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.

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.

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.

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.

Transcript

00:01 In this problem, you are asked to solve the given system of linear equations using matrices and row operations.

00:10 We are going to do this by taking the given system, write the augmented matrix for it, and then go through some various row operations to convert the augmented matrix into row echelon, reduced row echelon form.

00:28 Now if we can get the augmented matrix into a reduced row echelon form and assuming that the first column represents the coefficients of x, the second column y and then z and w, then we will know that this value of a is our x and then b is or y, c is our z and d would be our w.

00:56 Okay so let's take a look at the problem we're given and let's write the augmented matrix for that problem okay so the augmented matrix would start off looking like this from the first equation the coefficients of all of the variables are one and the constant is four okay the second equation we could write two negative one one zero with the constant being zero.

01:36 The third equation will have three, two, one, negative one with the constant being six, and then we have one, negative two, negative two, negative two, and negative one from our last equation.

01:56 And we're going to go through a series of row operations, and remember what we're trying to do is to convert this augmented matrix into this reduced row x -on -form.

02:07 It will take several row operations to go through.

02:12 There are different ways to work the problem, and i'm just going to go through one of the methods of what i see.

02:19 Now, the thing that i like to do is i like to go for as many zeros as i can get and try to never lose those zeros.

02:29 At the same time, when i see an opportunity to get the ones down these diagonals, i will also take that opportunity.

02:38 Again, i go for the zeros as much as i can, and once i get a zero, i try to never lose it.

02:46 And i also take whatever opportunities i can to get the ones down the diagonals.

02:51 And then usually i'll start off, and i will try to get the one up here.

02:54 If it's not given to me as a one, i try to get that.

02:58 And then my next row operations, i will get the zeros under it, and then go from there.

03:03 So that's what i'm going to do on this problem.

03:05 I start off and i can see i have the 1 right here at the front.

03:09 There's my 1.

03:11 So i'm going to go through some row operations to turn the numbers under it into 0.

03:17 All right, so the first thing i'm going to do is i'm going to get a new row 2 by taking row 1 times a negative 2 and add that to the current row 2.

03:29 So let's do that.

03:31 Since i'm using row 1, i'm not going to change it.

03:42 But i'm going to produce a new row 2 and again what i'm going to do is i'm going to take row 1 times a negative 2 and add it to row 2 so that will produce this okay, so i've got one of my zeros i'm going to produce a new row 3 by taking the current row 1 again times a negative 3 and add that to the current row 3 okay, so now i'm going to take a negative 3 here and add it to the row 3 and when i do that it will produce this row 3 and okay, and then i want to produce a new row 4.

04:36 So we're going to get those zeros down that first column under that 1 i will do that by taking my current row 1 times a negative 1 and add that to the current row 4 okay, so now i'm going to use a negative 1 here take a negative 1 times row 1 and when i do that it will produce this this new row 4 okay, so that's our first set of row operations okay, so now what i need to do, and i keep in mind, i'm trying for as many zeros as i can try to get.

05:19 So what i'm going to do now is i'm going to produce a new row one by taking row three and adding it to row one.

05:34 Okay, since i'm using row three, i'm not going to change it, so i'm going to go ahead and write row three down.

05:45 But i'm going to add row three to row one.

05:48 So i keep my one there where i need it.

05:50 But then this will turn into a zero.

05:54 We'll have a negative 1, negative 3, and then a negative 2.

06:01 Okay, i'm also going to produce a new row 2 by taking row 3 times a negative 2 and adding that to the current row 2...

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