Question

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ \begin{array}{l} A=\left[\begin{array}{ccccc} -1 & 3 & 1 & -3 & 4 \\ 3 & -3 & -1 & 1 & -4 \\ -2 & 3 & -2 & -3 & 1 \end{array}\right], \\ \vec{b}=\left[\begin{array}{c} 1 \\ 1 \\ -5 \end{array}\right] \end{array} $$

Short answer

Solutions involve parameterized vectors; specific examples include given particular solutions for each system.

Step-by-step solution

Introduction: Problem Setup

We have a matrix \(A\) and vectors \(\vec{b}\) and \(\vec{O}\). We need to solve both \(A \vec{x} = \vec{O}\) (the homogeneous system) and \(A \vec{x} = \vec{b}\) (the non-homogeneous system), finding two particular solutions for each when possible.

Solve the Homogeneous System

First, we tackle the equation \(A \vec{x} = \vec{O}\), where \(\vec{O} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}\). This involves finding the null space of matrix \(A\). We do this by performing row operations to bring \(A\) to its row-reduced echelon form (RREF).

Row Reduce Matrix A

Perform row operations on \(A\) to find its RREF.\[\left[\begin{array}{ccccc} 1 & -3 & -1 & 3 & -4 \ 0 & 0 & 1 & -2 & 3 \ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

Determine Solutions to Homogeneous System

From the RREF, deduce that solution vectors \(\vec{x}\) have free variables. The solution is of the form \[\vec{x} = \begin{bmatrix} 3t + 4s \ t \ 2s \ s \ 0 \end{bmatrix},\] where \(t\) and \(s\) are parameters. Two particular solutions could be \(\begin{bmatrix} 3 \ 1 \ 0 \ 0 \ 0 \end{bmatrix}\) and \(\begin{bmatrix} 4 \ 0 \ 2 \ 1 \ 0 \end{bmatrix}\).

Solve the Non-Homogeneous System

Now, solve \(A \vec{x} = \vec{b}\), where \(\vec{b} = \begin{bmatrix} 1 \ 1 \ -5 \end{bmatrix}\). Use augmented matrix:\[\left[\begin{array}{ccccc|c} -1 & 3 & 1 & -3 & 4 & 1 \ 3 & -3 & -1 & 1 & -4 & 1 \ -2 & 3 & -2 & -3 & 1 & -5 \end{array}\right]\]

Row Reduce Augmented Matrix

Transform the augmented matrix to its RREF:\[\left[\begin{array}{ccccc|c} 1 & -3 & 0 & 1 & 0 & 0 \ 0 & 0 & 1 & -2 & 3 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

Determine Solutions to Non-Homogeneous System

The RREF shows that the particular solution can include parameters for free variables. The general solution is of the form \[\vec{x} = \begin{bmatrix} 3 + 4s \ s \ 2s \ s \ 0 \end{bmatrix},\] such that \(s\) is a parameter. Two particular solutions could be \(\begin{bmatrix} 3 \ 0 \ 0 \ 0 \ 0 \end{bmatrix}\) and \(\begin{bmatrix} 7 \ 1 \ 2 \ 1 \ 0 \end{bmatrix}\).

Conclusion: Summary of Solutions

For \(A \vec{x} = \vec{O}\), the general solution involves parameters \(t\) and \(s\). For \(A \vec{x} = \vec{b}\), we found solutions using the parameter \(s\). Two particular solutions were provided for each case. These represent different vectors satisfying the respective equations.

Key concepts

Homogeneous System

A homogeneous system is a type of linear equation where the result is a zero vector. It's represented by the equation \( A \vec{x} = \vec{O} \), where \( \vec{O} \) is the zero vector. In this system, we're essentially looking for all vectors \( \vec{x} \) that, when multiplied by matrix \( A \), result in a vector of all zeros.
This system is important in linear algebra because it helps determine the null space of a matrix, which consists of all the vectors that, when multiplied by the matrix, yield the zero vector. The solutions to a homogeneous system offer insights into the properties of the matrix, such as its rank and whether or not it's invertible.
Solving homogeneous systems typically involves row-reducing the matrix to row-reduced echelon form (RREF) to discern the solutions.

Non-Homogeneous System

Compared to a homogeneous system, a non-homogeneous system involves a matrix equation where the result isn't a zero vector. It's given by \( A \vec{x} = \vec{b} \), with \( \vec{b} \) being a non-zero vector.
The solutions to this type of system rely on both the particular solutions to the specific vector \( \vec{b} \) and the homogeneous solution (the null space of \( A \)). Hence, any solution \( \vec{x_p} \) to the equation can be expressed as a sum of a particular solution and any solution from the null space: \( \vec{x} = \vec{x_p} + \vec{n} \), where \( \vec{n} \) is a solution to the homogeneous equation \( A \vec{x} = \vec{O} \).
Row-reducing an augmented matrix, which combines \( A \) and \( \vec{b} \), helps identify the particular solutions.

Row-Reduced Echelon Form (RREF)

Row-Reduced Echelon Form, or RREF, is a simplified version of a matrix achieved through row operations, making it easier to solve matrix equations.
A matrix in RREF satisfies several conditions:

• Any row with all zero elements is at the bottom of the matrix.
• The first non-zero element (pivot) in each non-zero row is 1, and it appears to the right of the pivots in any rows above it.
• Each pivot is the only non-zero element in its column.

By transforming a matrix into its RREF, we simplify systems of linear equations, revealing solutions or indicating the lack thereof. This process is integral for solving both homogeneous and non-homogeneous systems efficiently.

Free Variables

Free variables arise in solving matrix equations when there are more variables than pivot positions leading to infinite solutions. They are essential in expressing the solution set for systems of equations.
When a matrix is in RREF, any variable corresponding to a column without a pivot element is a free variable.

• They indicate dimensions in the solution space beyond basic constraints.
• In homogeneous systems, free variables result in solutions that form a vector space.
• For non-homogeneous systems, they contribute to the general solution in combination with a particular solution.

Identifying free variables is crucial for writing out general solutions, physically representing them by parameters that can vary freely, usually denoted by symbols like \( t \) or \( s \). This variability reveals the system's underlying geometry and structure.