant B which is
6 plus 3 so we have verified the result
but if you compute a plus B determinant
then it will not come out to be same as
a plus B let us see this we have already
computed a is equal to 6
we have computed determinant B s3 you
have computed a plus B as 9 but if you
have to compute a plus B bit remanent
then this comes out to be 2 into 36
minus 4 minus 2 into 18 minus 4 that
gives me 64 minus 28 and finally a plus
B is 26
clearly a plus B determinant is not the
same as a determinant plus B determinant
and that proves so in general this
result is not true in fact if we have to
add this and if the sum rows are some
are identical except one particular row
then we have the result given by earlier
theorem if we can generalize this then
alpha A plus beta B is not equal to
alpha A plus beta B actually is very
surprising that this reserve the event
hold for addition so this is a linear
property this must hold hold for
determinants but we have verified that
this is not true for determinants this
is very surprising and one has to be
very careful about it that alpha A plus
beta B is not equal to alpha A plus beta
B so that's the basic difference between
determinants and the numbers but if we
have two matrices a and B and of order
they are of order M then the product a B
and it's determinant is equal to
determinant of a into determinant of B
so this is not true this property does
not hold for addition but this property
holds for determinants so here one has
to be very careful the determinants are
apply determinants are applicable only
to square matrices so it may happen that
a B is a square matrix but a and B are
not square matrices so in that case
determinant of like in this case a is M
by K B is K by M a B will be M by M so a
B is a square matrix the determinant is
possible but if you want to apply this
then determinant of a M K is not defined
determinant of B is not defined so this
will not be defined for this set of
matrices this is this property is true
only when a and B are square matrices of
same order say M so in this case
determined a B is defined by determined
a and determined B are not defined so
this property is invalid
now to prove this actually the proof of
this property is really very complex
I will first try to prove it for M is
equal to two so for M is equal to two
let us consider two square matrices of
order to a as a b c d and b as X Y Z W
then determinant a is equal to ad minus
BC while determinant B is X W minus bi
set however the determinant a B will be
first take the product and then take the
determinant this means the product of
these two matrices the first row first
column element will be a X B's it this
element is obtained by multiplying a B
by by W and this element is seen X DS it
and finally the last element will be C y
DW so this is determinant a B so let us
see what is a multiplied by B if we
expand a B then it will be x + bz x cy +
GW - a VY plus BW into a Viper's BW if
you further expand then a B is equal to
ad minus BC x xw minus y said now i will
try to match the two things
so we start with ad minus BC xw - ad
minus BC white lies in my expansion and
then ad XW plus BC wise it - b CX w - a
divides it and Here I am trying to add
and subtract certain things so that the
final result can be obtained easily so
ax DW is with me I
ad C Y X and subtract c y XA similarly I
write down bz c vy + GW b z cy is with
me
i subtract b ZD w and add visa GW and
then these two terms now we can adjust
these quantities and we can write down x
plus ax plus BZ into DW plus ey minus CX
a y plus b w- d z BW + EI y and that
gives me the final result as this so
determinant of product is equal to
product of determinants now as far as
proof for general and that is that
cannot be done in half an hour so this
requires lot of computational effort so
i leave this is an exercise for the
viewers they just try and check that
this result hold for other values of m
also now we have computed determinants
of number of matrices some results may
be of worth noting that is the
determinant of identity matrix is 1 this
can be easily obtained for i 3 in
general i n is equal to 1 this identity
matrix happens to be a diagonal matrix
and all the diamond elements are 1 so i
n will be 1 and then of course we have
already proved the result that a b is
equal to determinant of a into B so we
can use it repeatedly to give the result
as the product of K matrices go they
have to be square matrices they have to
be of same order then this is equal to
product of K determinants one has to be
sure that all the determinants are of
same order all the matrices are of same
and they are off they are square only
then this result is valid what we can do
is we first take this term multiplied by
e 1 so it is a 1 into determinant of
this then we can apply and the same
result to this considering a 2 and thus
were terms can be taken together so
we'll be having a 1 a 2 into rest of the
terms and this can be repeatedly applied
to give the final result there are
certain remarks one thing is that for
product of matrices a B is equal to 0
does not imply that a is equal to 0 or B
is equal to 0 this is the result which
we have seen when we were discussing
determinants that it is quite possible
that a and B none of them is 0 but their
product a B is 0 now this is not true
for determinants like if I have
determinant of up product is 0 then this
simply means that either determinant a 0
or determinant B is 0 this can be proved
very easily like you can say that
determinant of this product is equal to
determinant a into determinant of B
determinant B and this simply means if
determinant a is 0 then what are
determine a and determinant B they are
simply number so one of the two numbers
have to be 0 so either will have
determined a is 0 or determine B is 0 so
this result was not in general true for
matrices but this result is true for
their determinants similarly this is
another rima the product of matrices a B
is not equal to B a we have proved that
the product of matrices is not
commutative that is a B is not equal to
B of course it is important that a B & B
a should be of same order and these two
products are possible but if you
consider the determinants a B of a
square matrix a and B of order M then
determinant a B is equal to determinant
of a into determinant B the same way
determinant of B
is equal to determine begin to determine
a but what are determined a be
determined B and a they are simply
numbers and the numbers can become
iterative so determine doing to
determine a can be written as determined
a in to determine B and that makes
determinant of a B is equal to
determinant of B a so this is again in
contrast with the theory of matrices
where a B is not equal to BA
but for determinants a B is equal to B
we have seen that if the matrix the
diagonal matrix then one can very easily
calculate the determinant by simply
multiplying the diagonal elements let us
see what happens if the matrix is a
block diagonal matrix by this I mean to
say that I have a matrix of orders say
four by four and this can be partitioned
into sub matrices like this this can be
partitioned and this is a sub matrix of
order 2 by 2 this is a sub matrix of
order 2 by 2 they lie in the diagonal
and these are the matrices which are
null matrices I denote them by Z so let
us say this is a matrix this is B matrix
and this is null matrix so we will have
a B Z ends it so if it is a block
diagonal matrix then how can we
calculate the determinant so if let us
say we have a 2 by 2 matrix and it is a
diagonal matrix then what will be a is a
simple singleton B will be a singleton
there will be 0 Z will be 0 and
determinant of this matrix will be
simply be the product of these two terms
that is a into B but in this case I am
having not a singleton but I am having a
matrix of order 2 by 2 or a matrix of
order 2 by 2 as the B matrix so how we
can calculate the determinant of this
matrix may be noted that this I have
taken for example it may happen that
this is about a 3 by 3 and this is of
order 2 by 2 so these men
mattresses may be rectangular but these
mattresses have to be square mattresses
like will have a fire back biometric so
it may be havoc 3x3 matrix n and 2x2
matrix a so this will be a rectangular
matrix and this will be a rectangular
matrix so we'll be having a block
diagonal matrix a B Z Z so how the
determinant of this can be calculated
this I'll show for C matrix which is the
determinant of a 1 1 a 1 2 0 0 a 2 1 a 2
2 0 0 0 0 B 1 1 B 1 2 0 0 b21 b22 now
the determinant of this can be easily
calculated what I can do is I can expand
it about the first column so it will be
a 1 1 multiplied by its cofactor plus a
2 1 multiplied by its cofactor this term
will not contribute this term will not
contribute that is the advantage of
having zeros in the particular row or
column in fact we select the particular
lower column which will have some of the
terms zeros now in this case every
column or row can will be having 2 0
terms so it does not make much
difference whether you use this column
or this column but for 10 millions I am
using this particular column now what is
a 1 1 a 1 1 is a cofactor of a 1 1 how
we obtain a 1 1 a 1 1 is obtained by
neglecting this row and this column that
means this particular determinant so
this particular to evaluate this
particular determinant I will again
expand by cofactors so for this I'll
consider this column so it is a 2 2
multiplied by this determinant is
determined I know is B this term and
this term will not contribute so a 1 1
is nothing but a 2 2 B similarly 8 a 1 a
2 a 2 1 a 2 1 that is cofactor a 2 1
will be
a 2 1 into a 1 2 let us see if we
calculate the cofactor of a 2 1 then
this row and this column has to be
deleted so the sub matrix will be having
a 1 2 0 0 0 B 1 1 B 1 2 0 b21 b22 and
but again by cofactor expansion this
will come out to be a 2 1 a 1 2
determinant B so determine B is common
in both the terms so we can be taken out
and what we have is a 1 1 a 2 2 minus a
2 1 a 1 2 what is this this is nothing
but the determinant of a matrix a11 a.22
minus a12 a21 that is nothing but
determinant a multiplied by determinant
B so what we can say is if you have a
block diagonal matrix consisting of a
matrix and B matrix in the diagonal then
the determinant of the diagonal matrix
is nothing but the determinant a and
determinant B that means if I have a
matrix C having
k-rock diagonal matrices on cape box of
diagonal matrices then we could be
having determinant see as determinant D
1 into determinant D 2 into determinant
D K so one can generalize this result
and this will help this will be helpful
in evaluating determinants of higher
order matrices cuz of the matrix of
cofactors of a IJ of the elements of a
is called the adjoint of a by this I
mean to say that if a IJ is the cofactor
of a IJ here the differences here is the
capital AI J small a IJ is the element
capital a I J represents the cofactor of
the IJ element of the matrix that is a
joint a is a 1 1 the cofactor of element
a 1 1 in the matrix a a 1 2 is the
cofactor of first row second column
element of the matrix a a 1 M and then
we take cofactor of each and every
element and then we take the transpose
of it
that defines the adjoint of a or adjoint
of a is transpose simply means the rows
and columns are interchanged so I am
writing a 1 1 a 1 2 A 1 in the first row
as the first column a 1 1 a 1 2 A 1 in
and this is the second row second column
which is here the second row and the
last row here will become the last
column here so if I have the matrix a as
this then cofactor a IJ is obtained as
by neglecting a by considering a sub
matrix deleting the iaith row and jth
column and what we have is an N by n
minus 1 by n minus 1 sub matrix take its
determinant and then multiplied it by
minus 1 raised to power I plus J that
becomes the determinant of that becomes
the cofactor of a IJ for example if you
have to find the determinant of the
matrix 1 2 3 2 5 2 2 1 0 or 3x3 matrix
then you have to find out the determined
you have to find out the cofactors of
elements of this matrix I call a 1 1 as
the cofactor of first element in the
first row and first column this I
obtained by neglecting this row and this
column and forming the determinant of
this sub matrix gjermund of this sub
matrix is 5 into 0 minus 2 into 1 that
means minus 2 so everyone comes out to
be minus 2 a 1 2 is this cofactor of
this element which is obtained by
neglecting this and this column and what
we have is 2 into 0
- two into four that comes out to be a 1
2 into 4 but actually this is determined
is -4 and I use the record patent for
getting the sign of the cofactor the
sign up for factor here is plus here and
minus here so this minus multiplied by
this - makes it 4 so a 1/2 comes out to
be 4 now a 1/3 is this determinant with
plus sign so it is 2 minus 10 it comes
out to be a 1/3 similarly one can
compute a 2 1 as 3 a 2 1 means cofactor
of this element which is 2 into 0 - 3
into 1 multiplied by minus 1 that makes
it 3 a 2 2 is minus 6 which is 1 into 0
minus 3 into 2 which is minus 6 and
cofactor of this element a2 3 is 3 1
minus 4 with a negative sign here
similarly a 3 1 is minus 11 a 3 1 is
this element it is 4 minus 15 so it is
minus 11 a 3 2 is 2 minus 6 with a
negative sign so it is for a 3 3 is 5
minus 4 comes out to be 1 so what is a
joint a a11 a21 a31 these are the
elements which are in the first column
but when we are packing up when we are
talking about adjoint a they have
they'll become in first row because it
is the transpose of matrix of cofactors
so what we have is minus 2 3 minus 11 is
a first row 4 - 6 4 is a second row
minus 8 3 1 is the third row so a joint
of a is equal to this matrix this 3 by 3
matrix
now on the basis of this we have a
theorem which states that a multiplied
by a joint of a is equal to a joint of a
multiplied by ay it is commutative and
not only this in equal to determinant of
a multiplied by identity this is an
important result so let us try to prove
this for this purpose let us consider
the product B as a into adjoint of a now
I denote a joint of a as C so this B is
equal to AC to calculate d is equal to
AC what I have to do is I have to
calculate the elements of the matrix B
which is the product of these two
matrices so bi is a typical element of
this matrix B is the product of this so
what we have is AI J multiplied by CI J
this is equal to summation over k AI k
ck j now this is a from the definition
of product of matrices now what is ck j
ck j is a joint of a so this is nothing
but a joint is C is equal to a joint of
a so it is nothing but cofactor of AI K
and its transpose so it becomes a JK so
C KJ is nothing but a JK so this means
bij is the summation of aik AJ K over K
now to compute this let us consider a
case when I is equal to J when we
consider I is equal to j then bij is
equal to summation over k aik a I K
because J is equal to I and we have
already proved that this is nothing but
determinant of a so when i is equal to j
bij is nothing but determinant
and this is true for all values of i
from 1 to in but what happens when i is
not equal to j to prove this result let
us consider bij is equal to AI k aj k
summed over k now we define a new matrix
d such that d RS is equal to a RS when R
is not equal to I for all values of s
lying between 0 and n but D is is equal
to edges for all values of n that means
the I get I get Rho of D matrix is the
same as J through of a matrix that is
how we define a new matrix D now this
means that bij is equal to AI ka JK this
is what we have here summed over k is
equal to di k by I am using di K because
aik is equal to di K by the definition
here into d JK aj k is equal to d JK
what is a JK a JK is cofactor matrix so
this is the same as d JK now one may
notice that since I F and K throes of D
are identical so all the determinants in
the all the cofactors d JK will be 0
because I attend K throes of D are
identical and that simply means that B
IJ is 0 whenever I is not equal to J and
this simply means that B IJ is equal to
determinant a when I is equal to J and
it is 0 when I is not equal to J by this
I mean to say that the determinant B
matrix will look like as this where
determined a will appear in the diagonal
because o for all the diagonal elements
I is equal to J and all the elements
which are lying on the upper side of the
diagonal or on the lower side of the
diagonal they are 0 because these are
the elements where eyes no
equal to J so basically B is a diagonal
matrix so if B is a diagonal matrix then
one can take determinant a outside so
number only we can take it outside and
what remains here is nothing but the
identity matrix and that proves the
result that a determinant adjoint a is
determinant a times I similarly one can
easily prove adjoint of a is equal to a
is the same as determinant a I now this
I leave as an exercise for the viewers
to verify this hence we can say that a
joint of a multiplied by a is equal to
adjoint of a multiplied by a is the same
as determinant of a multiplied by
identity matrix now on the basis of this
one can get some useful results
regarding non singular matrices so that
may introduce non singular matrix so we
say an on a square matrix of order n is
called non singular if there exists
another square matrix B of order n such
that a B when multiplied by the a
multiplied by B is the same as B
multiplied by a which gives me an
identity in that case I say that matrix
B is an inverse of matrix a so if such a
matrix B can be found then the matrix a
is called unknown singular matrix the
inverse of a is denoted by a inverse a
inverse so we say
B is equal to a inverse or we can say a
inverse is equal to B if no such B
exists then we say the matrix a is
singular matrix one can notice that this
much this product is commutative now we
have a result which says that if
determinant is not zero then a inverse
is equal to adjoint a divided by
determinant of E now this actually later
on once we prove this then becomes a
formula for finding out determinant of a
inverse that's the importance of adjoint
of e now to prove this let us consider
the product a adjoint e and we'll prove
that this is determinant a I so what we
can do is we multiply this result we
have already established so what we can
do is we pre multiply it by a inverse so
a inverse a is equal to a joint a is
equal to this becomes identity and a
joint a becomes determinant a these can
be taken outside determinant a is equal
to a inverse so what we have is a joint
a is equal to determinant a into a verse
or a inverse because determinant a is
simply a number you can always divided
by determinant a for the sides so a
inverse is equal to adjoint a divided by
determinant a
let us take this example a is equal to
two by two matrix a C B D then you have
to find the inverse of given matrix a
for this we consider first the
determinate a D minus BC so if ad minus
BC is not zero this is required so this
condition is true ad minus this is not
zero only then we can find out the
determinant only then we can apply this
formula because if ad minus BC becomes
zero then this is not possible and
inverse will not exist so a inverse is
equal to adjoint a divided by
determinant a and this gives a joint of
a as the adjoint of cofactor of a is d
cofactor of C is minus B cofactor of B
is minus C cofactor of D is a take the
transpose and what we have is D minus C
minus B a and then a inverse is equal to
determinant a in the denominator so ad
minus BC One upon ad minus BC multiplied
by D minus C minus B a becomes the
inverse of given matrix e let us verify
that this actually is an inverse so what
we can do is we multi we multiply with a
and let us see it comes out to be an
identity so if you multiply a by a
inverse will have this product and let
us see this product it is ad minus BC
which is one so we'll be having one here
but AC the second element if you
consider it is a my AC with negative
sign then CA they'll cancel out and what
we have is zero on the first row second
column similarly B D minus BD that gives
me 0 as the
element in the second row first column
the right-hand side and then a my B my B
minus C and D a that gives me one when
divided by ad minus BC so what we have
on the right hand side is the identity
matrix similarly one can prove that a
inverse a is also identity matrix user
can viewer can actually perform this
multiplication and get convinced that
this is actually comes out to be an
identity matrix so the result is
verified now this I have done for a
two-by-two matrix little more effort is
required for a 3x3 matrix so if I have a
3x3 matrix a as 1 2 3 2 5 2 2 1 0 then
its determinant is calculated as minus
81 into this 2 into this and finally 3
into this that comes out to be minus 80
a joint a is minus this I have already
done in one of the earlier exercises so
this I have already computed so simply a
inverse is 1 upon 18 1 upon minus 18 as
adjoint of a so that is a inverse so
given a and a inverse which I have
computed one can verify that this comes
out to be identity so this Row is
multiplied by this will be having minus
2 8 and minus 24 this comes out to be
minus 18 so the result is 1 this is 1
and you can similarly calculate other
terms and this comes out to be an
identity matrix of order 3
now this is an important theorem which
says that if a matrix is non-singular
and if determinant is not zero the proof
of this is that if determinant a is not
zero then the matrix is non-singular
this has actually two parts the
determinant a is not zero then
determined a inverse exists and if
determinant a inverse exists then
determinant a is not zero
now if determinant a is not zero then
one can find that determinant from the
previous result that pain was exists
because a joint a can very easily be
obtained the matrix of cofactors
determinant a is not zero so we can
divide it by determinant e and in the
earlier theorem we have proved that an
inverse is given by this result so if
determinant a is not zero then a inverse
exists and the matrix is non-singular
however if the matrix is non-singular
then it's inverse exists such that a a
inverse is one so the inverse matrix is
non-singular and its inverse exists then
we have to show that its determinant
will not be zero so let us consider a
inverse is equal to identity a a inverse
is equal to identity because inverse
exists in this the very definition of
inverse let us take the determinant of
this product so determinant of this
product comes out to be 1 because
determinant of identity is proved to be
1 so we apply the product determinant a
into a inverse which comes out to be 1
and this simply means determinant of 2
is equal to determinant of 2 matrices
its result is 1 so determinant a cannot
be 0 that simply means that for a non
singular matrix determinant a is not 0
now after proving this one can take this
result as a definition for a non
singular matrix previously we're
defining that non singular matrix is the
one which has in
which has inverse but if now we say that
if the matrix is having nonzero
determinant then it is non-singular
so this can be taken as a definition of
non singular matrices so let a be non
singular matrix then a a inverse is
identity or a inverse is equal to one so
a inverse is actually 1 upon determinant
e so if determinant a is not 0 then a
inverse is 1 upon determinant of a so if
you know that determinant of a one can
very easily find out the determinant of
its inverse so this result is helpful in
that sense so let us illustrate this
with this example we have a determinant
a as 2 determinant B s3 then you have to
compute the determinant of B inverse
multiplied by 3 ei into 2 B so what we
can do is we can write down the
determinant of this product as
determinant of B inverse into
determinant of 3a into determinant of 2
B now determine that the u.s. is the
same as 1 upon determinant of B so 6
into B inverse is nothing but 6 into 1
by 3 because determinant B is 3 then
determined of 3 a is 3 times the current
of a determinant of a is 2 times
determinant of B that is how this 6 is
coming so 3 a 2 is taken out so this
result is equal to 6 times B inverse
into a determinant of a into determinant
of B that is 6 into 1 by 3 into
determinant of a is 2 into determinant
of 3 which is 3 and final results come
out to be 12 so if determinants are
given then one can compute this type of
product
now I am considering the example in
which you have to find out the value C
such that the matrix a is singular the
matrix a is given as 3 minus C 0 3 0 C
plus 2 0 5 0 C plus 5 now I will be
using the theorem which I have just now
proved that the matrix is singular if
its determinant is 0 so determinant a is
equal to 3 minus C 0 3 0 C plus 2 0 5 0
C + 5 0 so basically this matrix
determinant is to be evaluated to
evaluate this determinant one can expand
it about this particular row so it is 3
minus C into c plus 2 into c plus 5
minus 0 into 3 times 5 that is 3 into c
plus 2 times 5 that is this term and
simplified comes out to be minus c c
plus 2 into c plus 2 is equal to 0 and
that simply means either save equal to 0
or C is equal to minus 2 or C is equal
to minus 2 so for these values of C the
matrix a is singular this matrix a is
singular viewers today we have discussed
some interesting properties of
determinants and we have defined adjoint
of a matrix and we have seen how this
can be used to evaluating inverse of a
square matrix thank you