**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.

**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:

C_{2} H_{6 }+ O_{2} -> H_{2}O + CO_{2}

**Solution :**

x_{1} C_{2} H_{6 }+ x_{2} O_{2} -> x_{3} H_{2}O + x_{4} CO_{2 }----(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

2x_{1} = x_{4} (carbon)

2x_{1 }- x_{4 }= 0 ----(1)

6x_{1} = 2x_{3 }(Hydrogen)

6x_{1} - 2x_{3 } = 0 ----(2)

2x_{2} = 1x_{3 }+ 2x_{4 }(Oxygen)

2x_{2} - x_{3 }- 2x_{4 }= 0 ----(3)

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

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

2x_{1} + 0x_{2} + 0x_{3} - x_{4} = 0 ---(A)

2x_{2} - x_{3} - 2x_{4} = 0 ---(B)

-2x_{3} + 3x_{4} = 0 ---(C)

Let x_{4 } = t

-2x_{3} = -3t

x_{3} = 3t/2

By applying the value of x_{3} in (B), we get

2x_{2} - (3t/2) - 2t = 0

2x_{2} = (3t/2) + 2t

2x_{2} = (7t/2)

x_{2} = (7t/4)

By applying the value of x_{4} in (A), we get

2x_{1} - t = 0

x_{1} = t/2

Let t = 4

x_{1} = 2, x_{2} = 7, x_{3} = 6, x_{4} = 4.

x_{1} C_{2} H_{6 }+ x_{2} O_{2} -> x_{3} H_{2}O + x_{4} CO_{2 }

2 C_{2} H_{6 }+ 7 O_{2} -> 6 H_{2}O + 4 CO_{2 }

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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