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Gaussian elimination is a mathematical process used to solve systems of linear equations. It systematically reduces the equations by performing operations on the rows of a matrix. These operations include swapping rows, multiplying a row by a nonzero scalar, and adding or subtracting multiples of one row from another to create zeros beneath the leading coefficients of each row.

• The ultimate goal of Gaussian elimination is to transform the matrix into row-echelon form. This makes it much easier to solve the system of equations.

By creating zeros below the main diagonal of the matrix, you prepare it for an even more systematic way to find solutions, known as back-substitution. Each use of elimination simplifies the matrix step by step, making the problem clearer to tackle by unveiling the relationship between variables.

Matrix representation is a powerful way to visualize and solve systems of linear equations. In this representation, each equation from the system is transformed into a matrix form. This involves creating three separate matrices: the coefficient matrix, the variable matrix, and the constant matrix.

• The coefficient matrix, often denoted as \( A \), includes the coefficients of the variables.
• The variable matrix, \( X \), contains the variables of the system \((x_1, x_2, x_3)\).
• The constant matrix, \( B \), consists of the constants from the right-hand side of the equations.

In our exercise, the system of equations was represented by: \[ A = \begin{bmatrix} 3 & 1 & -1 \ 1 & -1 & 1 \ 2 & 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \ -3 \ 0 \end{bmatrix} \] This representation simplifies the manipulation and understanding of the problem.

Reaching row-echelon form is one of the main objectives of Gaussian elimination. A matrix is in row-echelon form when each leading coefficient (the first nonzero number from the left, in a row) is to the right of the leading coefficient of the row above it. Also, all entries in a column below a leading coefficient must be zero.
Benefits of row-echelon form include:

• Streamlined solving of systems, making it easier to see how to apply back-substitution.
• Fewer calculations needed to solve the system compared to starting from scratch with the original equations.

In the given system, through strategic operations such as row subtraction and scaling, we transformed the original matrix to have a zero entry beneath the leading coefficient of the first row, preparing it to move readily towards back-substitution. This intermediary form is crucial as it sets the stage for the final solving steps.

Back-substitution is the final step in solving a system of linear equations that has been transformed into row-echelon form. Once the matrix is in row-echelon form, the last equation can usually be solved directly for one variable. This process works its way up the rows until all variables are solved.

• Start from the bottom row, which ideally provides a direct solution for one variable.
• Substitute this solution into the rows above, one at a time, gradually resolving more variables.
• Continue this step-by-step until all unknowns are found.

In our problem, after reducing to row-echelon form, we started with solving from the bottom row \( \left( \frac{4}{3}x_3 = -\frac{2}{3} \right) \) to find \( x_3 \). This solution was then substituted into the next row up, gradually leading to the solution for all variables: \( x_1 = -\frac{1}{3}, x_2 = 1.5, x_3 = -\frac{1}{2} \). The systematic nature of back-substitution verifies each step as you solve the system.