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. $$ A=\left[\begin{array}{ll} -4 & -1 \\ -3 & -2 \end{array}\right], \vec{b}=\left[\begin{array}{l} 1 \\ 4 \end{array}\right] $$

Short answer

Homogeneous: \(\begin{bmatrix} 1 \\ -4 \end{bmatrix}, \begin{bmatrix} 0.5 \\ -2 \end{bmatrix}\). Non-homogeneous: \(\begin{bmatrix} \frac{2}{5} \\ \frac{-13}{5} \end{bmatrix}\).

Step-by-step solution

Set Up the Homogeneous Equation

The homogeneous equation is given by \(A \vec{x} = \vec{O}\). In this form, the matrix equation is: \[ \begin{bmatrix} -4 & -1 \ -3 & -2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \] We will find \(\vec{x}\) such that this equation holds.

Solve the Homogeneous System using Matrix Methods

To solve, we set up the system of equations: 1. \(-4x_1 - x_2 = 0\) 2. \(-3x_1 - 2x_2 = 0\) From the first equation, solve for \(x_2\) as \(x_2 = -4x_1\). Substitute in the second equation to verify the consistency.Since both equations are multiples, we find one solution in vector form: \[ \vec{x} = \begin{bmatrix} t \ -4t \end{bmatrix} \quad \text{where } t \text{ is a parameter}\]

Find Two Particular Solutions for Homogeneous Equation

Choose specific values of the parameter \(t\) to find two particular solutions:1. Let \(t = 1\): \( \vec{x_1} = \begin{bmatrix} 1 \ -4 \end{bmatrix} \)2. Let \(t = 0.5\): \( \vec{x_2} = \begin{bmatrix} 0.5 \ -2 \end{bmatrix} \)

Set Up the Non-Homogeneous Equation

The non-homogeneous equation is given by \(A \vec{x} = \vec{b}\). In this form, the matrix equation is: \[ \begin{bmatrix} -4 & -1 \ -3 & -2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 1 \ 4 \end{bmatrix} \] We will solve this system by finding appropriate values for \(x_1\) and \(x_2\).

Solve the Non-Homogeneous System using Matrix Methods

Formulate the system of equations: 1. \(-4x_1 - x_2 = 1\) 2. \(-3x_1 - 2x_2 = 4\) Use substitution or elimination to find \(x_1\) and \(x_2\). Solving, we find:Multiply the first equation by 2: \(-8x_1 - 2x_2 = 2\)Then, use subtraction: \(-3x_1 - 2x_2 - (-8x_1 - 2x_2) = 4 - 2\) This simplifies to \(5x_1 = 2\).Thus, \(x_1 = \frac{2}{5}\) and substitute back to find \(x_2\):\(-4\left(\frac{2}{5}\right) - x_2 = 1\), solving gives \(x_2 = \frac{-13}{5}\) or \(-2.6\).

Find Two Particular Solutions for Non-Homogeneous Equation

The particular solution found is: \[ \vec{x} = \begin{bmatrix} \frac{2}{5} \ \frac{-13}{5} \end{bmatrix} \]To find another specific solution, modify slightly (e.g. check calculated variations or double-check parameters).Since the system is distinct, additional solutions are linear combinations based on coefficients.

Key concepts

matrix equations

Matrix equations help us compactly represent systems of linear equations. A matrix equation is of the form \( A\vec{x} = \vec{b} \), where \( A \) is a matrix, \( \vec{x} \) is a vector containing variables, and \( \vec{b} \) is a vector of constants. These matrix equations allow us to use operations from linear algebra, like matrix addition and multiplication, to solve complex systems succinctly. The elegance of matrix equations lies in their ability to represent many equations at once, which can be solved using various methods, including row reduction and matrix inversion.

• **Representation**: Use a single equation to encapsulate multiple linear equations.
• **Structured**: The coefficients of variables are stored in the matrix \( A \), simplifying system solving.
• **Solving**: Techniques include using augmented matrices, Gaussian elimination, or utilizing inverse matrices to find the values of \( \vec{x} \).

homogeneous system

A homogeneous system of linear equations is a type of matrix equation where the matrix and variables produce a zero vector, represented as \( A\vec{x} = \vec{O} \). In these systems, the constants in the vector \( \vec{b} \) are all zeros, making them homogeneous. Homogeneous systems are intriguing due to their consistent nature; they always have at least one solution, the zero vector \( \vec{x} = \vec{O} \).

• **Infinite Solutions:** Often reflective of systems with more variables than equations, leading to multiple solutions.
• **Zero Solution:** The trivial solution is always present, where all variables equal zero.
• **Parameterization:** Solutions can be represented with parameters, indicating infinite possibilities.

Exploring these systems reveals insights into vector spaces and linear dependence, helping us understand the dimensionality and basis of the solution set.

non-homogeneous system

Non-homogeneous systems are characterized by the matrix equation \( A\vec{x} = \vec{b} \), where \( \vec{b} \) is not equal to the zero vector. Unlike homogeneous systems, these do not guarantee an infinite number of solutions.

• **Existence of Solutions:** Depends on the consistency of equations. A system may have one, none, or many solutions depending on rank and the condition of the augmented matrix.
• **Unique Solution:** If the matrix \( A \) is invertible and the system has the same number of equations as variables, it typically leads to a unique solution.
• **General Solution:** May include the sum of a particular solution to the non-homogeneous equations and the general solution of the related homogeneous system.

Solving non-homogeneous systems provides insight into practical scenarios where systems differ slightly, and adjustments in theory help manage real-world variability.

matrix methods

Matrix methods provide structured approaches to solving linear systems efficiently. Popular methods include Gaussian elimination, LU decomposition, and matrix inverses.

• **Gaussian Elimination:** This method transforms matrices into a row-echelon form, simplifying the back-substitution process to find the solution.
• **LU Decomposition:** Splits a matrix into a product of a lower triangular matrix and an upper triangular matrix, easing the calculation of the solution.
• **Matrix Inversion:** Applicable when the matrix is square and invertible, allowing the solution to be found via \( \vec{x} = A^{-1}\vec{b} \).

Mastering these methods equips one with powerful tools for tackling not just theoretical linear systems but also real-world problems necessitating quick and accurate computation.