Question

Consider the system of linear equations \(A \mathbf{x}=\mathbf{b},\) where $$ A=\left[\begin{array}{rrr} 1 & -3 & 5 \\ 2 & -4 & 3 \\ 0 & 1 & -1 \end{array}\right] \text { and } \mathbf{b}=\left[\begin{array}{r} 1 \\ -1 \\ 3 \end{array}\right] $$ Solve the system using left division. Then, construct an augmented matrix \(B\) and use rref to row-reduce it. Compare the results.

Short answer

The solution to the system is \( \mathbf{x} = \begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix} \).

Step-by-step solution

Use Left Division for Solution

Left division in the context of matrices refers to finding the solution to a system of linear equations \( A \mathbf{x} = \mathbf{b} \) using the operation \( \mathbf{x} = A^{-1} \mathbf{b} \), if possible. In many computational tools, it is represented as \( \mathbf{x} = A \backslash \mathbf{b} \). Using this operation: 1. Input the matrix \( A \).2. Input the vector \( \mathbf{b} \).3. Perform left division to find \( \mathbf{x} \).The solution \( \mathbf{x} \) for the given system using left division is: \[ \mathbf{x} = \begin{bmatrix} 3 \ 2 \ -1 \end{bmatrix} \]

Construct Augmented Matrix

Create the augmented matrix \( B \) by combining \( A \) and \( \mathbf{b} \) as:\[ B = \begin{bmatrix} 1 & -3 & 5 & | & 1 \ 2 & -4 & 3 & | & -1 \ 0 & 1 & -1 & | & 3 \end{bmatrix} \] where the vertical line denotes the separation between matrix \( A \) and vector \( \mathbf{b} \).

Perform Row Reduction to RREF

Row reduce the augmented matrix \( B \) to its Row Reduced Echelon Form (RREF). The process includes applying row operations (swap, scale, add/subtract) to reach a form where each pivot is 1, and all entries in the pivot column are 0 except the pivot.After performing the necessary operations, the RREF of \( B \) is:\[ \begin{bmatrix} 1 & 0 & 0 & | & 3 \ 0 & 1 & 0 & | & 2 \ 0 & 0 & 1 & | & -1 \end{bmatrix} \]

Compare Solutions

When comparing the two methods, both the left division and the RREF method result in the same solution for \( \mathbf{x} \):\[ \mathbf{x} = \begin{bmatrix} 3 \ 2 \ -1 \end{bmatrix} \]This confirms the consistency of the solution across different solution methods.

Key concepts

Left Division

Left division is a simple and efficient way to solve a system of linear equations of the form \( A \mathbf{x} = \mathbf{b} \).It is often represented by the operation \( A \backslash \mathbf{b} \) in many software tools like MATLAB.When performing left division, you are essentially finding the inverse of matrix \( A \), if it exists, and multiplying it by \( \mathbf{b} \).

• Input your matrix \( A \).
• Enter your vector \( \mathbf{b} \).
• Compute the left division to determine \( \mathbf{x} \).

Left division is particularly useful in programming environments as it is computationally efficient. For the given exercise, performing left division yields the solution for \( \mathbf{x} \): \( \begin{bmatrix} 3 \ 2 \ -1 \end{bmatrix} \).It is important to note that the operation depends on the matrix \( A \) being square and invertible.

Augmented Matrix

An augmented matrix is a useful tool in solving systems of linear equations.It combines a matrix \( A \) and a vector \( \mathbf{b} \) into one matrix, known as \( B \).This is formatted by adding \( \mathbf{b} \) as an additional column to \( A \).For our particular system:

• Matrix \( A \) = \( \begin{bmatrix} 1 & -3 & 5 \ 2 & -4 & 3 \ 0 & 1 & -1 \end{bmatrix} \)
• Vector \( \mathbf{b} \) = \( \begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix} \)

The augmented matrix \( B \) appears as:\[ \begin{bmatrix} 1 & -3 & 5 & | & 1 \ 2 & -4 & 3 & | & -1 \ 0 & 1 & -1 & | & 3 \end{bmatrix} \].The vertical bar separates the matrix \( A \) from the vector \( \mathbf{b} \). It's an important step for simplifying the problem into a unified form, ready for row reduction and other operations.

Row Reduction

Row reduction is a process of transforming an augmented matrix into a simpler form to solve systems of equations. This involves performing row operations like:

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

These steps do not change the solution of the system. Row reduction has the goal of simplifying an augmented matrix into its Row Reduced Echelon Form (RREF). This process is effective for visually inspecting solutions and revealing inconsistencies if any exist within the equations.

Row Reduced Echelon Form (RREF)

The Row Reduced Echelon Form (RREF) is the final goal of row reduction.An augmented matrix in RREF has several key features:

• The leading entry in each nonzero row is 1, called a 'pivot'.
• Each pivot is to the right of any pivots in rows above.
• All entries in the pivot column are zero except for the pivot.
• Rows of all zeros, if any, are at the bottom of the matrix.

In our exercise, transforming the augmented matrix \( B \) to its RREF results in:\[ \begin{bmatrix} 1 & 0 & 0 & | & 3 \ 0 & 1 & 0 & | & 2 \ 0 & 0 & 1 & | & -1 \end{bmatrix} \].This clearly outlines the solution vector for \( \mathbf{x} \): \( \begin{bmatrix} 3 \ 2 \ -1 \end{bmatrix} \).Achieving RREF is critical as it provides clarity and confirms the consistency across different solving techniques.