Unit - 1
Linear Algebra- Matrices
- Reduce the following matrix into normal form and find its rank,
Let A =
Apply we get
A
Apply we get
A
Apply
A
Apply
A
Apply
A
Hence the rank of matrix A is 2 i.e. .
2. Reduce the following matrix into normal form and find its rank,
Let A =
Apply and
A
Apply
A
Apply
A
Apply
A
Apply
A
Hence the rank of the matrix A is 2 i.e. .
3. Reduce the following matrix into normal form and find its rank,
Let A =
Apply
Apply
Apply
Apply
Apply and
Apply
Hence the rank of matrix A is 2 i.e. .
4. Examine whether the following vectors are linearly independent or not.
and .
Solution:
Consider the vector equation,
i.e. … (1)
Which can be written in matrix form as,
R12
R2 – 3R1, R3 – R1
R3 + R2
Here Rank of coefficient matrix is equal to the no. Of unknowns. i.e. r = n = 3.
Hence the system has unique trivial solution.
i.e.
i.e. vector equation (1) has only trivial solution. Hence the given vectors x1, x2, x3 are linearly independent.
5. At what value of P the following vectors are linearly independent.
Solution:
Consider the vector equation.
i.e.
Which is a homogeneous system of three equations in 3 unknowns and has a unique trivial solution.
If and only if Determinant of coefficient matrix is non zero.
consider .
.
i.e.
Thus for the system has only trivial solution and Hence the vectors are linearly independent.
6. Determine the eigen values of eigen vector of the matrix.
Solution:
Consider the characteristic equation as,
i.e.
i.e.
i.e.
Which is the required characteristic equation.
are the required eigen values.
Now consider the equation
… (1)
Case I:
If Equation (1) becomes
R1 + R2
Thus
independent variable.
Now rewrite equation as,
Put x3 = t
&
Thus .
Is the eigen vector corresponding to .
Case II:
If equation (1) becomes,
Here
independent variables
Now rewrite the equations as,
Put
&
.
Is the eigen vector corresponding to .
Case III:
If equation (1) becomes,
Here rank of
independent variable.
Now rewrite the equations as,
Put
Thus .
Is the eigen vector for .
7. Find the eigen values of eigen vector for the matrix.
Solution:
Consider the characteristic equation as
i.e.
i.e.
are the required eigen values.
Now consider the equation
… (1)
Case I:
Equation (1) becomes,
Thus and n = 3
3 – 2 = 1 independent variables.
Now rewrite the equations as,
Put
,
i.e. the eigen vector for
Case II:
If equation (1) becomes,
Thus
Independent variables.
Now rewrite the equations as,
Put
Is the eigen vector for
Now
Case II:-
If equation (1) gives,
R1 – R2
Thus
independent variables
Now
Put
Thus
Is the eigen vector for .