DISCUSS THE NATURE OF SOLUTIONS OF LINEAR EQUATIONS IN THREE VARIABLES

Discuss the Nature of Solutions of Linear Equations in Three Variables :

Here we are going to see, how to discuss the nature of solutions of linear equations in three variables.

Discuss the nature of solutions 

Each equation in the system represents a plane in three dimensional space and solution of the system of equations is precisely the point of intersection of the three planes defined by the three linear equations of the system.

The system may have only one solution, infinitely many solutions or no solution depending on how the planes intersect one another.

  • By solving linear equations, if we get different values for x, y and z, then we may decide that they have unique solutions.
  • If you obtain a false equation such as 0 = 1, in any of the steps then the system has no solution.
  • If you do not obtain a false solution, but obtain an identity, such as 0 = 0 then the system has infinitely many solutions.

Discuss the Nature of Solutions of Linear Equations in Three Variables - Examples

Question 1 :

Discuss the nature of solutions of the following system of equations

(i) x + 2y −z = 6 ; −3x − 2y + 5z = −12 ; x −2z = 3

Solution :

x + 2y − z = 6 -----(1)

-3x − 2y + 5z = −12  -----(2)

x −2z = 3   -----(3)

(1) + (2)

     x + 2y − z = 6 

 -3x − 2y + 5z = −12

  ---------------------

   -2x + 4z  =  -6  -----(4)

From (3),

x  =  3 + 2z

By applying the value of x in (4)

-2(3 + 2z) + 4z  =  -6

-6 - 4z + 4z  =  -6

-6 + 6  =  0

  0  =  0

So, the system has infinitely many solution.

(ii) 2y +z = 3(−x +1) ; −x + 3y −z = −4 ; 3x + 2y + z  =  -1/2

Solution :

2y + z = 3(−x +1)

2y + z  =  -3x + 3

3x + 2y + z  =  3  ------(1)

 −x + 3y −z = −4  ------(2)

3x + 2y + z  =  -1/2

2(3x + 2y + z)  =  -1

6x + 4y + 2z  =  -1 ------(3)

(1) + (2)

3x + 2y + z  =  3

 −x + 3y −z = −4 

---------------------

2x + 5y  =  -1  ---(4)

2(2) + (3)

     −2x + 6y − 2z  = −8

       6x + 4y + 2z  =  -1

     ----------------------

     4x + 10y  =  -9  ---(5)

2(4) - (5)

  4x + 10y  =  -2

  4x + 10y  =  -9

  (-)   (-)      (+)

 ------------------ 

       0  =  7

Hence the system has no solution.

(iii)   (y + z)/4  =  (z + x)/3  =  (x + y)/2 ; x + y + z  = 27

Solution :

(y + z)/4  =  (z + x)/3

3(y + z)  =  4(z + x)

3y + 3z  =  4z + 4x

4x - 3y + 4z - 3z  =  0

4x - 3y + z  =  0 ---(1)

(z + x)/3  =  (x + y)/2 

2(z + x)  =  3(x + y)

2z + 2x  =  3x + 3y

3x - 2x + 3y - 2z  =  0

x + 3y - 2z  =  0 ----(2)

x + y + z  = 27  ----(3)

(1) + (2)

      4x - 3y + z  =  0

      x + 3y - 2z  =  0 

    --------------------

      5x - z  =  0 ----(4)

(1) - 3(3)

     4x - 3y + z  =  0 

     3x + 3y + 3z  = 81

     --------------------

      7x + 4z  =  81  ---(5)

4(4) + (5)

    20x - 4z  =  0

    7x + 4z  =  81

   ---------------

   27x  =  81

  x =  81/27  =  3

Hence the equations will have unique solution. By finding the solution further, we get

(4)  ==> 7x + 4z  =  81 

  7(3) + 4z  =  81

21 + 4z  =  81

4z  =  81 - 21

4z  =  60

z  =  15

(1) ==> 4x - 3y + z  =  0

  4(3) - 3y + 15  =  0

  12 + 15 - 3y  =  0

-3y  =  27

y  =  -9

After having gone through the stuff given above, we hope that the students would have understood, "Discuss the Nature of Solutions of Linear Equations in Three Variables". 

Apart from the stuff given in this section "Discuss the Nature of Solutions of Linear Equations in Three Variables"if you need any other stuff in math, please use our google custom search here.

Widget is loading comments...