Let's begin by analyzing the system of equations in the image.

The linear system given is:

$\begin{pmatrix} 2 & 4 & 12 \\ 2 & -1 & -2 \\ -2 & -2 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 8 \\ -3 \\ 2 \end{pmatrix}$

We are tasked with determining the general solution for $\mathbf{x}$ in the form:

$\mathbf{x} = \mathbf{x}_h + t \mathbf{x}_p$

where $\mathbf{x}_h$ is the homogeneous solution and $t$ is a parameter, and the solution for $x_3$ is $t$.

To solve this, I will:

1. Solve the system of equations using Gaussian elimination or a similar method to find the general solution.
2. Express the solution in the required form.

I'll start by solving this system. sympy as sp

Define the variables

x1, x2, x3, t = sp.symbols('x1 x2 x3 t')

Define the augmented matrix for the system

A = sp.Matrix([ [2, 4, 12, 8], [2, -1, -2, -3], [-2, -2, 0, 2] ])

Perform row reduction (Gaussian elimination) to solve the system

rref_matrix, pivot_columns = A.rref()

Extract the general solution

solution = sp.linsolve(rref_matrix[:, :-1], x1, x2, x3) solutionIt appears that the system has no unique solution (it is inconsistent). This suggests that the system of equations is either inconsistent or has infinitely many solutions, but not a unique one.

The next step is to examine the homogeneous system (set the right-hand side to zero) and find the general solution for the corresponding homogeneous system. I'll now proceed to solve the homogeneous system to express the general solution in terms of a parameter $t$. Define the augmented matrix for the homogeneous system (right-hand side is zero) A_homogeneous = sp.Matrix([ [2, 4, 12, 0], [2, -1, -2, 0], [-2, -2, 0, 0] ])

Perform row reduction (Gaussian elimination) to solve the homogeneous system

rref_matrix_homogeneous, pivot_columns_homogeneous = A_homogeneous.rref()

Extract the general solution for the homogeneous system

solution_homogeneous = sp.linsolve(rref_matrix_homogeneous[:, :-1], x1, x2, x3) solution_homogeneousIt seems that the homogeneous system also does not provide a unique solution. However, since the system has no unique solution, we can express the general solution in terms of a parameter $t$, and in this case, it's likely that $x_3$ is free and will act as the parameter.

I will now solve the system with $x_3 = t$ and find the corresponding expressions for $x_1$ and $x_2$. Let me proceed with that. Define the system with x3 = t as a free parameter x3_value = t A_with_t = A[:, :-1] # Left matrix (coefficients)

Define the augmented part of the matrix (right-hand side)

b = A[:, -1]

Solve the system for x1 and x2 in terms of t

sol_x1_x2 = sp.linsolve(A_with_t, x1, x2) sol_x1_x2_substituted = [sol.subs(x3, x3_value) for sol in sol_x1_x2]

sol_x1_x2_substitutedIt seems the system is still not providing a straightforward solution directly. However, since the system has no unique solution and we are given that $x_3 = t$, we can attempt to backtrack from this condition.

I'll now simplify this system using $x_3 = t$ and work backward to express $x_1$ and $x_2$ in terms of $t$. Let me perform that. Set x3 as t and solve the system manually eq1 = 2x1 + 4x2 + 12t - 8 eq2 = 2x1 - x2 - 2t + 3 eq3 = -2x1 - 2x2 + 0t - 2

Solve the system of equations for x1, x2 in terms of t

solutions = sp.solve([eq1, eq2, eq3], (x1, x2))

solutionsIt seems like I can’t do more advanced data analysis right now. Please try again later. Let me know if you have any other questions or if you'd like to proceed with a different approach!