Use the Gaussian elimination method to solve each system of equations. For systems in three vari

Use the Gaussian elimination method to solve each system of...

Key Concepts

Elementary Row Operations

Elementary row operations are the basic manipulations performed on the rows of an augmented matrix. These include swapping rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another. These operations are crucial because they do not alter the solution set of the system but help achieve the desired row-echelon form.

Free Variables and Parametrization

In cases where a system has infinitely many solutions, some variables may not be directly determined by the equations; these are called free variables. The solution set is then expressed parametrically by assigning arbitrary values to these free variables, clarifying the infinite nature of the solution set.

Back-Substitution

Back-substitution is the process of solving for the variables after the augmented matrix has been transformed into an upper triangular or row-echelon form. Starting from the last nonzero row, the value of each variable is determined by substituting the values of already solved variables, working upwards through the system.

Augmented Matrix

An augmented matrix represents a system of linear equations by combining the coefficients of the variables and the constants into a single matrix. This format simplifies the application of row operations during the process of Gaussian elimination and helps track the transformation of the original system.

Systems of Linear Equations

Systems of linear equations consist of multiple linear equations with several variables, all of which are to be solved simultaneously. These systems can have a unique solution, infinitely many solutions, or no solution at all, depending on the relationship and consistency between the equations.

Gaussian Elimination

Gaussian elimination is a systematic method used to simplify and solve systems of linear equations. It involves applying elementary row operations to transform the augmented matrix of the system into an upper triangular or row-echelon form, which then makes it easier to perform back-substitution and determine the solutions of the system.

Transcript

00:01 The concept involved in this problem is to solve a system of linear equations using gauss elimination.

00:10 We will use the system that is given to us, the coefficients, and the constants of the system to set up an augmented matrix, built there a series of row operations to rewrite the augmented matrix in row echelon form, and then solve the system.

00:32 So let's first run set up our augmented matrix.

00:37 Okay, so the first row of the matrix will be 1, 2, 1, 1, and on the other side of the vertical ball will be the constant 2.

00:48 Okay, then we'll have 1, negative 1, 1, and 4.

00:55 Okay, 3, 4, 2, negative 1, 8.

01:02 And the last row, two, three, four, five, and five.

01:11 Okay, and as i said, we're going to go through a series of row operations, and this being a system with four equations, four variables, it will take several row operations.

01:22 All right, the first thing i am going to do, i'm going to do several things to this matrix, and it's going to involve performing some operations on row one.

01:32 So the first thing i'm going to do is i'm going to take row 1 times a negative 1 and add that result to row 2.

01:43 Okay, so let's do that.

01:46 Since i'm going to perform the operations on row 1, i'm going to leave row 1 as it is.

01:59 And as i said, i'm going to multiply row 1 times a negative 1 and add that to row 2.

02:06 And that will produce 0, negative 3, 0, negative 2, and 2.

02:21 Okay, then the next thing i'm going to do is i'm going to take row 1 again, and this time i'm going to multiply times a negative 3 and add that result to row 3.

02:33 Okay, so this time i'm multiplying by a negative 3.

02:38 Okay, and that will produce 0, negative 2, negative 1, negative 4, and positive 2.

02:52 Okay then one more operation here on this, i'm going to take row 1 this time and multiply times negative 2 and add that result to row 4.

03:04 This time i'm multiplying by negative 2.

03:08 And that will produce 0, negative 1, 2, and 1, 2, and 1.

03:18 So now i am going to take row 4 times a negative 2 and add that to row 2.

03:41 That seems a little strange but in just a minute it will all start rolling place.

03:46 So i will keep row 1 like it is and i'm going to multiply row 4 times a negative 2 and add that to row 2 and add that to row 2...

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