Unit -5 MATRICES
elibraryrit.blogpot.com
RANK OF A MATRIX
Example 1.1
Find the rank of the matrix
Solution:
Let A=
Order of A is 2 × 2 ∴ ρ(A)≤ 2
Consider the second order minor
There is a minor of order 2, which is not zero. ∴ρ (A) = 2
Example 1.2
Find the rank of the matrix
Solution:
Let A=
Order of A is 2 × 2 ∴ρ(A)≤ 2
Consider the second order minor
Since the second order minor vanishes, ρ(A) ≠ 2
Consider a first order minor |−5| ≠ 0
There is a minor of order 1, which is not zero
∴ ρ (A) = 1
Example 1.3
Find the rank of the matrix
Solution:
Let A=
Order Of A is 3x3
∴ ρ (A) ≤ 3
Consider the third order minor 
There is a minor of order 3, which is not zero
∴ρ (A) = 3.
Example 1.4
Find the rank of the matrix
Solution:
Let A=
Order Of A is 3x3
∴ ρ (A) ≤ 3
Consider the third order minor
Since the third order minor vanishes, therefore ρ(A) ≠ 3
Consider a second order minor
There is a minor of order 2, which is not zero.
∴ ρ(A) = 2.
Example 1.5
Find the rank of the matrix
Solution:
Let A =
Order of A is 3 × 4
∴ ρ(A)≤ 3.
Consider the third order minors
Since all third order minors vanishes, ρ(A) ≠ 3.
Now, let us consider the second order minors,
Consider one of the second order minors
There is a minor of order 2 which is not zero.
∴ρ (A) = 2.
Echelon form and finding the rank of the matrix (upto the order of 3×4) : Solved Example Problems
Example 1.6
Find the rank of the matrix A=
Solution :
The order of A is 3 × 3.
∴ ρ(A) ≤ 3.
Let us transform the matrix A to an echelon form by using elementary transformations.
The number of non zero rows is 2
∴Rank of A is 2.
ρ (A) = 2.
Note
A row having atleast one non -zero element is called as non-zero row.
Example 1.7
Find the rank of the matrix A=
Solution:
The order of A is 3 × 4.
∴ ρ (A)≤3.
Let us transform the matrix A to an echelon form
The number of non zero rows is 3. ∴ ρ(A) =3.
Example 1.8
Find the rank of the matrix A=
Solution:
The order of A is 3 × 4.
∴ ρ(A) ≤ 3.
Let us transform the matrix A to an echelon form
The number of non zero rows is 3.
∴ ρ (A) =3.
Testing the consistency of non homogeneous linear equations (two and three variables) by rank method : Solved Example Problems
Example 1.9
Show that the equations x + y = 5, 2x + y = 8 are consistent and solve them.
Solution:
The matrix equation corresponding to the given system is
AX=B
Number of non-zero rows is 2.
ρ (A )= ρ ([ A, B]) = 2 = Number of unknowns.
The given system is consistent and has unique solution.
Now, the given system is transformed into
x + y = 5
y = 2
∴ (1) ⇒ x + 2 = 5
x = 3
Solution is x = 3, y = 2
Example 1.10
Show that the equations 2x + y = 5, 4x + 2 y = 10 are consistent and solve them.
Solution:
The matrix equation corresponding to the system is
ρ ( A ) = ρ ([ A, B]) = 1 < number of unknowns
∴ The given system is consistent and has infinitely many solutions.
Now, the given system is transformed into the matrix equation.
Let us take y = k, k ∈R
⇒ 2x + k = 5
x = 1/2 ( 5 − k)
x = 1/2 ( 5 − k) , y = k for all k ∈R
Thus by giving different values for k, we get different solution. Hence the system has infinite number of solutions.
Example 1.11
Show that the equations 3x − 2 y = 6, 6x − 4 y = 10 are inconsistent.
Solution:
The matrix equation corresponding to the given system is
∴The given system is inconsistent and has no solution.
Example 1.12
Show that the equations 2x + y + z = 5, x + y + z = 4, x − y + 2z = 1 are consistent and hence solve them.
Solution:
The matrix equation corresponding to the given system is
Obviously the last equivalent matrix is in the echelon form. It has three non-zero rows.
ρ( A ) = ρ( [A, B] )= 3 = Number of unknowns .
The given system is consistent and has unique solution.
To find the solution, let us rewrite the above echelon form into the matrix form.
x + y + z = 4 (1)
y + z = 3 (2)
3z = 3 (3)
(3)⇒ z = 1
(2)⇒ y = 3 − z = 2
(1) ⇒ x = 4 − y − z
x=1
∴ x = 1, y = 2, z = 1
Example 1.13
Show that the equations x + y + z = 6, x + 2 y + 3z = 14, x + 4 y + 7z = 30 are consistent and solve them.
Solution:
The matrix equation corresponding to the given system is
Obviously the last equivalent matrix is in the echelon form. It has two non-zero rows.
∴ ρ ( [A, B] ) = 2, ρ ( A) = 2
ρ ( A ) = ρ ( [A, B] ) = 2 < Number of unknowns.
The given system is consistent and has infinitely many solutions.
The given system is equivalent to the matrix equation,
x + y + z = 6 (1)
y + 2z = 8 (2)
(2)⇒ y = 8 − 2z,
(1)⇒ x = 6 − y − z = 6 − (8 − 2 z) − z = z – 2
Let us take z = k, k ∈R , we get x = k − 2, y = 8 − 2k , Thus by giving different values for k we get different solutions. Hence the given system has infinitely many solutions.
Example 1.14
Show that the equations x − 4 y + 7z = 14, 3x + 8 y − 2z = 13, 7x − 8 y + 26z = 5 are inconsistent.
Solution:
The matrix equation corresponding to the given system is
The last equivalent matrix is in the echelon form. [A, B] has 3 non-zero rows and [A] has 2 non-zero rows.
The system is inconsistent and has no solution.
Example 1.15
Find k, if the equations x + 2 y − 3z = −2, 3x − y − 2z = 1, 2x + 3y − 5z = k are consistent.
Solution:
The matrix equation corresponding to the given system is
For the equations to be consistent, ρ ( [A, B] ) = ρ ( A)= 2
∴21 + 7k = 0
7k = −21 .
k = −3
Example 1.16
Find k, if the equations x + y + z = 7, x + 2 y + 3z = 18, y + kz = 6 are inconsistent.
Solution:
The matrix equation corresponding to the given system is
For the equations to be inconsistent
ρ ( [A, B] ) ≠ ρ ( A)
It is possible if k − 2 = 0 .
K=2
Example 1.17
Investigate for what values of ‘a’ and ‘b’ the following system of equations
x + y + z = 6, x + 2 y + 3z = 10, x + 2 y + az = b have
(i) no solution (ii) a unique solution (iii) an infinite number of solutions.
Solution:
The matrix equation corresponding to the given system is
Case (i) For no solution:
The system possesses no solution only when ρ ( A )≠ ρ ([ A, B]) which is possible only when a − 3 = 0 and b − 10 ≠ 0
Hence for a = 3, b ≠ 10 , the system possesses no solution.
Case (ii) For a unique solution:
The system possesses a unique solution only when ρ ( A ) = ρ ([ A, B]) = number of unknowns.
i.e when ρ ( A ) = ρ ([ A, B]) = 3
Which is possible only when a − 3 ≠ 0 and b may be any real number as we can observe .
Hence for a ≠ 3 and b ∈R , the system possesses a unique solution.
Case (iii) For an infinite number of solutions:
The system possesses an infinite number of solutions only when
ρ ( A )= ρ ([ A, B]) < number of unknowns
i,e when ρ ( A)= ρ ([ A, B])= 2 < 3 ( number of unknowns) which is possible only when a − 3 = 0, b − 10 = 0
Hence for a = 3, b =10, the system possesses infinite number of solutions.
Example 1.18
The total number of units produced (P) is a linear function of amount of over times in labour (in hours) (l), amount of additional machine time (m) and fixed finishing time (a)
i.e, P = a + bl + cm
From the data given below, find the values of constants a, b and c
Estimate the production when overtime in labour is 50 hrs and additional machine time is 15 hrs.
Solution:
We have, P = a + bl + cm
Putting above values we have
6,950 = a + 40b + 10c
6,725 = a + 35b + 9c
7,100 = a + 40b + 12c
The Matrix equation corresponding to the given system is
∴ The given system is equivalent to the matrix equation
∴ The production equation is P = 5000 + 30l + 75m
Pat l = 50, m=15 = 5000 + 30(50) + 75(15)
=7625 units.
∴The production = 7,625 units.
Exercise 1.1 : Rank of a Matrix - Problem Questions with Answer, Solution | Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail |
Chapter: 12th Business Maths and Statistics : Applications of Matrices and Determinants
Exercise 1.1 : Rank of a Matrix
Exercise 1.1
1. Find the rank of each of the following matrices
2. If A = 
3. Solve the following system of equations by rank method
x + y + z = 9, 2x + 5y + 7z = 52, 2x − y − z = 0
4. Show that the equations 5x + 3y + 7z = 4, 3x + 26 y + 2z = 9, 7x + 2 y + 10z = 5 are consistent and solve them by rank method.
5. Show that the following system of equations have unique solution:
x + y + z = 3, x + 2 y + 3z = 4, x + 4 y + 9z = 6 by rank method.
6. For what values of the parameter λ , will the following equations fail to have unique solution: 3x − y + λ z = 1, 2x + y + z = 2, x + 2 y − λ z = −1 by rank method.
7. The price of three commodities X,Y and Z are x,y and z respectively Mr.Anand purchases 6 units of Z and sells 2 units of X and 3 units of Y. Mr.Amar purchases a unit of Y and sells 3 units of X and 2units of Z. Mr.Amit purchases a unit of X and sells 3 units of Y and a unit of Z. In the process they earn ₹5,000/-, ₹2,000/- and ₹5,500/- respectively Find the prices per unit of three commodities by rank method.
8. An amount of ₹5,000/- is to be deposited in three different bonds bearing 6%, 7% and 8% per year respectively. Total annual income is ₹358/-. If the income from first two investments is ₹70/- more than the income from the third, then find the amount of investment in each bond by rank method.
Answers:
Related Topics
Related Topics