So, today we shall see this system of linear equations, system 
system of linear equations. This is also an
important concept in linear algebra. Here,
we shall discuss about consistency of a system
of linear equations; and also we shall discuss,
how to find solution of this. A system of
a system of m linear equations 
in n variables can be written as a 1 1 x 1
plus a 1 2 x 2 plus a 1 n x n is equal to
b 1, a 2 1 x 1 plus a 2 2 x 2 plus a 2 n x
n is equal to b 2. Like this m equations a
m 1 x 1 plus a m 2 x 2 plus a m n x n is equal
to b m.
Here the coefficients, where this a i j that
belong to a field; i from 1 to m, j from 1
to n, b i also belong to this filed, and x
1, x 2 to x n are variables or unknown that
we have to find out, or in other words solving
the system of linear equations means; we have
to find unknowns x 1, x 2 to x n, so that
this equations will be true. Here we are considering
this coefficients a i j are from a field F,
and also the elements b 1, b 2 to b m are
also elements of this field F. Here in fact,
we say that the system of linear equations
is a system over this field F. In place of
this field F, one can also consider the real
field or complex field also.
So, this system of m linear equations in n
variables can also be written as; this system
can be represented as Ax is equal to b, where
A is this matrix that a 1 1 a 1 2 a 1 n a
2 1 a 2 2 a 2 n a m 1 a m 2 a m n, and this
x is.. So, this A is m by n matrix and x is
this column vector of this matrix x 1, x 2
to x n; this is of size n by 1, and b is also
a column vector, or this matrix b 1 up to
b m; this is m by 1 matrix. So, this matrix
is called the co-efficient matrix of the system,
A is called the co-efficient matrix of the
system.
Further, if further if b i is equal to 0 for
all i, then the system is called homogeneous.
Called homogeneous Otherwise, it is non-homogeneous
Otherwise, it is not homogeneous. So, here
we shall discuss about consistency of the
system; that means, when solutions exit for
the system, and also we shall see how to find
solutions. So, the following theorem gives
the condition for consistency of the system
as well as existence of number of solutions
of the system. So, let Ax equal to b be a
system of linear equations be a system of
be a system of linear equations in n variables,
then we have the following results: The system
is consistent, if rank of the co-efficient
matrix A is equal to rank of the augmented
matrix A tilta, where A tilta is this augmented
matrix A b.
Second condition says that if rank of A is
equal to rank of A tilta; that means, the
system is consistent, and if this rank is
equal to n, then the system has a unique solution.
So, third result here is that: If rank of
A is equal to rank of A tilta; and this is
equal to k, and that is strictly n less than
n that is, number of variables variables strictly
less than the number of variables, then the
system has infinite number of solutions infinite
number of solutions. So, we are not proving
this theorem, but will use it and solve problems.
One immediate corollary is that; So, In case
of a homogeneous system of equations, we get
that the rank of the co-efficient matrix is
equal to rank of the augmented matrix, because
all b i are equal to 0. Therefore, a homogeneous
system a homogeneous system Ax equal to 0
is always consistent, or always a solution
exists for the system. Notice that for Ax
equal to 0, this 0 0 0; that means, all x
i(s) are 0 is always a solution. So, we will
have in fact, the following lemma that tells
that the collection of all solution of a homogeneous
system forms a subspace. Let S be the set
of all solutions of this homogeneous system
Ax is equal to 0, then S is a subspace of
this vector space F n. This is not difficult
to proof.
We can take any two solutions in a S. Let
x, y belongs to S that means; x and y be 2
solutions of this system Ax equal to 0, and
alpha, beta be any two scalar or elements
of this field F. Then we shall see whether
this alpha x plus beta y is a solution for
the system or not. So, this can be written
as alpha Ax plus beta Ay, and this is equal
to zero. Since Ax equal to 0 and Ay equal
to zero we get this combination which also
zero. This implies that alpha x plus beta
y belongs to x. Hence S is a subspace of this
F n. And therefore, this set of all solution
of the system Ax equal to 0 is called the
solution space of this homogeneous system.
So, we have the following definition; the
set S of all solutions of Ax equal to 0. This
homogeneous system is called the solution
space of the system is called the solution
space of the system.
So next, we shall see some examples of system
of linear equations; and we shall apply the
results in theorem one, and check whether
that system consistent; and if the system
is consistent, then we shall find all solutions
of that system.
So, we have following examples. So, here we
will ask that find all solutions all solutions
of the following systems of the following
systems if consistent. So, first system is
this x plus 2 y minus 3 z is equal to minus
1 , 3 x minus y plus 2 z is equal to 7, 5
x plus 3 y minus 4 z is equal to 2. So, we
shall see whether the system is consistent
first. Here the co-efficient matrix is co-efficient
matrix A is this matrix that is, 1 2 minus
3 3 minus 1 2 5 3 minus 4. Augmented matrix
is Augmented matrix A tilta is the matrix:
1 2 minus 3 minus 1 3 minus 1 2 7 5 3 minus
4 2. So, next we shall find out rank of these
two matrices, and check whether they are equal
or not. So, for this we consider the augmented
matrix only.
So, now, we consider this consider the augmented
matrix, and find its echelon form. So, the
matrix is: 1 2 minus 3 minus 1 3 minus 1 2
7 5 3 minus 4 2. So, for finding this echelon
form, here we replace this row 2 by row 2
minus 3 row 1 and also this row 3 we replace
by row 3 minus 5 times row 1. So, the matrix
we get is like this: 1 2 minus 3 minus 1 0
minus 7 11 10 0 minus 7 11 7. So next, again
we apply elementary row operations. And here
we replace this row 3 by row 3 minus row 2.
So, the resultant matrix will be 1 2 minus
3 minus 1 0 minus 7 11 10 0 0 0 minus 3. So,
here in this echelon form of the augmented
matrix, we can see that first 3 columns of
this matrix is the echelon form of matrix
A.
So, therefore, the rank of here first three
columns is the echelon form of the co-efficient
matrix A of the matrix A. So, here in this
echelon form of A we are a getting zero row.
So, rank of this co-efficient matrix is 2
and rank of this augmented matrix A tilta
is equal to 3. So therefore, rank of A is
not equal to rank of this augmented A tilta.
And hence, the system is inconsistent the
system is inconsistent; that means, the system
has no solutions.
So, next we will see another example; that
example two. Here the system is like this;
2 x plus y minus 2 z is equal to 10, 3 x plus
2 y plus 2 z is equal to 1, 5 x plus 4 y plus
3 z is equal to 4. So, here again we will
find rank of the co-efficient matrix and rank
of the augmented matrix for checking the consistency
condition. So therefore, we find echelon form
of the augmented matrix. So, here we find
echelon form of the augmented matrix. This
is given by 2 1 minus 2 10 3 2 2 1 5 4 3 4.
So, here we make the following operation:
That elementary row operations that row 2
we replace by 2 row 2 minus 3 row 1 3 3 times
row 1 and R 3 we replace by 2 R 3 minus 5
times row 1. And we get the matrix be like
this that is: 2 1 minus 2 10 0 1 10 minus
28 0 3 16 minus 42.
So next, we apply the following elementary
row operation that is row 3 replace we replace
by row 3 minus 3 times row 2. And get the
matrix be like this that is 2 1 minus 2 10
0 1 10 minus 28 0 0 minus 14 42. So, in the
echelon form of the augmented matrix, the
first 3 columns is the echelon form of the
coefficient matrix. So, notice that rank of
this coefficient matrix is equal to 3 as well
as the rank of the augmented matrix both are
equal to 3, and that is equal to the number
of this is equal to number of variables, and
hence the system is consistent hence the system
is consistent So, here we shall see how to
find solutions of the system.
So, from this echelon form of the augmented
matrix we get following system that is the
echelon form of the given system. So, echelon
form echelon form of the augmented matrix
gives that gives the system; 2 x plus y minus
2 z is equal to 10, y plus 10 z is equal to
minus 28 and minus 14 z is equal to 42. So,
from the last equation; from third equation
we get z equal to minus 3; putting this in
equation 2 We get y is equal to 2 and finally
from equation one we get x equal to 1. So
therefore, therefore the unique solution of
the system of the system is x equal to 1,
y equal to 2 and z equal to minus 3. So, next
we see another equation. System of equations
in this third systems is like this; x plus
2 y minus 3 z equal to 6, 2 x minus y plus
4 z is equal to 2, and 4 x plus 3 y minus
2 z is equal to 14. So, for this system also
we find echelon form of the augmented matrix.
And here the augmented matrix is like this.
So, here we find echelon form of the augmented
matrix. So, the augmented matrix is 1 2 minus
3 6 2 minus 1 4 2 4 3 minus 2 14. So, we apply
this elementary row operations, that we replace
row 2 by row 2 minus 2 row 1, and row 3 we
replace by row 3 minus 4 times row 1, and
we get the matrix be like this. 1 2 minus
3 6 0 minus 5 10 minus 10 0 minus 5 10 minus
10, then applying this elementary row operation
that, replacing row 3 by row 3 minus row 2,
we get the matrix be 1 2 minus 3 6 0 minus
5 10 minus 10 and 0 0 0 0. So, here you see
that the rank of the coefficient matrix A
is same as rank of the augmented matrix and
that is equal to 2, but that is strictly less
than 3 the number of variables. Hence the
system is consistent, and has infinite number
of solutions, infinite number of solutions.
So, next we will find the solutions of this
system. So, again from the echelon form of
the augmented matrix we get the system be
like this. So, here the here the echelon form
of echelon form of augmented matrix matrix
gives the system
So, this system is like this that is 2 x plus
y minus 2 z is equal to 10, y plus 10 z is
equal to minus 2. Sorry, that we will get
this; sorry, we will not get this one. The
system is a like this: x plus 2 y minus 3
z is equal to 6, minus 5 y plus 10 z is equal
to minus 10. So this system we get from the
echelon form of the augmented matrix, or we
can write this system be like this or x plus
2 y minus 3 z equal to 6 and y minus 2 z is
equal to that is 2. So, here we have three
variables and two equations. So, one variable
we take as free variable. We take this variable
z as the free variable. In fact, the variables
which do not occur at the beginning of any
equation in the echelon form is called a free
variable. So, z is a free variable here free
variable here; that means, it can take any
value of the field.
So, let z is equal to alpha and alpha belongs
to this real numbers. So, from equation two
we get, then this value of y is equal to 2
plus 2 alpha, and value of x will be 2 minus
alpha. So, the set of all solutions the set
of all solutions the of the system is like
this: 2 minus alpha, 2 plus 2 alpha, alpha,
such that alpha belongs to the set of all
real numbers. Here we are considering the
system over this set of real numbers. That
is why this alpha will vary over all the elements
of this real field. So, next we see another
example, and that is a system homogeneous
system of linear equations. So, next example
is a homogeneous system of linear equations.
So, here find all possible solutions of the
system below. Also find the dimension of the
solution space. So, here the system is like
this: x plus 2 y minus z equal to 0, 2 x plus
5 y plus 2 z is equal to 0, x plus 4 y plus
7 z is equal to 0, x plus 3 y plus 3 z equal
to 0. So, this is the system of linear equations,
and this system is a homogeneous system, because
all this b i(s) are equal to 0. So, here we
shall find out rank of the co-efficient matrix
to check whether the system has unique solution
or infinite number of solutions. So, echelon
form of echelon form of the co-efficient matrix
is matrix.
So, the co-efficient matrix is given by: 1
2 minus 1 2 5 2 1 4 7 1 3 3. So, here we apply
the following elementary row operations, that
row 2 we replace by row 2 minus 2 row 1, row
3 we replace by row 3 minus row 1, row 4 we
replace by row 4 minus row 1. And we get this
matrix here that: 1 2 minus 1 0 1 4 0 2 4
0 2 this will be 8 and 0 1 4. So here, we
next apply the following elementary row operation
that we replace row 3 by half of of row 3,
and get the matrix be like this: 1 2 minus
1 0 1 4 0 1 4 0 1 4. So here all this 3 rows;
row 3 row 2 row 3 and row 4 are identical.
So, here we replace this row 3 by row 3 minus
row 2, and row 4 also we replace by row 4
minus row 2. And we get the matrix be: 1 2
minus 1 0 1 4 0 0 0 0 0 0. So, from here we
get that the rank of coefficient matrix A
is equal to 2, and this is strictly the number
of variables that is 3.
So, the system has infinite number of solutions
infinite number of solutions. And these solutions,
we get like this. We get the system below
system below from the echelon of the coefficient
matrix that that is also called echelon form
of the system. And it is x plus 2 y minus
z equal to 0, and y plus 4 z is equal to 0.
So, again we consider this z as free variable
that is variable which do not appear at the
beginning of any of the equations in echelon
form. So, z is the free variable. So, let
z equal to alpha; that belongs to R. Then
we get y is equal to minus 4 alpha, and x
is equal to 9 alpha. So, this set of all solution
are the solution space the solution space
of the given system is, this set S that is
consists of all vectors like this 9 alpha,
minus 4 alpha, alpha, where alpha belongs
to this set of all real numbers.
Next, we find out the dimension of the solution
space. So, that dimension of the solution
space S is also related with rank of the matrix
A.
Here, if we consider matrix A is a linear
transformation, then this S will be the nullity
of this linear transformation A. Therefore,
from the rank nullity theorem also we get
this result that, if we think A as a linear
transformation linear transformation, then
S is the null space of A. So, from the rank
nullity theorem rank nullity theorem, we get
that dimension of the solution space S is
equal to total number of variables that is
3 minus rank of A. So, that is 3 minus 2 that
is equal to 1. So, this is how we solve system
of linear equations; both homogeneous and
non homogeneous. And we shall come across
the system of equations; latter on while finding
eigenvalue and eigenvectors of matrices.
So, next we shall discuss about this another
important concept in linear algebra is this
eigenvalue and eigenvector of matrices. This
is an important concept of linear algebra,
and this is very useful in engineering and
sciences sciences. Engineers and scientist,
they come across this concept of eigenvalue
and eigenvectors called open. So, first we
shall give definition of an eigenvalue. So,
here basically we find eigenvalue and eigenvectors
of square matrices. So, let A be an n by n
matrix over a field F, then an element of
F; an element lambda in F is called an eigenvalue
of A if there exist a non-zero vector x in
F n, such that this equations holds A x is
equal to lambda x. Here, we consider this
x as a column vector that is, here we consider
x as an n cross 1 matrix are a column vector.
So, if this lambda is an eigenvalue of A,
if lambda is an eigenvalue of A and x is a
non-zero vector, non-zero vector such that
Ax is equal to lambda x , then x is called
an eigenvector corresponding an eigenvector
corresponding to the eigenvalue lambda. Notice
that eigenvectors corresponding to lambda
need not be unique. There may be several eigenvectors
corresponding to an eigenvalue. So, here in
fact, if we we have this following lemma that
says that if x 1 and x 2 are eigenvalues to
eigenvalue to eigenvectors corresponding to
an eigenvalue, then their sum is also an eigenvector
corresponding to this same eigenvalue, and
also scalar multiple of any eigenvector will
be again an eigenvector.
So, in other words, if we include the zero
vectors to the set of all eigenvectors corresponding
to lambda, then that forms a sub space. So,
here we consider that, let A be an n by n
matrix 
and lambda be an eigenvalue of A. let S be
the set of all eigenvectors all eigenvectors
corresponding to lambda to lambda, then this
S union the zero vector is a subspace of this
field F n. So, here this proof is not difficult,
it is trivial. So let us consider let x and
y be two eigenvectors corresponding to lambda
and alpha, beta be two scalar from F. So,
here this alpha x plus beta y will be an eigenvector
corresponding to this eigenvalue lambda or
not; that is what we check. S, A of alpha
x plus beta y is equal to alpha Ax plus beta
Ay, and this is equal to alpha into lambda
x plus beta into lambda y or this can be written
as lambda times alpha x plus beta y. So, this
S union 0 is a subspace. And this subspace
is called eigenspace corresponding to the
eigenvalue.
We give this definition of an eigenspace.
So, if S is S is the set of all eigenvectors
corresponding to corresponding to an eigenvalue
lambda of A, then the subspace S union 0 is
called the eigenspace eigenspace corresponding
to corresponding to this eigenvalue lambda.
So, let us see an example quickly. Here we
consider this matrix consider this matrix
A be like this: 5 4 1 2 over this real numbers
of course. And a vector x be like this: 4
1. Now, this Ax that is 5 4 1 2 that is multiplied
by 4 1 that gives 24 6 that is 6 times 4 1,
so that is equal to 6 times x. So, this implies
that 6 is an eigenvalue of A an eigenvalue
of A; and 4 1 is an eigenvector corresponding
to the eigenvalue 6. So, we stop this lecture
here. And in the next lecture, we will discuss
about method to find eigenvalues and eigenvectors
of matrices. So, that is all for this lecture.
Thank you. .
