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

Gaussian Elimination

Gaussian elimination is a method that uses elementary row operations to transform the augmented matrix into an upper triangular (or row echelon) form. Once this form is achieved, back-substitution can be applied to efficiently determine the values of the variables.

Elementary Row Operations

Elementary row operations include swapping two rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another row. These operations are the building blocks of matrix manipulation techniques such as Gaussian elimination, ensuring that the solution set of the system is preserved while transforming the matrix.

Consistency and Inconsistency

A system of equations is termed consistent if it has at least one solution, and inconsistent if no solution exists. In the matrix context, inconsistency is typically identified by the emergence of a row in the augmented matrix that represents a contradiction (such as an equation like 0 = nonzero).

Matrix Representation

Matrix representation involves organizing the coefficients of the variables from the system of equations into a rectangular array. This compact form makes it easier to apply algebraic operations, particularly when solving multiple equations simultaneously using systematic approaches.

System of Linear Equations

A system of linear equations consists of multiple equations involving the same set of variables. The goal is to find values for the variables that satisfy all the equations simultaneously. This concept is fundamental in linear algebra and appears in many areas of mathematics and applied sciences.

Augmented Matrix

An augmented matrix combines the coefficient matrix and the constants from the right-hand side of the equations into one matrix. This structure is particularly useful for applying row operations and is a standard format in methods such as Gaussian elimination.

Transcript

00:01 The concept, the idea behind working this problem is to take a system of linear equations and solve the system using matrices.

00:11 Now as we use matrices, we will have to use some various row operations.

00:17 Our goal is we're going through the row operation is to try to transform the matrix from the system into what is known as a reduced row echelon form matrix.

00:33 Now i have here a template for a reduced row echelon form matrix for a system that was what's termed a three by three system, one that has three equations and three variables.

00:47 Now with this reduced row echelon form, remember that the first column represents the coefficients of x, the second of y and the third of z, and then we have our constants so if you can get your matrix into the row echelon the reduced row echelon form then basically you are able to say that x equals a where a is some real number that y equals b and that z is equal to c and you would have your order triple that is a solution to that system so that's what now we are going to take the system that we're given, we will write a matrix for that system.

01:39 We'll go through some row operations and try to change the matrix into this reduced row echelon form.

01:48 Now it does often take several steps to work through.

01:53 As i'm walking through this problem, i will have some notations such as lowercase r1, lowercase r2, and lower case r3.

02:06 And what these are going to indicate is the row of a current matrix that i'm working with.

02:14 Okay, the row of a current matrix, like row one, row two, row three, the current matrix that i'm working with.

02:21 Now when i'm using uppercase r1, r2, r3, this is going to indicate the row operation that i will perform.

02:36 Okay, so let's get started.

02:39 I've got this system that my first step is to write a matrix associated with it.

02:46 So using the first equation, the coefficients and the constant, i will have 1, negative 2, 3, vertical bar 7.

02:58 Second equation, 2, 1, 1, 4.

03:03 The third equation will give me this third row of negative three, two, negative two, and negative ten.

03:13 Now i'm going to perform some row of operations on this matrix to produce other matrices and again my goal is to try to get the matrix in this form.

03:27 There are often different ways to start a problem.

03:34 I'm going to go through what kind of strikes me first as i'm going through this.

03:39 And one of the things that you do want to try to do is get the ones in spots where they need to be and the zeros in the spots that they need to be.

03:50 And as soon as you can get one of the rows with the zeros and ones where they need to be, that's going to be your main road to work with.

03:58 So keep in mind the goal that you're going for so that you're not just going in a bunch of circles doing one thing or another and not accomplishing anything.

04:06 So when i look at this matrix, i have a 1 right here.

04:12 I like it there.

04:14 That's the way it's supposed to be with my reduced row echelon.

04:18 So i'm going to work with that row a little bit.

04:21 But i'm going to make a new row 2.

04:26 So my operational row 2 is going to be to take the current row 1, multiply by negative 2, and add that to row 2.

04:39 I'm also on this one i can do two row operations in the same step.

04:45 So i'm also going to make a new row 3 by taking row 1 times 3 and adding that to row 3.

04:56 Okay so this is what my new matrix is going to look like.

05:00 I will keep my first row as it is.

05:03 I'm not changing it.

05:06 Not now anyway...