Solving Systems of Linear Equations Using Matrices: A Comprehensive Guide

In this guide, we will explore how to solve a system of linear equations using matrices. Specifically, we will demonstrate the process through a practical example and provide detailed steps for readers to follow.

Introduction to Linear Equations and Matrices

Linear equations are fundamental in mathematics, especially when dealing with real-world problems that involve multiple variables. Matrices offer a powerful way to represent and solve these equations systematically. A system of linear equations can be represented in matrix form as (AX B), where (A) is the coefficient matrix, (X) is the variable matrix, and (B) is the constant matrix.

Problem Statement

Consider the following system of linear equations:

2x - 3y 5z 11

3x 2y - 4z 5

x y - 2z 3

Step-by-Step Solution Using Matrix Method

Step 1: Write the System in Matrix Form

The system can be represented in matrix form as:

[begin{bmatrix}2 -3 5 3 2 -4 1 1 -2 end{bmatrix}begin{bmatrix}x y z end{bmatrix}begin{bmatrix}11 -5 -3 end{bmatrix}]

Step 2: Form the Augmented Matrix

The augmented matrix is formed by appending the constant matrix to the coefficient matrix:

[begin{bmatrix}2 -3 5 | 11 3 2 -4 | -5 1 1 -2 | -3 end{bmatrix}]

Step 3: Row Operations to Reduce to Row Echelon Form

We will perform a series of row operations to bring the augmented matrix to row echelon form:

-Eliminate the first element of Row 2 and Row 3: -Row 2 - 3R1: [R_2 [3 - 31 2 - 31 -4 - 3-2 -5 - 3-3] [0 -1 2 4]] -Row 3 - 2R1: [R_3 [2 - 21 -3 - 21 5 - 2-2 11 - 2-3] [0 -5 9 17]] -The matrix becomes: [begin{bmatrix}1 1 -2 | -3 0 -1 2 | 4 0 -5 9 | 17 end{bmatrix}] -Simplify Row 2 by multiplying by -1: [begin{bmatrix}1 1 -2 | -3 0 1 -2 | -4 0 -5 9 | 17 end{bmatrix}] -Eliminate the second element of Row 3 using Row 2: -Row 3 5R2: [R_3 [0 50 -5 51 9 5-2 17 5-4] [0 0 -1 -3]] -The final matrix is: [begin{bmatrix}1 1 -2 | -3 0 1 -2 | -4 0 0 -1 | -3 end{bmatrix}]

Step 4: Back Substitution

From Row 3:

[-z -3 Rightarrow z 3]

From Row 2:

[y - 2z -4 Rightarrow y - 2(3) -4 Rightarrow y - 6 -4 Rightarrow y 2]

From Row 1:

[x y - 2z -3 Rightarrow x 2 - 2(3) -3 Rightarrow x 2 - 6 -3 Rightarrow x - 4 -3 Rightarrow x 1]

Final Solution:

[x 1, y 2, z 3]

In vector form, the solution is:

[begin{bmatrix}x y z end{bmatrix} begin{bmatrix}1 2 3 end{bmatrix}]

Using the Matrix Inverse Method

Alternatively, we can use the inverse matrix method to solve the system. Given the equation (AX B), we can find (X) by multiplying both sides by the inverse of (A):

First, calculate the determinant of matrix (A):

[text{det}(A) 2(-4) - (-3)(-4) (5)(3) - (3)(-4) - (-2)(2) (-5)(1) -6 - 12 15 12 - 4 - 5 -1]

The inverse of (A) is given by:

[text{adj}(A) begin{bmatrix}0 -1 2 2 -9 23 1 -5 13 end{bmatrix}]

Thus,

[text{adj}(A) -begin{bmatrix}0 -1 2 2 -9 23 1 -5 13 end{bmatrix}]

Finally,

[begin{bmatrix}x y z end{bmatrix} -begin{bmatrix}0 -1 2 2 -9 23 1 -5 13 end{bmatrix}begin{bmatrix}11 -5 -3 end{bmatrix} begin{bmatrix}-(-1) -2 -3 end{bmatrix} begin{bmatrix}1 2 3 end{bmatrix}]

This confirms our solution.

By following these detailed steps, you can solve any system of linear equations using matrices effectively.