Solve each system of equations using matrices (row operations). If the system has no solution, s

Solve each system of equations using matrices (row...

Key Concepts

Inconsistency

Inconsistency in a system of linear equations means that no set of values exists that simultaneously satisfies all equations. This is often identified in a row-reduced augmented matrix when a row reduces to a form leading to an impossible statement, such as 0 = nonzero value. Recognizing inconsistency is crucial in determining whether a system has any solution.

Row Operations

Row operations are methods used to manipulate the rows of a matrix to achieve a simpler form, such as row echelon form or reduced row echelon form. These operations include swapping rows, multiplying a row by a nonzero constant, and adding or subtracting multiples of one row to another. They are essential for systematically solving linear systems through elimination methods.

Systems of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. Solving these systems involves finding the common solution(s) for the variables that satisfy all equations simultaneously. This technique is fundamental in linear algebra and appears in numerous applications across science and engineering.

Matrices

A matrix is a rectangular array of numbers or expressions arranged in rows and columns. In solving systems of linear equations, matrices provide a compact way to represent the system, allowing the use of algebraic operations to simplify and solve the equations. Matrices facilitate techniques such as row reduction and determinants.

Augmented Matrix

An augmented matrix represents a system of linear equations where the coefficients of the variables and the constants from the right-hand side of the equations are combined into one matrix. This format is particularly useful when applying row operations to simplify the system and, in turn, easily identify the solution or determine the nature of the system, such as consistency.

Transcript

00:01 This problem involves using the concept of matrices and row operations to solve a system of linear equations.

00:11 In this problem, we will take a system of linear equations.

00:16 We will write a matrix for that system.

00:21 Then we will perform various row operations on that matrix to try to put it in what is termed reduced row echelon form.

00:32 In green, the matrix i have here in green is the format for a reduced row echelon form matrix.

00:40 Now keep in mind in your system the first column represents a coefficient of x from the system, second column the coefficients of y, and then you have your constants.

00:54 And that is assuming that your equations in your system are in standard form.

01:02 So if you have a matrix in reduced row echelon form, then from that you can form two equations where x equals a, a mean some real number, and y equals b, again b some real number.

01:21 Then that's telling you that the solution to your system is the ordered pair a b.

01:30 So that's the goal that we want to go for in this problem.

01:34 Again, we want to take the system we're given, write a matrix for it, perform various row operations on that matrix to transform it into this reduced row echelon form, so that we can get the solutions to the system.

01:52 Now, as i'm working through the problem, i will have notations such as lowercase r1, lowercase r2.

02:01 And what those are, those are they indicate the rows of my current matrix.

02:12 So i'll say row of current.

02:15 Now the uppercase r1 or 2 are going to involve my row operations to produce a new matrix.

02:29 Okay, so let's get started with this problem.

02:33 I've got this matrix, i'm sorry, this system given to us.

02:38 So we are going to write a matrix for it.

02:42 The first row of our matrix will be 3, negative 5, we've got the vertical bar to separate the coefficients from the constant, and then 3.

02:55 The second row will be 15, 5, and 21.

03:04 Now as you're working with matrices and row operations, you will find that there are many times very different ways to start a problem.

03:15 So the way that i work through this might not be exactly like the way another person would work through it, but what i'm going to do is going to work fine and get us where we need to be.

03:27 Okay, so my first row operation that i'm going to do is i'm going to form a new row two.

03:36 And i'm going to do that by taking a negative 5 times my current row 1 and adding that to row 2.

03:49 Okay, so i'm not going to change row 1.

03:54 I'm going to keep it as i have it right now.

03:58 This step will not change row 1...

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