Answer:
(x, y, z) = [x, (-3.5x - 0.5), (4x - 1)]
K = -2.
The solution to the extended equation isn't unique; there are infinite solutions.
But, the value of K = - 2 makes it all consistent.
Explanation:
Note that
푥 = x
푦 = y
푧 = z
푘 = K
So, the expression becomes,
3푥+2푦+푧=1−푥+2푦+2푧=−1−4푥+푧=푘
3x+2y+z = 1−x+2y+2z = −1−4x+z = K
We're told to solve this simultaneous equation, using Gaussian Elimination
We first obtain the simultaneous equation.
3x+2y+z = 1−x+2y+2z
4x - z = 1 (eqn 1)
1−x+2y+2z = −1−4x+z
3x + 2y + z = -2 (eqn 2)
3x+2y+z = −1−4x+z
7x + 2y = -1 (eqn 3)
4x - z = 1
3x + 2y + z = -2
7x + 2y = -1
We then put the 3 equations in an augmented matrix with the RHS of the equation as the augmented part
[4 0 -1 | 1]
[3 2 1 | -2]
[7 2 0 | -1]
We then perform row reduction methods on the matrix to obtain its Echelon form
This reduction is done more clearly in the attached image to this answer.
First step, R₁ --> R₁/4
[1 0 -0.25 | 0.25]
[3 2 1 | -2]
[7 2 0 | -1]
Second step, R₂ --> R₂ - 3R₁; R₃ --> R₃ - 7R₁
[1 0 -0.25 | 0.25]
[0 2 1.75 | -2.75]
[0 2 1.75 | -2.75]
Third and final step, R₃ --> R₃ - R₂
[1 0 -0.25 | 0.25]
[0 2 1.75 | -2.75]
[0 0 0.00 | 0.00]
The matrix ends up with all zeros in the last row. Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, we return to one of the original equations and solve for y and z
4x - z = 1
3x + 2y + z = -2
7x + 2y = -1
From eqn 1; z = 4x - 1
From eqn 3; y =( -7x - 1)/2
y = -3.5x - 0.5
So, the solution to this system is
(x, y, z) = [x, (-3.5x - 0.5), (4x - 1)]
The values of K that make the equation obtained consistent
From the first extended equation,
3x+2y+z = 1−x+2y+2z = −1−4x+z = K
3x + 2y + z = K
But from eqn 2
3x + 2y + z = -2
So,
3x + 2y + z = K = -2
K = -2.
The solution to the extended equation isn't unique; there are infinite solutions.
But the value of K = - 2 makes it all consistent.
Hope this Helps!!!