Welcome to this session. Now, we will be continuing
whatever the example we are talking about
in the last lecture, we have taken up one
example of a set of non-homogeneous algebraic
equation for the solution using the Eigen
value Eigen vector method. So, we took up
a two-dimensional problem, 2 into 2 matrix
for demonstration purpose, the whole procedure
can simply be extended for higher dimensional
problem numerically, that we have already
discussed in the last class.
Now, let us come back to the specific example
that we are talking about. So, in this example
it is a two-dimensional problem. So, we followed
the various steps to be adopted to solve this
problem. The first step, one has to get the
evaluation of Eigen values. So, we have identified
the, we have evaluated the Eigen values for
the matrix A in this particular case, and
these Eigen values turned out to be minus
4 and minus 1.
.
Next step is to evaluate the Eigen vectors
of the corresponding Eigen values, and we
have already evaluated the Eigen vectors for
this matrix A, X 1 and X 2. Next step is evaluation
of Eigen values Eigen 
vectors A transpose. We have already proved
the Eigen values will be the same for A and
A transpose, that we have already proved earlier.
So, this is step number two.
So, evaluation of Eigenvectors of A transpose.
So, Eigen values for this A transpose are
identical, A transpose are same as minus 4
and minus 1. So, for lambda 1 is equal to
minus 4, let us evaluate the Eigen vectors
of this A transpose matrix. So, minus 2 minus
lambda 1 1 2 minus 3 minus lambda 1, X 1 X
2 are the elements of this vector is equal
to 0. So, if we put it as lambda 1 is equal
to minus 4 will be getting 2 1 2 1 X 1 X 2
equal to 0.
So, we will be getting 2 x 1 plus x 2 is equal
to 0 from this. And 2 x 1 plus x 2 equal to
0. So, they are identical. So, if you select
X 1 is equal to 1, then X 2 has to be minus
2. So therefore, Y 1 is equal is 1 minus 2
is an Eigen vector for the Eigen value lambda
1 is equal to minus 4. So, in case of lambda
2 is equal to minus 1, we will be having minus
1 1 2 minus 2, I am just writing from these
equations I am just writing this one. So,
I am just omitted one more one step here,
they are identical, by now it has to be clear
to you and you also know how to evaluate Eigen
values or Eigen vectors given a matrix.
So, this will be minus X 1 plus X 1 is equal
to 0 and 2 X 1 minus 2 X 2 is equal to 0.
They are identical. So, if you select X 1
is equal to 1, then X 2 turns out to be 1.
So therefore, Y 2 is equal to 1 1 is an Eigen
vector 
to Eigen value lambda 1 is equal to minus
1.
So therefore, this is the same method, this
is the same way we have evaluated the Eigen
vectors of matrix A, we can evaluate the Eigen
vectors of matrix A transpose. Now, let us
write down this the solutions in one place.
So, we have the Eigen vectors, X 1 is 1 minus
1 and X 2 is 2 1 and Y 1 is 1 minus 2 and
Y 2 is 1 1. So, this corresponds to lambda
1 is equal to minus 4, these two corresponds
to lambda 1 is equal to minus 1. So, they
are Eigen vectors of A and they are Eigen
vectors of A transpose.
So, an Eigen values we know, lambda 1 is equal
to minus 4 and lambda 2 is equal to minus
1. Next step, that is step three. We have
already finished step one and step two, that
is evaluation of Eigen vectors of A and evaluation
of Eigen vectors A transpose. So, next step
is to get the solution of b is equal to summation
beta i X i and in this case i is going to
1 to 2. So, beta 1, if you remember beta 1
is nothing but, Y 1 transpose b divided by
Y 1 transpose X 1. So, if we write it down
it becomes Y 1 transpose, so it will be 1
minus 2, this will be simply 1 minus 2, the
column becomes rows. So, it becomes 1 minus
2, rows becomes columns, and b is minus 1
0, and Y 1 transpose is same as 1 minus 2,
and X 1 is 1 minus 1.
So, if you compute, this is 1 into minus 1
minus 1 plus minus 2 into 0, that is 0 divided
by 1 into 1 plus minus 2 into minus 1, so
this will be plus 2, so this will be minus
one-third. So, beta 1 is minus one-third.
Similarly, we calculate beta 2, beta 2 will
be Y 2 transpose b divided by Y 2 transpose
X 2. And Y 2 transpose will be 1 1 minus 1
0 1 1 and X 2 will be 2 1, X 2 will be 2 1.
And we get this, this will be minus 1 plus
0 into 1, that will be 0 2 into 1 plus 1 into
1. So, this will be minus 1 upon 3.
.
Next, we evaluate alpha 1 and alpha 2. If
you remember that alpha i was equal to beta
i by lambda i. So, once you get the beta's,
then you will be able to and you already know
the lambdas, so you can get the alpha i's.
So, alpha 1 will be nothing but, beta 1 by
lambda 1. So, beta 1 will be minus 1 upon
3 and lambda 1 is minus 4. So, this will be
plus 1 by 12 and alpha 2 is minus 1 upon 3
divided by minus 1. So, this will be plus
1 upon 3.
So, now since the alpha's are known and x
i's ,the Eigen vectors of A are already determined,
we will be in a position to get the solution
matrix. So, the solution matrix is or solution
vector is given as, vector is nothing but,
a special matrix, so solution vector is given
as X is equal to summation alpha i X i. So,
you will be having two elements of the solution
vector, X 1 and X 2. So, this will be alpha
1 X 1 plus alpha 2 X 2. Alpha 1 is 1 by 12
and X 1 we have already found out 1 minus
1 plus, alpha 2 is 1 upon 3, so this will
be 2 into 1, 2 and 1. So, this will be equal
to.
So, we will be getting the solution matrix,
solution of this problem, solution elements
X 1 is equal to 1 by 12 into 1 plus 1 by 3
into 2, so it will be 2 by 3. So therefore,
12 1 8, so this will be 9 by 12, so it will
be three by four. Similarly, the other solution
is 1 by 12 with a minus sign plus 1 upon 3.
So, this will be 1 upon 4. So, three by 4
and 1 over 4 is the solution of this particular
problem.
So, the solution vector x will be comprised
of two elements, 3 by 4 and 1 by 4, and that
gives the complete solution to this problem.
So, that is how this problem clearly demonstrates
the solution using the Eigen value problem,
how to get the solution using the Eigen value
and Eigen vector method, without taking a
request to the matrix inversion that is necessary
for utilizing Gauss Seidel algorithm or Gauss
Elimination method.
So, we will be taking up one more example
in this type of problem, simply because the
Eigen values in that particular problem becomes
imaginary, it becomes imaginary. So, you need
not to worry about that, depending on the
situation, depending on the definition of
the problem, one can get real valued Eigen
values or imaginary Eigen values. But the
final solution will be a real solution, if
it is a real valued problem, because all chemical
engineering systems are real valued problem.
So, finally, you will be getting a real solution
even if the Eigen values are imaginary. That
is the reason I will be taking up this particular
example, where the Eigen values you will be
getting as imaginary.
.
So, the next problem is example two. And we
are going to solve this equation non-homogenous
set of algebraic equation, A X is equal to
b, where the matrix A is 1 2 minus 2 1 and
the vector b is 3 and 4. So, let us look into
the solution. Step number one is Eigen values
of A, that is nothing but, determinant of
A minus lambda i is equal to 0. So, we compute
that 1 minus lambda 2 minus 2 1 minus lambda,
that will be equal to 0. Open up this determinant.
If you open up this determinant, this becomes
1 minus lambda square plus 4 is equal to 0.
So, you will be getting lambda square minus
2 lambda plus 5 is equal to 0. So, it is a
quadratic, the took solutions correspond to
the two Eigen values. So, lambda 12 will be
minus b, so it will be plus 2 plus minus under
root b square 4 minus 4 a c. So, 20 divided
by 2. So, it will be 2 plus minus 4 i divided
by 2. So, you will be having imaginary Eigen
values in this particular problem. So, lambda
1 becomes 1 plus 2 i and lambda 2 is equal
to 1 minus 2 i. And this roots, they occur
as complex conjugate.
So, let us talk about the first root and we
will be evaluating the corresponding Eigen
vectors. So, corresponding to Eigen values,
next we evaluate the Eigen vectors of the
matrix A.
So, for Eigen value lambda 1 is equal to 1
plus 2 i, we will be having A minus lambda
1 i X 1 is equal to 0. So, X 1 is the Eigen
vector 
corresponding to Eigen value lambda 1. So
therefore, we will be getting this as 1 minus
lambda 1 2 minus 2 1 minus lambda 1, X 1 X
2, these are the elements of the Eigen vector
X 1. So, you put the value of lambda 1 as
1 plus 2 i. So, this becomes minus 2 i 2 minus
2 plus and 1 minus lambda 1, sgain it is minus
2 i. So, you formulate the two equations.
So, minus 2 i X 1 plus 2 X 2 is equal to 0.
Minus 2 X 1 minus 2 i X 2 is equal to 0. These
two equations are identical because if we
multiplied these equation by i, what we will
be getting is that, minus 2 i X 1 i square
is minus 1, so minus minus plus, 2 x 2 is
equal to 0. So, these two equations are identical
and therefore, if we select 
X 1 is equal to 1, then X 2 will be i. So
therefore, X 1 is 1 i is the Eigen vector
corresponding to Eigen value lambda 1 is equal
to 1 plus 2 i. So, X 1 is the Eigen vector,
1 plus 2 i is the corresponding Eigen value.
.
So, next we compute the Eigen vector corresponding
to the Eigen value lambda 2. So, for lambda
2 is equal to 1 minus 2 I, again we get the
corresponding Eigen vector by solving this
equation A minus lambda 2 i multiplied by
X 2 is equal to 0, where X 2 is the corresponding
Eigen vector. So, therefore, this becomes
1 minus lambda 2 2 minus 2 1 minus lambda
2. And X 2 will be having two elements, X
1 and X 2 that will be equal to 0. So therefore,
if you put lambda 2 as 1 minus 2 i here, so
you will be getting 2 i 2 minus 2 plus 2 i
and X 1 and X 2 is equal to 0. So, the first
equation we will be getting 2 i X 1 plus 2
X 2 is equal to 0 and minus 2 X 1 plus 2 i
X 2 is equal to 0. Again both are identical
because you multiply this equation by i, you
will be getting back the other one. So, you
will be getting finally i X 1 plus X 2 is
equal to 0.
So therefore, if we select X 1 is equal to
1, then X 2 will be is equal to minus i. So,
if X 1 is equal to 1, then X 2 will be minus
i. Therefore, X 2 is equal to 1 minus i is
the Eigen vector corresponding to lambda 2
is equal to 1 minus 2 i. So, we evaluated
the Eigen vectors for the corresponding Eigen
values. Next step is to evaluate the Eigen
vectors for A transpose.
.
So, that is step two, Eigen vectors of A transpose.
So, if you look into the A transpose, this
becomes, this is a transpose matrix 1 2 minus
2 and 1. So, lambda 1 and lambda 2 remain
the same because you have already proved the
Eigen values of A and Eigen value of A transpose
are identical. So, they remain same, only
the transpose matrix elements will be changed,
inter changed, rows becomes columns, lambda
2 becomes 1 minus 2 i, they are identical.
So therefore, for lambda 1 is equal to 1 plus
2 i, we have A transpose minus lambda 1 i
Y 1. So, Y 1 is the corresponding Eigen vector
of A transpose for the Eigen value 1 plus
2 i. Therefore, it becomes just put the values
here. So, it becomes 1 minus lambda 1 A transpose
1 minus lambda 1 minus 2, then 2 and 1 minus
lambda 1, Y 1 Y 2 are the elements of the
Eigen vector Y 1. So therefore, minus 2 i
minus 2 2 minus 2 i Y 1 Y 2 is equal to 0.
Now, if we construct the equations, these
equations becomes Y 1 is equal to i Y 2. So
therefore, if we select Y 1 is equal to 1,
then Y 2 has to be is equal to minus i.
So, 1 minus i is the first Eigen vector for
lambda 1 is equal to 1 plus 2 i. Similarly,
1 can evaluate the corresponding Eigen vectors
for lambda 2. So, Y 1 is equal to 1 minus
i for lambda 1 is equal to 1 plus 2 i. Next,
we evaluate the Eigen vector for next lambda
values, the Eigen value 1 minus 2 i, we formulate
the same thing A transpose minus lambda 2
i multiplied by Y 2 should be equal to 0.
So, this becomes 1 minus lambda 2 minus 2
2 1 minus lambda 2 Y , Y 2 will be having
the elements, let us say Y 1 and Y 2 is equal
to 0. So, it becomes put the value of lambda
there, lambda 2, it becomes 2 i minus 2 2
plus 2 i Y 1 Y 2 will be equal to 0. So, we
will be having two identical equations and
this equations in this case will be Y 1 is
equal to minus i Y 2. So therefore, if we
select Y 1 is equal to 1, then Y 2 will be
plus i.
So, 1 plus i Y 2 is equal to1 and i is Eigen
vector 
for lambda 2 is equal to 1 minus 2 i. So therefore,
we have evaluated the Eigen vectors for A,
we have evaluated the Eigen vectors of A transpose.
Next step is step 3. For step 3, we evaluate
the coefficients beta in the equation b is
equal to summation beta i X i. So, if you
remember beta 1 is equal to Y 1 transpose
b divided by Y 1 transpose X 1. So, this becomes
1 minus i and b is nothing but, 3 4 1 minus
i and X 1 is nothing but, 1 i. So, you will
be having 3 minus 4 i divided by 1 plus, i
square is minus, minus minus plus 1, so it
becomes 1 by 2 3 minus 4 i. So, that is beta
1. Similarly, we can evaluate beta 2, beta
2 is Y 2 transpose b divided by Y 2 transpose
X 2. So therefore, this becomes 1 i 3 4 and
it is 1 i 1 minus i. So therefore, you will
be getting 3, minus minus plus, 3 plus 4 i
divided by 1, minus i square, so minus minus
plus 1, so this becomes half 3 plus 4 i. So,
you will be getting alpha 1 as beta 1 by lambda
1. So, alpha 1 becomes 3 minus 4 i divided
by 2 and lambda 1 becomes 1 plus 2 i. So,
lambda 1, you just put the value of lambda
1 there. And similarly, we can get alpha 2
as beta 2 by lambda 2 and this becomes 1 by
2 3 plus 4 i divided by 1 minus 2 i.
.
So, next, so we have evaluated the betas,
we have evaluated the alphas. Next and final
step is the solution X is equal to summation
alpha i X i. So, the solution, this becomes
a solution vector. Already determined 
the Eigen vectors, already evaluated, so we
just put the solution vector, these are the
elements of the solution, vector X 1 and X
2. So, you just open up this summation, this
summation runs from index 1 to 2. So, 3 minus
4 i divided by 2 into 1 plus 2 i, that is
the alpha 1. Next is X 1, X 1 is 1 and i plus
3 plus 4 i divided by 2 into 1 minus t2 i,
that is the alpha 2 and X 2, X 2 is 1 minus
i.
So, you will be getting, just simplify the
whole thing, just make it, you just multiply
both side, multiply both numerator and denominator
by complex conjugate of 1 plus 2 i. So, multiply
by 1 minus 2 i on top and the bottom. So,
it becomes 3 minus 4 i multiplied by 1 minus
2 i divided by 2 into 1, minus minus plus,
4 1 and i plus 3 plus 4 i 1 plus 2 i divided
by 2 1 plus 4 1 minus i.
So, this becomes, you just keep on simplifying,
this becomes minus 1 minus 2 i, I just omitted
one more step, divided by 2 1 and i. Next,
we will be getting minus 1 plus 2 i divided
by 2 1 minus i. So, X 1 we will be getting
as minus 1 minus 2 i divided by 2 multiplied
by 1 plus minus 1 plus 2 i divided by 2 multiplied
by 1, and if you simplify this whole thing
this becomes minus 1. So, you will be getting
a real valued solution. Similarly, the X 2
will be i by 2 minus 1 minus 2 i plus minus
i by 2 minus 1 plus 2 i. So, this will become
2.
So, minus 1 and 2 is the solution vector for
this particular problem. And if you see that,
although the Eigen values are not real, they
are imaginary but, the solution is a real
solution. So, this example demonstrates that
even the Eigen values become imaginary, one
we will get a real valued solution for this
particular problem.
So, this gives a demonstration of how the
Eigen values and Eigen vector methods can
be implemented to solve the set of algebraic
equations, which are typically occurring for
the steady state problem in any chemical engineering
application, and how one will get an elegant
solution using the Eigen value and Eigen vector
method. Next, I will talk about solution of,
so we have looked into the solution of set
of algebraic equation. Next, I will be talking
about solution of set of ordinary differential
equation.
.
Solution to 
set of ODEs. Typically, this will be occurring,
ODEs will be occurring if we will be talking
about a transient problem. If we have a transient
problem, then the ordinary differential equations
may be the governing equations of the chemical
engineering process.
So, if there is a set of ODEs, they can be
written in a compact matrix notation. One
can use a compact matrix notation to detect
this, to write it down d x d t is equal to
A X, in this case X is a vector and A is a
matrix and the initial conditions are given.
This is vector at t is equal to 0 is X i.
So, that will be the initial condition specified
for this set of governing equations, which
are in the form of ordinary differential equation.
Now, in order to utilize the Eigen value and
Eigen vector method, we have to assume the
solution form 
as X is Z e to the power lambda t. So, just
lambda is a scalar, Z is a constant vector
and lambda being scalar, t is the time. So
therefore, we just assume the solution in
this form. Now, what do we do? Next, we substitute
this into this equation and see what we get.
.
So, if we do that we will be getting d x d
t. So, lambda Z e to the power lambda t should
be is equal to A X. So, what A X is? A X is
nothing but, lambda X because X is nothing
but, Z e to the power lambda t. So, if lambda
is this scalar given in this form, you will
be finally, getting A X is equal to lambda
X. Now, I think you will be able to get the,
identify the form of this equation, this is
a standard Eigen value problem. So, this is
an Eigen value problem, where lambda's are
nothing but, Eigen values of matrix A.
So, again we write A X is equal to d x d t.
So, X we write Z e to the power lambda t.
So, it becomes A Z e to the power lambda t
and d x d t we write as lambda Z e to the
power lambda t. So, A Z is equal to lambda
Z. So, what is Z? So, Z is nothing but, Eigen
vectors of A. So, if we now see that, what
we have assumed the form of the solution,
we have assumed as X is equal to Z e to the
power lambda t. By doing these algebraic manipulation,
we have identified what these lambdas are.
These lambdas are nothing but, the Eigen values
of A. And what this Z is? These Z's are nothing
but, the Eigen vectors of A. So, we write
it, lambda i's are the Eigen values of A,
the corresponding set of zeta's Z i's are
nothing but, the Eigen vectors of A.
Therefore, the solution form whatever we are
assuming, we identified this unknowns, lambda's
are the Eigen values of A, Z's are the corresponding
Eigen vectors. So therefore, the solution
vector X t can be expressed 
as a linear combinations of Z i's the Eigen
vectors. Because we have already proved earlier
that Eigen vectors are always independent
vectors. So, Eigen vectors form a basis set
vector, that is the reason the solution vector,
that will be belonging to the same space,
can be expressed as a linear combination of
the Eigen vectors,
So, therefore, we will be able to write X
as a function of time as summation i is equal
to 1 to n c i some constant multiplier, which
may be a function of t, multiplied by Z i,
this let us say equation number A. Now, in
this case since A is a real constant matrix,
then Z i are time independent because they
are the Eigen vectors, they do not depend
on the time. Because the matrix itself is
a constant matrix, time independent constant
matrix. So therefore, the Eigen vectors will
also be time independent.
So therefore, X in the equation A must also,
this is the solution vector, the solution
vector must satisfy the original equation,
must satisfy original governing equation.
And what was that? That was d x d t should
be is equal to A X. So, we write d d t of
x, x is summation C i Z i. So, it will be
summation C i Z I, this summation is over
i, A times X, X is nothing but, summation
C i Z i.
So therefore, since Z is independent of time,
so this will be nothing but, summation d C
i by d t Z i is equal to A summation C i Z
i. So, this will be just C i is a scalar multiplier,
it may be time dependent. So, C i will be
out. So, A Z i. So, we have already seen that
A Z i is nothing but, lambda Z i. We have
already seen in a few minutes back that A
Z i is nothing but, lambda i Z i. So, we substitute
that here. So, what we will be getting is
that, d C i d t Z i is minus summation C i
lambda i Z i is equal to 0. So, we can take
the summation series out, i is equal to 1
to n d C i d t minus lambda i C i Z i is equal
to 0.
.
Now, you remember Z i's are the Eigen vectors
of the matrix A, that means they always form
a, they belong to a basis set. And what the
basis set is? The basis set is a set of independent
vectors. So therefore, in order to satisfy
this equation, all the corresponding coefficients
should be equal to 0, we have already proved
this theorem earlier.
So, therefore, to satisfy this equation, all
the corresponding coefficients should be is
equal to 0. So, d C i d t minus lambda i C
i should be equal to 0. So, what is the solution
to this? The solution is very simple. So,
C i as a function of t will be C i at time
t is equal to 0, that is C i at t time t is
equal to 0 initial condition e to the power
lambda i t. And therefore, you can write down
the complete solution X as a function of t
is nothing but, summation 
C i e to the power lambda i t times Z i, where
Z i at the Eigen vectors at time t is equal
to 0. We have X 0 that is known to us, this
initial condition set is known to us. This
is nothing but, summation C i 0 Z i. So, let
us say this is equation number B. Now, let
us consider that Y i are the Eigen vectors
of A transpose. So, these are all talking
about the Eigen vectors of A.
.
Now, let us say Y i forms a set of Eigen vectors
of the matrix A transpose, then what do you
do? You take inner product 
of equation B with respect to Y 1 and let
us see what we get. So, inner product of equation
B with respect to Y 1 will give you inner
product of X naught Y 1 should be is equal
to C 1 naught inner product of Z 1 and Y 1.
So, C 1 naught we can get as inner product
of X naught Y 1 divided by inner product of
Z 1 Y 1. So, likewise we can generically you
can write as C i 0 is inner product of X naught
Y i 0 divided by Z i Y i.
So, solution becomes the complete solution
now becomes X t, the time dependent solution
vector is nothing but, summation i is equal
to 1 to n C i naught e to the power lambda
i t Z I, where lambda i's are the Eigen values
of A known to us, Z i's are the corresponding
Eigen vectors of A matrix, they are already
known to us. And C i naught will be obtained
from this. And if you look into the expression
of C i naught, this is inner product of X
naught and Y i. So, X naught is known to us,
that is the initial value vector and Y i is
the Eigen vector of A transpose. So, since
we know, we have already found out the Eigen
vectors of A transpose. So, this is known
to us. Z i are the Eigen vectors of A, Y i
are the Eigen vectors of A transpose. So,
they are known to us. So, we can compute completely
what C naught C i naught is, that will be
put here, lambda is unknown, Z i unknown.
So, one can we get the complete solution matrix.
That is how one can obtain the solution of
set of ordinary differential equation in using
Eigen value Eigen vector method.
.
Next, I will take up an example to demonstrate
this method. Again I will be taking up a two-dimensional
problem. So, that this can be solved in the
class here but, one can do the similar thing
for a multivariable, multi skilled problem,
multi dimensional problem using numerical
techniques.
So, d X one d t is equal to minus X 1 plus
2 X 2, d x 2 d t is equal to X 1 minus 2 X
2. Suppose these two are the corresponding,
the governing equation of a transient chemical
engineering problem, so we write these two
equation into a compact matrix notation d
x d t is equal to A X, where X is the solution
vector, X 1, X 2 and A is the coefficient
matrix, that is minus 1 2 1 minus 2.
Now, let us go step by step. Step one is evaluation
of Eigen values. Eigen values of A will be,
I am not, we have already seen how to evaluate
Eigen values. So, the Eigen values are in
this case is lambda 1 equal to 0 and lambda
2 is equal to minus 3. Now, you can evaluate
the corresponding Eigen vectors X 1 capital
X 1 and capital X 2, which are basically Eigen
vectors of A. So, we have already seen how
to do that in detail in the earlier examples,
I am just writing the solution.
So, for lambda 1 is equal to 0, the corresponding
Eigen vector is 2 1 transpose, specifically
2 1. And corresponding to lambda 2 is equal
to minus 3, we will be getting Z 2 is 1 minus
1 transpose. So, these are Eigen vectors 
of matrix A.
.
Next, we evaluate that is step number 2. Next,
we evaluate in step 2 the Eigen values Eigen
vectors of A transpose. And the corresponding
Eigen vectors are for lambda 1 is equal to
0, the Eigen vector is 1 1 transpose and for
lambda 1, for lambda 2 is equal to minus 3
Y 2 is 1 minus 2 transpose. So therefore,
one can construct the solution vector as X
t is summation of C i 0 e to the power lambda
i t Z i, where i goes from 1 to 2. So, once
we get that, then next step is how to obtain
the C i's
Now, in this particular problem, let us say
the initial value matrix is given as X naught
2 and 3, that means at time t is equal to
0, X 1 is equal to 2 and at the time t is
equal to 0, my X 2 was 3. So, initial value
matrix is given by 2 and 3, then you will
be in a position to compute C i naught's.
So, this C i naught's, because now Z i's are
all known, you have already determined, lambdas
we have already determined, so next is C i
naught. C i naught can be determined by inner
product of initial value matrix 2 2 and 3,
inner product between X naught and Y I, that
is the Eigen vector of A transpose and divided
by inner product between Z i and Y I, Z i
is basically the Eigen vector of A and Eigen
vector of A transpose. So, we can and all
these evaluated Y 1 and Y 2. So, we will be
in a position to evaluate C 1 naught and C
2 naught.
.
So, if you remember just put the values of
C 1 naught is nothing but, inner product of
X naught and Y 1 divided by inner product
of Z 1 and Y 1. And C 2 naught is nothing
but, the inner product of X naught and Y 2
and divided by Z 2 and Y 2. So, we know the
initial value matrix, we know the Eigen vector
initial value vector, we know the Eigen vectors
of A transpose, we know the Eigen vectors
of A, so we will be in a position to compute
C 1 and C 2. So, C 1 turns out to be 2 3 1
1. So, basically you can write it as 
X naught t Y 1 Z 1 t multiplied by Y 1, this
will be X naught t Y 2 Z 2 t Y 2.
So, C naught C 1 naught can be written as
2 3 1 1, this will be 2 1 1 1. So, if you
compute this thing, this will be 2 into 1
plus 3 into 1, so this will be 2 into 1 plus
1 into 1, this becomes 5 by 3. And C 2 naught
will be simply 2 3 1 minus 2 divided by 1
minus 1 1 minus 2. So, this becomes 2 minus
6 divided by 1 plus 2. So, this will be minus
4 by 3.
So, once we have computed C 1 naught and C
2 naught, then we will be able to construct
the complete solution. And in the next class
we will be looking into the complete solution
and finish of this problem, and that will
give you a clear demonstration of how Eigen
value Eigen vector method can be utilized
in order to solve the homogeneous ordinary
differential equation, set of ordinary differential
equation, by using the using this particular
method quite elegantly. And we also, in the
next class we will be looking into what is
the solution of set of non-homogeneous ordinary
differential equation by using the Eigen value
Eigen vector method. So, I stop this class
at this point and continue in the next class
from this point onwards. Thank you very much
for your kind attention.
