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 H6 + O2 -> H2O + CO2
Solution :
x1 C2 H6 + x2 O2 -> x3 H2O + x4 CO2 ----(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 - x4 = 0 ----(1)
6x1 = 2x3 (Hydrogen)
6x1 - 2x3 = 0 ----(2)
2x2 = 1x3 + 2x4 (Oxygen)
2x2 - x3 - 2x4 = 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 x4 = 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 H6 + x2 O2 -> x3 H2O + x4 CO2
2 C2 H6 + 7 O2 -> 6 H2O + 4 CO2
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Aug 31, 24 09:05 PM
Aug 31, 24 08:51 PM
Aug 31, 24 06:59 PM