# SOLVING EQUATIONS FOR TRIVIAL AND NON TRIVIAL SOLUTIONS

Solving Equations for Trivial and Nontrivial Solutions :

Here we are going to see some example problems of solving equations for trivial and non trivial solutions.

## Solving Equations for Trivial and Nontrivial Solutions - Questions

Question 1 :

Solve the following system of homogenous equations.

(i) 3x + 2y + 7z = 0, 4x − 3y − 2z = 0, 5x + 9y + 23z = 0

Solution :

rank of (A) is 2 and rank of (A, B) is 2 < 3.

3x + 2y + 7z  =  0 -----(1)

-17y - 34z  =  0 -----(2)

Let z = t

-17y  =  34t

y  =  34t/(-17)  =  -2t

By applying the value of z in (1), we get

3x + 2(-2t) + 7t  =  0

3x - 4t + 7t  =  0

3x  =  -3t

x  =  -t

Hence the solution is (-t, -2t, t)

(ii) 2x + 3y − z = 0, x − y − 2z = 0, 3x + y + 3z = 0

Solution :

Rank of A is 3 and rank of (A, B) is 3.

Since rank of A and rank of (A, B) are equal, it has trivial solution.

Question 2 :

Determine the values of λ for which the following system of equations x + y + 3z = 0, 4x + 3y + λz = 0, 2x + y + 2z = 0 has (i) a unique solution (ii) a non-trivial solution.

Solution :

(i) a unique solution

If λ ≠ 8, then rank of A and rank of (A, B) will be equal to 3.It will have unique solution.

(ii) a non-trivial solution.

If λ = 8, then rank of A and rank of (A, B) will be equal to 2.It will have non trivial solution.

Question 3 :

By using Gaussian elimination method, balance the chemical reaction equation:

C2 H+ O2 -> H2O + CO2

Solution :

x1 C2 H+ x2 O2 -> x3 H2O + x4 CO----(1)

The number of carbon atoms on the left-hand side of (1) should be equal to the number of carbon atoms on the right-hand side of (1). So we get a linear homogenous equation

2x1   =  x4  (carbon)

2x1 - x=  0  ----(1)

6x1   =  2x(Hydrogen)

6x1 - 2x =  0  ----(2)

2x2  =  1x2x4 (Oxygen)

2x2 - x- 2x=  0  ----(3)

rank of A is 3 = rank of (A, B) = 3 < 4

Then the system is consistent and it has infinitely many solution.

2x1 + 0x2 + 0x3 - x4  =  0   ---(A)

2x2 - x3 - 2x4  =  0   ---(B)

-2x3 + 3x4  =  0   ---(C)

Let x =  t

-2x3  = -3t

x3  = 3t/2

By applying the value of x3 in (B), we get

2x2 - (3t/2) - 2t  =  0

2x2 =  (3t/2) + 2t

2x2 =  (7t/2)

x2 =  (7t/4)

By applying the value of x4 in (A), we get

2x1 - t  =  0

x1  =  t/2

Let t = 4

x1  =  2, x2 =  7, x3  = 6, x4  =  4.

x1 C2 H+ x2 O2 -> x3 H2O + x4 CO

2 C2 H+ 7 O2 -> 6 H2O + 4 CO

After having gone through the stuff given above, we hope that the students would have understood, "Solving Equations for Trivial and Nontrivial Solutions".

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