Solving Linear Equations With Matrices A Comprehensive Guide

Jul 9, 2025 by ADMIN 61 views

This article delves into the methods for solving linear equations using matrices, a fundamental concept in linear algebra. We will explore how to find general solutions for homogeneous equations, particular solutions for non-homogeneous equations, and ultimately, the general solution for non-homogeneous systems. This guide will walk you through the process step-by-step, providing clear explanations and examples to solidify your understanding.

a) Finding the General Solution of the Homogeneous Equation AZ = 0

In this section, our primary goal is to determine the general solution Z for the homogeneous equation AZ = 0, where A is the given matrix:

A = \begin{pmatrix} 1 & 1 & -1 \\ 1 & -1 & 2 \end{pmatrix}

The homogeneous equation AZ = 0 represents a system of linear equations where the right-hand side is a zero vector. To find the general solution, we need to identify all possible vectors Z that, when multiplied by A, result in the zero vector. This solution set forms the null space (or kernel) of the matrix A. To solve this, we'll employ the method of Gaussian elimination to transform the matrix A into its reduced row-echelon form. This form simplifies the system of equations, making it easier to identify the free variables and express the solution in parametric form.

First, we set up the augmented matrix by appending a zero vector to the right of matrix A:

\begin{pmatrix} 1 & 1 & -1 & | & 0 \\ 1 & -1 & 2 & | & 0 \end{pmatrix}

Next, we perform row operations to bring the matrix into reduced row-echelon form. We start by subtracting the first row from the second row (R2 = R2 - R1):

\begin{pmatrix} 1 & 1 & -1 & | & 0 \\ 0 & -2 & 3 & | & 0 \end{pmatrix}

Now, we divide the second row by -2 (R2 = R2 / -2):

\begin{pmatrix} 1 & 1 & -1 & | & 0 \\ 0 & 1 & -3/2 & | & 0 \end{pmatrix}

Finally, we subtract the second row from the first row (R1 = R1 - R2):

\begin{pmatrix} 1 & 0 & 1/2 & | & 0 \\ 0 & 1 & -3/2 & | & 0 \end{pmatrix}

This is the reduced row-echelon form of the matrix. Now we can rewrite the system of equations:

z_1 + (1/2)z_3 = 0 \\
z_2 - (3/2)z_3 = 0

Here, z_1 and z_2 are leading variables, and z_3 is a free variable. We can express z_1 and z_2 in terms of z_3:

z_1 = -(1/2)z_3 \\
z_2 = (3/2)z_3

Let's denote z_3 as the parameter t. Then the general solution Z can be written in vector form:

Z = \begin{pmatrix} z_1 \\ z_2 \\ z_3 \end{pmatrix} = \begin{pmatrix} -(1/2)t \\ (3/2)t \\ t \end{pmatrix} = t \begin{pmatrix} -1/2 \\ 3/2 \\ 1 \end{pmatrix}

To eliminate fractions, we can multiply the vector by 2, so the general solution becomes:

Z = t \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}

Where t is any scalar. This means that any vector Z that is a scalar multiple of the vector (-1, 3, 2) is a solution to the homogeneous equation AZ = 0. The set of all such vectors forms a line in three-dimensional space passing through the origin.

b) Finding a Particular Solution of AX = b

Now, let's shift our focus to finding a particular solution of the non-homogeneous equation AX = b, where

A = \begin{pmatrix} 1 & 1 & -1 \\ 1 & -1 & 2 \end{pmatrix} \quad \text{and} \quad b = \begin{pmatrix} 1 \\ 2 \end{pmatrix}

Unlike the homogeneous equation, here we have a non-zero vector b on the right-hand side. A particular solution is any vector X that satisfies this equation. There might be infinitely many solutions, but we only need to find one for this part. To find a particular solution, we can again use Gaussian elimination, but this time we'll augment the matrix A with the vector b:

\begin{pmatrix} 1 & 1 & -1 & | & 1 \\ 1 & -1 & 2 & | & 2 \end{pmatrix}

We proceed with row operations to bring the matrix into reduced row-echelon form. Subtract the first row from the second row (R2 = R2 - R1):

\begin{pmatrix} 1 & 1 & -1 & | & 1 \\ 0 & -2 & 3 & | & 1 \end{pmatrix}

Divide the second row by -2 (R2 = R2 / -2):

\begin{pmatrix} 1 & 1 & -1 & | & 1 \\ 0 & 1 & -3/2 & | & -1/2 \end{pmatrix}

Subtract the second row from the first row (R1 = R1 - R2):

\begin{pmatrix} 1 & 0 & 1/2 & | & 3/2 \\ 0 & 1 & -3/2 & | & -1/2 \end{pmatrix}

Now we have the reduced row-echelon form. The system of equations is:

x_1 + (1/2)x_3 = 3/2 \\
x_2 - (3/2)x_3 = -1/2

To find a particular solution, we can set the free variable x_3 to a convenient value, such as 0. This simplifies the equations:

x_1 = 3/2 \\
x_2 = -1/2 \\
x_3 = 0

Thus, a particular solution X_p is:

X_p = \begin{pmatrix} 3/2 \\ -1/2 \\ 0 \end{pmatrix}

We have found one solution to the non-homogeneous equation. This vector, when multiplied by A, will result in the vector b.

c) Finding the General Solution of AX = b

Now that we have a particular solution X_p for the non-homogeneous equation AX = b and the general solution Z for the homogeneous equation AZ = 0, we can construct the general solution for the non-homogeneous equation. The general solution X is the sum of the particular solution and the general solution of the homogeneous equation:

X = X_p + Z

We found that:

X_p = \begin{pmatrix} 3/2 \\ -1/2 \\ 0 \end{pmatrix}

and

Z = t \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}

Where t is any scalar. Therefore, the general solution X is:

X = \begin{pmatrix} 3/2 \\ -1/2 \\ 0 \end{pmatrix} + t \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 3/2 - t \\ -1/2 + 3t \\ 2t \end{pmatrix}

This represents the set of all solutions to the equation AX = b. It's a line in three-dimensional space, parallel to the solution space of the homogeneous equation (the null space of A) and passing through the particular solution X_p.

The general solution X includes a particular solution and a linear combination of the solutions to the homogeneous equation. This structure highlights a crucial concept in linear algebra: the general solution to a non-homogeneous linear equation is the sum of a particular solution and the general solution to the corresponding homogeneous equation.

By varying the parameter t, we obtain an infinite number of solutions, each satisfying the equation AX = b. This comprehensive solution encapsulates all possible vectors X that, when multiplied by the matrix A, produce the vector b.

Conclusion

In this article, we've systematically tackled the problem of solving linear equations using matrices. We started by finding the general solution to the homogeneous equation AZ = 0, which involved reducing the matrix to its row-echelon form and expressing the solution in terms of free variables. Then, we determined a particular solution to the non-homogeneous equation AX = b using a similar approach. Finally, we combined these results to obtain the general solution to the non-homogeneous equation, illustrating the fundamental principle that the general solution is the sum of a particular solution and the general solution to the corresponding homogeneous equation. Understanding these methods is crucial for anyone working with linear systems, as they provide a powerful and systematic way to find all possible solutions.