How To Find Non Trivial Solution Of A Matrix

SOLVING EQUATIONS FOR Footling AND NON Fiddling 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 two < iii.

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 go

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, ten − y − 2z = 0, 3x + y + 3z = 0

Solution :

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

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

Question ii :

Determine the values of λ for which the post-obit organization of equations x + y + 3z = 0, 4x + 3y + λz = 0, 2x + y + 2z = 0 has (i) a unique solution (ii) a not-trivial solution.

Solution :

(i) a unique solution

If λ ≠ eight, then rank of A and rank of (A, B) will be equal to three.Information technology will have unique solution.

(ii) a non-little solution.

If λ =  viii, and so rank of A and rank of (A, B) will be equal to 2.It will accept not trivial solution.

Question 3 :

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

C₂ H₆+ O₂ -> H₂O + CO₂

Solution :

10₁ C ₂  H ₆ + 10_ii O ₂  -> x₃ H ₂ O + ten_four CO ₂ ----(1)

The number of carbon atoms on the left-manus side of (1) should exist equal to the number of carbonatoms on the correct-paw side of (one). And then we get a linear homogenous equation

2x ₁    =  10₄  (carbon)

2x_i - x_four =  0  ----(1)

6x₁   =  2x_iii (Hydrogen)

6x₁ - 2x₃ =  0  ----(2)

2x₂  =  one x ₃+2x ₄ (Oxygen)

2x_two -10₃- 2x₄ =  0  ----(3)

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

So the system is consequent and it has infinitely many solution.

2x_i + 0x_ii + 0x₃ - ten₄  =  0   ---(A)

2x₂ - x_iii - 2x₄  =  0   ---(B)

-2x₃ + 3x₄  =  0   ---(C)

Permit x₄ =  t

-2x ₃   = -3t

ten₃  = 3t/2

By applying the value of 10 ₃  in (B), we get

2x_two - (3t/2) - 2t  =  0

2x_two =  (3t/two) + 2t

2 x _two  =  (7t/2)

ten₂ =  (7t/4)

Past applying the value of ten_four in (A), nosotros get

2x_ane - t  =  0

10₁  =  t/2

Allow t = four

10 ₁   =  2, 10 ₂  =  7,ten₃  = 6, x₄  =  four.

ten₁ C₂ H₆+ ten_two O_ii -> x₃ H₂O + 10₄ CO₂

ii C₂ H₆+ vii O₂ -> 6 H₂O + 4 CO₂

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How To Find Non Trivial Solution Of A Matrix,

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