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. $$ A=\left[\begin{array}{lll} -4 & 3 & 2 \\ -4 & 5 & 0 \end{array}\right], \vec{b}=\left[\begin{array}{l} -4 \\ -4 \end{array}\right] $$

Short answer

For (a): Solutions are \( t\begin{bmatrix} \frac{5}{4} \\ 1 \\ 1 \end{bmatrix} \); two examples: \( \begin{bmatrix} \frac{5}{4} \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} \frac{5}{2} \\ 2 \\ 2 \end{bmatrix} \). For (b): General solution \( \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + t\begin{bmatrix} \frac{5}{4} \\ 1 \\ 1 \end{bmatrix} \); examples: \( \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} \frac{9}{4} \\ 1 \\ 1 \end{bmatrix} \).

Step-by-step solution

Solving the Homogeneous System

Start by solving the homogeneous system \( A \vec{x} = \vec{O} \). Write down the augmented matrix and use Gaussian elimination to find the solutions. The system is: \[\begin{align*}-4x_1 + 3x_2 + 2x_3 &= 0 \-4x_1 + 5x_2 + 0x_3 &= 0\end{align*}\]Perform row operations to achieve row-echelon form. Subtract the first row from the second:\[\begin{align*}-4x_1 + 3x_2 + 2x_3 &= 0 \0x_1 + 2x_2 - 2x_3 &= 0\end{align*}\]Next, simplify the second equation by dividing by 2:\[x_2 - x_3 = 0\] From this, we have \( x_2 = x_3 \). Choose free variable \( x_3 = t \), then \( x_2 = t \). Now substitute into the first equation:\[-4x_1 + 3t + 2t = 0 \Rightarrow -4x_1 + 5t = 0 \Rightarrow x_1 = \frac{5}{4}t\]Thus, a general solution is \( \vec{x} = t \begin{bmatrix} \frac{5}{4} \ 1 \ 1 \end{bmatrix} \). Two particular solutions are \( t = 1 \) and \( t = 2 \): - Particular Solution 1: \( \vec{x}_1 = \begin{bmatrix} \frac{5}{4} \ 1 \ 1 \end{bmatrix} \)- Particular Solution 2: \( \vec{x}_2 = \begin{bmatrix} \frac{5}{2} \ 2 \ 2 \end{bmatrix} \)

Solving the Non-Homogeneous System

For the non-homogeneous system \( A \vec{x} = \vec{b} \), write and solve the matrix equation:\[\begin{align*}-4x_1 + 3x_2 + 2x_3 &= -4 \-4x_1 + 5x_2 + 0x_3 &= -4\end{align*}\]Perform the same operation as before to achieve row-echelon form:- Subtract the first equation from the second: \[0x_1 + 2x_2 - 2x_3 = 0 \Rightarrow x_2 = x_3\]Substitute \( x_2 = x_3 = t \) into the first equation:\[-4x_1 + 3t + 2t = -4 \Rightarrow -4x_1 + 5t = -4 \Rightarrow -4x_1 = -4 - 5t \Rightarrow x_1 = 1 + \frac{5}{4}t\]Thus, a general solution is \( \vec{x} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} + t \begin{bmatrix} \frac{5}{4} \ 1 \ 1 \end{bmatrix} \). Two particular solutions for this system (considering \( t = 0 \) and \( t = 1 \)) are:- Particular Solution 1: \( \vec{x}_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} \)- Particular Solution 2: \( \vec{x}_2 = \begin{bmatrix} \frac{9}{4} \ 1 \ 1 \end{bmatrix} \)

Key concepts

Homogeneous Systems

In linear algebra, a **homogeneous system** of equations is one where all the constant terms are zero. In our exercise, we had the matrix equation \( A \vec{x} = \vec{O} \), where \( \vec{O} \) is a zero vector. Such systems always have at least one solution called the trivial solution, where every variable is zero. However, they can also have infinitely many solutions depending on the matrix's rank.

To solve a homogeneous system, we often use matrix transformations like **Gaussian elimination** to reduce the system to a simpler form. Once simplified, free variables are assigned parameters, allowing us to express solutions in terms of these parameters. The provided example demonstrates that the solution can be written as \( \vec{x} = t \begin{bmatrix} \frac{5}{4} \ 1 \ 1 \end{bmatrix} \). Here, \( t \) is a free parameter that can take any real value.

Such a solution indicates a line of vectors in the variable space, each vector corresponding to a different choice of \( t \). This demonstrates the uniqueness of homogeneous systems and their ability to reveal relationships between variables in linear algebra.

Non-Homogeneous Systems

A **non-homogeneous system** introduces a constant vector, making the system broader in application compared to a homogeneous system. It can be mathematically expressed as \( A \vec{x} = \vec{b} \), where \( \vec{b} eq \vec{O} \). In our task, the vector \( \vec{b} \) is given, bringing additional complexity into solving the system.

Unlike homogeneous systems, these don't always have solutions. The existence of a solution depends on the relationship between the matrix \( A \) and vector \( \vec{b} \). If a solution exists, the solutions often take the form of a particular solution plus any solution to the corresponding homogeneous system. Our exercise neatly demonstrates this by first using Gaussian elimination to find a particular solution, then adding the general solution of the homogeneous system.

In this specific problem, the general solution was expressed as \( \vec{x} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} + t \begin{bmatrix} \frac{5}{4} \ 1 \ 1 \end{bmatrix} \), where \( t \) is a free variable. This solution highlights how non-homogeneous systems can be built upon the backbone of homogeneous solutions.

Gaussian Elimination

**Gaussian elimination** is a critical technique in solving systems of linear equations. It transforms any given matrix into row-echelon form, making it easier to solve the system via back substitution. The process involves using row operations such as row swapping, row scaling, and adding multiples of one row to another to simplify the matrix.

In both parts of our exercise, Gaussian elimination was performed to bring the system of equations into a simpler, more manageable form. By subtracting and simplifying equations, you can isolate certain variables, progressively revealing solutions. For example, our exercise used elimination to transform the original system:

• -4x_1 + 3x_2 + 2x_3 = 0
• -4x_1 + 5x_2 + 0x_3 = 0

into a simpler form, from which relations between variables were directly deduced. These steps are vital in understanding not just how to reach a solution, but also why the solution works, which deepens comprehension of linear algebra concepts.

Particular Solutions

**Particular solutions** represent specific solutions for a system of equations influenced by initial conditions or specific constraints. When dealing with non-homogeneous systems, it is essential to find at least one particular solution. The particular solution does not carry free parameters, offering immediate values for each variable that satisfy the system.

In the exercise discussed, two particular solutions were found for each equation system. For the non-homogeneous system, particular solutions like \( \vec{x}_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} \) were derived through careful manipulation and evaluation of equations, tailored to satisfy these specific system conditions. These preferred solutions provide concrete examples of how a system can have multiple valid answers, penetrable via both analytical understanding and calculation stacks.

Moreover, combining these solutions with the homogeneous ones augments the system's full solution set. This process illustrates how solutions interlink in linear structures, reflecting wider applications in mathematical modeling and problem-solving.