System Of Linear Equations Using Matrices

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Solving Systems of Linear Equations: A full breakdown Using Matrices

A system of linear equations is a collection of two or more linear equations involving the same set of variables. Plus, this is where the power of matrices comes into play. In practice, using matrices provides a systematic, efficient, and scalable method for solving systems of any size, forming a cornerstone of linear algebra with vast applications in engineering, computer science, economics, and data science. Practically speaking, while simple systems with two or three variables can often be solved by graphing or substitution, these methods become incredibly cumbersome for larger systems. This guide will walk you through the fundamental concepts and step-by-step techniques for solving these systems using matrix algebra No workaround needed..

What is a Matrix Representation of a System?

Before diving into solution methods, it's crucial to understand how a system of equations is translated into matrix form. Consider a system of three linear equations with three variables (x, y, z):

1. 2x + y - z = 5
2. x - y + 2z = -3
3. 3x + y + z = 4

This system can be represented compactly using a matrix equation of the form A X = B.

• Coefficient Matrix (A): This matrix contains only the coefficients of the variables. A = [ [2, 1, -1], [1, -1, 2], [3, 1, 1] ]

• Variable Matrix (X): This is a column matrix containing the variables. X = [ [x], [y], [z] ]

• Constant Matrix (B): This is a column matrix containing the constants from the right-hand side of each equation. B = [ [5], [-3], [4] ]

The entire system is now condensed into the single matrix equation: [ [2, 1, -1], [ [x], [ [5], [1, -1,2], * [y], = [-3], [3, 1, 1] ] [z] ] [4] ]

This representation, A X = B, is the foundation for all matrix-based solution methods.

Method 1: Gaussian Elimination (Row Reduction)

Gaussian elimination is a direct method for solving the matrix equation A X = B. The goal is to transform the augmented matrix [A|B] into a form from which the solution can be easily found through back-substitution. The augmented matrix simply combines the coefficient matrix A and the constant matrix B.

For our example, the augmented matrix is: [A|B] = [ [2, 1, -1 | 5], [1, -1, 2 | -3], [3, 1, 1 | 4] ]

We perform elementary row operations to simplify this matrix. On the flip side, the allowed operations are:

1. Practically speaking, swap any two rows. 2. In practice, multiply a row by a non-zero scalar. Because of that, 3. Add or subtract a multiple of one row to/from another row.

The target is Row Echelon Form (REF), where:

• The first non-zero element (the leading coefficient) in each row is 1. But * Each leading 1 is to the right of the leading 1 in the row above it. * Rows with all zeros are at the bottom.

Let's solve our system step-by-step.

Step 1: Get a leading 1 in the first row, first column. It's often easier to swap rows to get a 1 in the top-left position. Swap Row 1 (R1) and Row 2 (R2): New Matrix: [ [1, -1, 2 | -3], [2, 1, -1 | 5], [3, 1, 1 | 4] ]

Step 2: Create zeros below the leading 1 in the first column.

• To eliminate the 2 in R2, perform: R2 -> R2 - 2*R1 (2 - 21 = 0), (1 - 2(-1) = 3), (-1 - 22 = -5), (5 - 2(-3) = 11)
• To eliminate the 3 in R3, perform: R3 -> R3 - 3*R1 (3 - 31 = 0), (1 - 3(-1) = 4), (1 - 32 = -5), (4 - 3(-3) = 13)

Matrix becomes: [ [1, -1, 2 | -3], [0, 3, -5 | 11], [0, 4, -5 | 13] ]

Step 3: Get a leading 1 in the second row, second column. Divide the second row by 3: R2 -> R2 / 3 Matrix becomes: [ [1, -1, 2 | -3], [0, 1, -5/3 | 11/3], [0, 4, -5 | 13] ]

Step 4: Create a zero below the leading 1 in the second column. To eliminate the 4 in R3, perform: R3 -> R3 - 4*R2 (0 - 40 = 0), (4 - 41 = 0), (-5 - 4*(-5/3) = -5 + 20/3 = 5/3), (13 - 4*(11/3) = 13 - 44/3 = -5/3)

Matrix becomes: [ [1, -1, 2 | -3], [0, 1, -5/3 | 11/3], [0, 0, 5/3 | -5/3] ]

This is in Row Echelon Form. That said, x - y + 2z = -3 2. The corresponding system of equations is now:

1. y - (5/3)z = 11/3

Step 5: Back-Substitution Now we solve from the bottom up.

• From equation (3): (5/3)z = -5/3 => z = -1
• Substitute z = -1 into equation (2): y - (5/3)(-1) = 11/3 => y + 5/3 = 11/3 => y = 6/3 = 2
• Substitute y = 2 and z = -1 into equation (1): x - 2 + 2(-1) = -3 => x - 2 - 2 = -3 => x - 4 = -3 => x = 1

The solution is x = 1, y = 2, z = -1.

Method 2: Matrix Inversion

The matrix inversion method is an elegant approach that directly solves A X = B by multiplying both sides by the inverse of matrix A, denoted as A⁻¹. If A⁻¹ exists, the solution is simply **X =