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Gaussian elimination is a popular method for solving systems of linear equations. It helps us transform a matrix into an easier form to work with, typically achieving what's known as row-echelon form or even further into reduced row-echelon form. This transformation is accomplished through a series of row operations:

• Swapping rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row to another

These operations simplify the process of solving systems of equations by making the coefficients more straightforward to handle.
In the context of our problem, Gaussian elimination aided in eliminating the variable \( x_1\) from the equations, leading to a simpler system where \( x_2\) and \( x_3\) could be solved directly. This set of linear equations was reduced step-by-step to forms that allowed straightforward back-substitution, helping us find the unique solution for the vector \( \mathbf{x} \).

A system of equations is a collection of two or more equations with the same set of unknowns. In the context of linear transformations and matrices, this often involves representing the transformation as a matrix and formulating the equations in terms of the vector components.
In our case, substituting the matrices and vectors into the equation \( T(\mathbf{x}) = A \mathbf{x} = \mathbf{b} \) led to a system of three linear equations:

• \( 1x_1 + 0x_2 - 2x_3 = -1 \)
• \( -2x_1 + 1x_2 + 6x_3 = 7 \)
• \( 3x_1 - 2x_2 - 5x_3 = -3 \)

These equations describe the relationships between the components of the vector \( \mathbf{x} \) that map to \( \mathbf{b} \) under transformation by \( A \). Solving these simultaneously tells us the specific vector \( \mathbf{x} \) that corresponds to \( \mathbf{b} \).

The rank of a matrix is a concept that tells us about the number of linearly independent rows or columns in a matrix. It provides us with insights about the solutions of a system of equations.
For a square matrix, like the one in our problem, the full rank implies that every row (or column) is linearly independent, which signifies that the system of equations has a unique solution.
In our example, the matrix \( A \) has three rows, and it was found to have full rank (3). This means there are exactly three linearly independent equations, and thus, no extra degrees of freedom. As a result, the solution vector \( \mathbf{x} \) that we determined is unique, without any free variables left open-ended.

A vector space is a collection of vectors that can be scaled and added together, following specific arithmetic rules. These spaces are foundational to understanding linear equations and transformations, since solutions often form a vector space themselves.
In linear algebra, vector spaces provide a framework where operations like vector addition and scalar multiplication are always well-defined.
In the exercise, finding the vector \( \mathbf{x} \) whose image under transformation is \( \mathbf{b} \) involves understanding the space in which \( \mathbf{x} \) resides. Specifically, the rank of the matrix \( A \) ensured that the space was of full dimension, allowing for the unique transformation of \( \mathbf{x} \) into \( \mathbf{b} \). Each component of \( \mathbf{x} \) and its unique representation within the solution set is rooted in the vector space defined by the matrix system.