System of linear equations
Dr. S. RAMESH
e-mail: rsiriset@gitam.edu
Aim
To find the solution of the system of linear equations
Tools
Row-echelon form a matrix.
Types of linear equations
We have two types of linear equations
1. Homogeneous linear equations
2. Non-homogeneous linear equations
System of linear homogeneous equations
Consider the system of homogeneous linear equations
a11 x1 + a12 x2 + · · · + a1n xn = 0
a21 x1 + a22 x2 + · · · + a2n xn = 0
···
···
···
am1 x1 + am2 x2 + · · · + amn xn = 0
containing m-equations and n-unknowns (x1 , x2 , · · · , xn ).
Matrix form
The above system of equations we can represent them in matrix form
that is
x1 0
a11 a12 · · · a1n x2 0
a21 a22 · · · a2n
·
= ·
· · · · · ·
· ·
· · · · · ·
· ·
am1 am2 · · · amn
xn 0
That is AX = O
where A is called co-efficient matrix, X is called unknown matrix.
Consistence
Given system of equations are said to be consistent if there is a non-zero
(non-trivial) solution. Otherwise it is called inconsistent.
Remarks
1. A system AX = 0 is consistent if and only if det(A) = 0.
2. A system AX = 0 has only zero-solution if and only if det(A) 6= 0.
3. The above system has a non-zero solution if ρ(A) = r < m ≤ n. In
this case, system has (n − r ) number of variables are independent.
How to find the solution
1. zero solution is always exist.
2. To find non-zero solution, we have to reduce the matrix A into
row-echelon form and then rewrite the system, hence we can get the
non-zero solution.
Problem
Find the solution of the system of equations
x + 2y + 3z = 0; 3x + 4y + 4z = 0; 7x + 10y + 12z = 0.
Solution
1 2 3 x 0
We can rewrite the system as 3 4 4 y = 0
7 10 12 z 0
1 2 3 1 2 3
Now, 3 4 4 → R2 − 3R1 , R3 − 7R1 0 −2 −5
7 10 12 0 −4 −9
1 2 3
→ R3 − 2R1 0 −2 −5
0 0 1
Therefore the above matrix is in row-echelon form.
We can rewrite the given system using row-echelon form, that is
z = 0; −2y − 5z = 0 and x + 2y + 3z = 0. Therefore we get
z = 0, y = 0, x = 0.
Another way of conclusion
From the row-echelon form the given system, we can say that ρ(A) = 3.
Therefore det(A) 6= 0 and hence the given system has only zero solution.
Test for consistency and solve the following system of
equations.
1. x1 − x2 = 0; 2x1 + x2 = 0.
2. 2x1 + x2 = 0; x1 + x2 = 0.
3. x − z = 0; y + 3z = 0.
4. 4x + 2y + z + 3w = 0; 6x + 3y + 4z + 7w = 0; 2x + y + w = 0.
5. x + y − 3z + 2w = 0; 2x − y + 2z − 3w = 0;
3x − 2y + z − 4w = 0; −4x + y − 3z + w = 0.
6. x1 + 2x3 − 2x4 = 0; 2x1 − x2 − x4 = 0;
x1 + 2x3 − x4 = 0; 4x1 − x2 + 3x3 − x4 = 0.
7. x + y − z + t = 0; x − y + 2z − t = 0; 3x + y + t = 0.
8. −x + y + 4z = 0; x + 3y + 8z = 0; 12 x + y + 52 z = 0.
1
9. + 2y − 6z = 0; −4x + 5z = 0;
2x
−3x + 6y − 13z = 0; − 13 x + 2y − 83 z = 0.
Another problems
1 Determine the values of λ for which the set of equations may
possess non-trivial solution
3x + y − λz = 0; 4x − 2y − 3z = 0; 2λx + 4y + λz = 0. (Hint:
λ = 1, −9). Obtain the solution in each case.
Solution
Rewrite the given system in matrix form, that is
3 1 −λ x 0
4 −2 −3 y = 0
2λ 4 λ z 0
The determinant of the matrix A is λ2 + 8λ − 9 = 0
2 Determine the values of k for which the set of equations may
possess non-trivial solution
(k − 1)x + (4k − 2)y + (k + 3)z = 0;
(k − 1)x + (3k + 1)y + 2kz = 0; 2x + (3k + 1)y + 3(k − 1)z = 0.
(Hint: k = 0, 3). Obtain the solution in each case.
3 Find the values of λ, for the system of equations
2x − 2y + z = λx; 2x − 3y + 2z = λy ; −x + 2y = λz
can possess a non-trivial solution only if λ = 1, −3. Obtain the
solution in each case.
System of linear non-homogeneous equations
Consider the system of non-homogeneous linear equations
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
···
···
···
am1 x1 + am2 x2 + · · · + amn xn = bm
containing m-equations and n-unknowns (x1 , x2 , · · · , xn ).
Matrix form
The above system of equations we can represent them in matrix form
that is
x1 b1
a11 a12 · · · a1n
a21 a22 · · · a2n x2
b2
·
= ·
· · · · · ·
· ·
· · · · · ·
· ·
am1 am2 · · · amn
xn bm
That is AX = B
where Am×n is called co-efficient matrix, Bm×1 , Xn×1 is called unknown
matrix.
Augment matrix
The matrix K = [A : B]m×(n+1) is called argument matrix of the given
system.
Consistence
Given system of non-homogeneous equations are said to be consistent if
there is a solution. Otherwise it is called inconsistent.
Rouche’s Remarks
1. A system AX = B is consistent if and only if ρ(A) = ρ(K ).
2. A system AX = B is inconsistent if and only if ρ(A) 6= ρ(K ).
3. A system AX = B has a unique solution if ρ(A) = ρ(K ) = n = the
number of unknowns.
4. A system AX = B has an infinite number of solutions if
ρ(A) = ρ(K ) < n = the number of unknowns. In this case the given
system has (n − r ) number of variables are independent.
How to find the solution
To find solution, we have to reduce the matrix K into row-echelon form
and then rewrite the system, hence we can get the solution.
Test for consistency and solve the following
1. x + y + z = 4; 2x + 5y − 2z = 3; x + 7y + 7z = 5.
2. x + 2y + z − 2; 3x + y − 2z = 1; 4x − 3y − z = 3; 2x + 4y + 2z = 4.
3. x − 4y + 7z = 8; 3x + 8y − 2z = 6; 7x − 8y + 26z = 31
4. x + 2y − z = 3, 3x − y + 2z = 1; 2x − 2y + 3z = 2; x − y + z = −1.
5. x + y + z = 3; 3x − 5y + 2z = 8; 5x − 3y + 4z = 14.
6. 5x + 3y + 7z = 4; 3x + 26y + 2z = 9; 7x + 2y + 10z = 5.
7. x − y + 2z + t − 2 = 0; 3x + 2y + t − 1 = 0; 4x + y + 2z + 2t − 3 = 0.
8. 2x + 6y + 11 = 0; 6x + 20y − 6z + 3 = 0; 6y − 18z + 1 = 0.
9. x − y + 2z = 1; 2x + 2z = 1; x − 3y + 4z = 2.
10. x − 2y + z + 2t = 1; x + y − z + t = 2; x + 7y − 5z − t = 3.
11. 2x1 − 3x2 − 7x3 + 5x4 + 2x5 = −2; x1 − 2x2 − 4x3 + 3x4 + x5 = −2;
2x1 − 4x3 + 2x4 + x5 = 3; x1 − 5x2 − 7x3 + 6x4 + 2x5 = −7.
Another problems
1 Obtain for what values of λ and µ the equations
x + y + z = 6; x + 2y + 3z = 10; x + 2y + λz = µ
have (i) no solution (ii) a unique solution and (iii) an infinite number
of solutions. Find the solution in each case.
2 Obtain for what values of λ and µ the equations
2x + 3y + 5z = 9; 7x + 3y − 2z = 8; 2x + 3y + λz = µ
have (i) no solution (ii) a unique solution and (iii) an infinite number
of solutions. Find the solution in each case.
Problems continued
3 For what values of λ, the equations
x + y + z = 1; x + 2y + 4z = λ; x + 4y + 10z = λ2 .
have a solution and solve them completely in each case.
4 If a + b + c 6= 0, show that the system
−2x + y + z = a; x − 2y + z = b; x + y − 2z = c
has no solution. If a + b + c = 0, show that it has infinitely many
solutions.
5 Find the values of λ for which the equations
(2 − λ)x + 2y + 3 = 0; 2x + (4 − λ)y + 7 = 0; 2x + 5y + (6 − λ) = 0
are consistent and find the values x and y corresponding to each of
these cases.
THANK YOU
