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. $$ \begin{array}{l} A=\left[\begin{array}{cccc} 1 & 5 & -4 & -1 \\ 1 & 0 & -2 & 1 \end{array}\right] \\ \vec{b}=\left[\begin{array}{c} 0 \\ -2 \end{array}\right] \end{array} $$

Short answer

For (a), solutions are linear combinations of two particular solutions: \([ -5, 1, 0, 0 ]^T\) and \([ 4, 0, 1, 0 ]^T\). For (b), two particular solutions are \([ \frac{14}{5}, -\frac{2}{5}, 0, 0 ]^T\) and \([ \frac{9}{5}, -\frac{4}{5}, 1, 0 ]^T\).

Step-by-step solution

Understand the Given Matrix and Vector

You are provided with a matrix \( A \) and a vector \( \vec{b} \). Here, \( A = \begin{bmatrix} 1 & 5 & -4 & -1 \ 1 & 0 & -2 & 1 \end{bmatrix} \) and \( \vec{b} = \begin{bmatrix} 0 \ -2 \end{bmatrix} \). We need to solve two different equations: \( A \vec{x} = \vec{0} \) and \( A \vec{x} = \vec{b} \).

Solve the Homogeneous Equation (a) \(A \vec{x} = \vec{0}\)

For the homogeneous equation, we need to find the solutions of the system given by the augmented matrix:\[\left[\begin{array}{cccc|c}1 & 5 & -4 & -1 & 0 \1 & 0 & -2 & 1 & 0 \\end{array}\right]\]Eliminate the first column of the second row.Subtract the first row from the second:\[\left[\begin{array}{cccc|c}1 & 5 & -4 & -1 & 0 \0 & -5 & 2 & 2 & 0 \\end{array}\right]\]Now, solve for the variables. Express the system:1. \( x_1 + 5x_2 - 4x_3 - x_4 = 0 \)2. \( -5x_2 + 2x_3 + 2x_4 = 0 \)Set \( x_2 = s, x_3 = t, x_4 = u \), the solutions are: \( x_1 = -5s + 4t + u \), \( x_2 = s \), \( x_3 = t \), \( x_4 = u \).

Present Particular Solutions for \(A \vec{x} = \vec{0}\)

Find two particular solutions. Let one variable vary at a time:1. Let \( s = 1, t = 0, u = 0 \). Then \(\vec{x} = \begin{bmatrix} -5 \ 1 \ 0 \ 0 \end{bmatrix}\).2. Let \( s = 0, t = 1, u = 0 \). Then \(\vec{x} = \begin{bmatrix} 4 \ 0 \ 1 \ 0 \end{bmatrix}\).

Solve the Non-Homogeneous Equation (b) \(A \vec{x} = \vec{b}\)

For the non-homogeneous equation, work with the augmented matrix:\[\left[\begin{array}{cccc|c}1 & 5 & -4 & -1 & 0 \1 & 0 & -2 & 1 & -2 \\end{array}\right]\]Subtract the first row from the second:\[\left[\begin{array}{cccc|c}1 & 5 & -4 & -1 & 0 \0 & -5 & 2 & 2 & -2 \\end{array}\right]\]Solve the system:1. \( x_1 + 5x_2 - 4x_3 - x_4 = 0 \)2. \( -5x_2 + 2x_3 + 2x_4 = -2 \)Express \( x_2 \) as \( x_2 = \frac{2 + 2x_4 - 2x_3}{5} \) and solve for \( x_1 \).

Present Particular Solutions for \(A \vec{x} = \vec{b}\)

Find at least two particular solutions, setting free variables:1. Set \( x_3 = 0, x_4 = 0 \). Then \( x_2 = -\frac{2}{5} \) and \( x_1 = \frac{14}{5} \), giving \( \vec{x} = \begin{bmatrix} \frac{14}{5} \ -\frac{2}{5} \ 0 \ 0 \end{bmatrix} \).2. Set \( x_3 = 1, x_4 = 0 \). Then \( x_2 = -\frac{4}{5} \) and \( x_1 = \frac{9}{5} \), giving \( \vec{x} = \begin{bmatrix} \frac{9}{5} \ -\frac{4}{5} \ 1 \ 0 \end{bmatrix} \).

Key concepts

Matrix Algebra

Matrix algebra is a powerful mathematical tool used to solve systems of linear equations. In our exercise, we are dealing with two matrices: the coefficient matrix \( A \) and a vector \( \vec{b} \). Matrix algebra involves performing operations such as addition, subtraction, multiplication, and finding inverses. These operations help us solve equations of the form \( A\vec{x} = \vec{0} \) (homogeneous system) and \( A\vec{x} = \vec{b} \) (non-homogeneous system). By transforming matrices into a simpler form, we make it easier to solve for the unknowns. The matrix \( A \) contains the coefficients of the variables, which we can manipulate through matrix algebra to find solutions for \( \vec{x} \).
Understanding matrix operations can remind us of operations with real numbers. For instance:

• Matrix addition involves adding corresponding elements.
• Matrix multiplication requires finding the dot product of rows and columns.
• Also, matrices need to have specific dimensions to perform certain operations effectively.

These concepts simplify the process of solving a system with multiple equations and unknowns.

Homogeneous System

In mathematics, a homogeneous system of linear equations is one in which all constants are zero. The problem stated as \( A \vec{x} = \vec{0} \) is a classic example of a homogeneous system. Homogeneous systems are interesting because they always have at least one solution, the trivial solution, where all variables are zero.
In our exercise, we transformed the given matrix into an augmented matrix that represents the system of equations.

• We set up an augmented matrix by appending a zero vector to matrix \( A \).
• Then, we used row operations to solve for the variables.

The system described only by zero constants often provides infinitely many solutions, incorporating free variables. In our exercise, these free variables were denoted by \( s, t, \) and \( u \), showing that the solution set is a vector that can vary by adjusting these parameters.

Non-Homogeneous System

In contrast to a homogeneous system, a non-homogeneous system of equations has constants that are not all zero, as seen in the equation \( A \vec{x} = \vec{b} \). This system can have a unique solution, infinitely many solutions, or no solutions, depending on the relationships among the equations in the system.
To solve the non-homogeneous equation in our exercise, we created an augmented matrix with matrix \( A \) and vector \( \vec{b} \) together. We used similar steps of row reduction to simplify the problem and solve the system. In our solution:

• Free variables were allowed to vary, demonstrating how solutions can be generated.
• We expressed the solutions for dependent variables in terms of these free variables.

This step shows the flexibility and the methodical approach in finding solutions to complex systems of equations.

Augmented Matrix

An augmented matrix is a convenient way to represent a system of linear equations. It not only displays the coefficients of the variables but also includes the constant terms, forming a full picture of the system and allowing for row operations to simplify and solve the system.
In our problem, we formed an augmented matrix by appending a column of zeros in the homogeneous case and the vector \( \vec{b} \) for the non-homogeneous case to matrix \( A \). The layout can be visualized as:

• Matrix \( A \) on the left side of the vertical line, which consists of the coefficients of the system.
• The vector or constant column (either zero vector or vector \( \vec{b} \)) on the right side.

This arrangement is vital because it allows the use of row operations to achieve row-echelon form or reduced row-echelon form, making it easier to find solutions to the system.

Particular Solution

Finding particular solutions is an important step in solving both homogeneous and non-homogeneous systems. Particular solutions are specific instances of solutions to the system, usually found after free variables are provided specific values.
The particular solutions can exhibit the range of possibilities within the solution sets, showcasing how different combinations of free variable values lead to distinct solutions.
In our exercise, for the homogeneous system, various values for \( s, t, \) and \( u \) provided particular solutions like \( \begin{bmatrix} -5 \ 1 \ 0 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 4 \ 0 \ 1 \ 0 \end{bmatrix} \). For the non-homogeneous system, variations in free variables provided solutions such as \( \begin{bmatrix} \frac{14}{5} \ -\frac{2}{5} \ 0 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} \frac{9}{5} \ -\frac{4}{5} \ 1 \ 0 \end{bmatrix} \). These results highlight the flexibility inherent in solving such systems, where different settings of parameters bring out a spectrum of particular solutions.