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

Consistent vs. Inconsistent Systems

A consistent system of linear equations is one that has at least one solution, while an inconsistent system has no solution because the equations contradict each other. During the row reduction process, if a row is found that corresponds to an impossible equation (for example, 0 = nonzero constant), it indicates that the system is inconsistent. Recognizing the nature of the solution set is crucial when using matrix methods to solve systems.

Gaussian Elimination

Gaussian elimination is an algorithmic process used to solve systems of linear equations by reducing the augmented matrix into an upper triangular or row echelon form through a sequence of elementary row operations. Once in this simpler form, the system can be solved by back substitution, which systematically determines the values of the variables starting from the last equation.

Systems of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. The goal is to find values for these variables that satisfy all equations simultaneously. These systems can have a unique solution, infinitely many solutions, or no solution at all, and they are fundamental in understanding relationships among linear functions and in solving practical problems in various fields.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations, where the coefficients of the variables are placed in a matrix along with the constants from the right-hand side of the equations as a separate column. This format facilitates the application of matrix operations and is essential for methods like Gaussian elimination, allowing the equations to be manipulated systematically.

Elementary Row Operations

Elementary row operations are basic manipulations that can be applied to the rows of an augmented matrix without changing the solution set of the corresponding system. They include swapping two rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another row. These operations form the foundation of the Gaussian elimination process and help transform the matrix into a simpler form for solving the system.

Transcript

00:01 This is a problem in which the concept behind it is to solve a system of linear equations using matrices.

00:09 We will take a system of linear equations, write an augmented matrix for it, and we will try to get that augmented matrix going through some different row operations.

00:20 We will try to get that augmented matrix into what is called reduced row echelon form.

00:26 Now this reduced row echelon form that you are looking at is for a system that involves four equations and four variables.

00:37 Now if you have your matrix in this form, and let's say your variables are x, y, z, and w, then you would be able to conclude that x would be equal to some real number a, that y would equal some real number b, that z would equal some real number c, and that w would equal some real number d.

01:02 And then you would have the solution to your system.

01:06 Now as i'm walking through the problem, there are some notations i'm going to use.

01:11 The lowercase rs are sub 1, r sub 2, r sub 3, or so 4 are going to refer to the row that i'm working with in the current matrix.

01:27 And then the uppercase rs r sub 1 or so 2 or so 3 or so 4 will indicate the row operation that i will perform.

01:41 Okay, so with that in mind, let's start this problem.

01:45 We have a system of equations.

01:47 It has four equations, four variables.

01:50 So i'm going to set up an augmented matrix for it.

01:55 The first equation will give me this first row.

01:58 1, 1, 1, 1, that's using the coefficients of the variables, x, y, z, and w.

02:08 And then i have a constant 4.

02:11 Okay, the second equation will give me this row.

02:17 There's no c.

02:19 The third equation will give me this row.

02:27 And then the fourth equation will give us this fourth row.

02:34 Okay, now keep in mind, our goal is to go through some various row operations to get this reduced row echelon form.

02:43 Now, the fact that we have four equations, four variables, it will take several.

02:50 Row operations to go through.

02:52 And as i go through the various row operations, sometimes i might look like i'm not getting anywhere, but keep in mind, we're trying to get these zeros and ones in the places that they are in the reduced row echelon form.

03:08 I'm going to start off, and since i already have this first entry in my augmented matrix as a one, i'm going to use it to try to get some zeros below it.

03:19 So i'm going to get a new row 2 by taking row 1 times a negative 2 and add that to row 2.

03:32 I'm going to get a new row 3 by taking row 1 times a negative 3 and added that to row 3.

03:46 And then i'm going to have a new row 4 in which i take row 1 times a negative 1 and add that to row 4.

03:55 So i'm going to do several operations in one step.

04:01 We can do that that helps save a little bit of all the matrices you would have to go through.

04:07 Okay, since i'm using row one, i'm not going to change it any, so i'm write it down.

04:13 Okay, now i'm going to take row one times a negative two and add that to row two to produce a new row two.

04:21 And this is what it would look like.

04:25 Just going to go ahead and do the computations, silent, you can do those also work with me okay so there's my new row 2 okay for row 3 i'm gonna multiply row 1 by negative 3 and add that to row 3 and this is what i would come up with and then for my new row 4 i'm gonna take row 1 by negative 1 and add that down to row 4 and i will get this new row 4 okay, so there's my new matrix that i will work off of.

05:17 And one thing that i did, i did accomplish getting the zeros down that first column.

05:23 Okay, so my next row operation, i'm going to form a new row two by taking row three times a negative two and add that to row two.

05:41 And i'm going to get a new row four by taking row three times a negative three and adding that to row four.

05:53 So i'll perform a couple of row operations on this one.

05:56 Now i'm going to go ahead and write down row one since i'm not doing anything too yet.

06:01 I'll make sure i keep it.

06:04 And i'm not going to change row three.

06:09 Not yet.

06:10 So i'm going to go ahead and write it down.

06:14 But i'm going to get a new row two by taking row three times a negative two and adding that to row two.

06:23 And that will produce this row two.

06:35 And i will get a new row 4 by taking row 3 times a negative 3 and adding that to the current row 4.

06:45 Okay, so we'll produce this row.

06:54 Okay, now i've got another 0 in there, and again, what you want to do is try to get a lot of those zeros...

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