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

Inconsistent Systems

An inconsistent system is one in which no solution exists because the equations contradict each other. In the matrix form, inconsistency becomes evident when a row reduces to a form where all the coefficients are zero but the constant term is nonzero, leading to an impossible statement like 0 = 5.

Gaussian Elimination

Gaussian elimination is a method that uses row operations to convert an augmented matrix into an upper triangular or row echelon form. This simplified form of the matrix makes it possible to solve the system by back-substitution. The process is also valuable for checking if a system has a unique solution, infinitely many solutions, or no solution at all.

Parametric or Infinite Solutions

When a system of equations has fewer independent equations than variables, the solution set may be infinite, expressed in terms of free parameters. Identifying free variables through row reduction indicates that there is not a unique solution, and instead, the variables can be expressed in terms of one or more parameters, representing an entire family of solutions.

Matrix Representation

Writing a system of linear equations in the form of an augmented matrix organizes the coefficients and constants into rows and columns. This structured format simplifies the process of applying systematic methods, such as row operations, to solve the system. It also provides a clear visualization of the relationships among the variables and the equations.

System of Linear Equations

A system of linear equations consists of multiple equations involving the same set of variables. The goal is to determine values for the variables that satisfy all equations simultaneously. Such systems are fundamental in various fields of study, including mathematics, engineering, and economics, as they can model diverse problems ranging from simple relationships to complex real-world phenomena.

Row Operations

Row operations are the manipulations performed on the rows of an augmented matrix to simplify it. These include swapping rows, multiplying a row by a nonzero constant, and adding a multiple of one row to another row. Applying these operations allows us to systematically reduce the matrix, making it easier to solve for the variables or to identify inconsistencies.

Transcript

00:01 We have a system of equations 4x plus y equals 5 2x minus y plus z equals or minus w equals 5 and z plus w equals 4.

00:09 We want to solve for this system of equations.

00:12 The way we're going to do that is by using matrices and raw operations on those matrices.

00:18 So the first step is going to be figuring out what size of matrix we need.

00:23 Since there are four variables x, y, z, and w, we know there are going to be four columns.

00:29 X, y, z, and w.

00:34 And we have the constants here.

00:36 So we have a row for the constants.

00:39 Then we have three rows since there are three equations, right? we have row one, row two, and row three.

00:48 So now we can fill this in.

00:50 Negative four, two, zero, because there's no x.

00:56 We have y.

00:57 So 1, negative 1, 0.

01:00 We have 0, 1, 1, 0, 0, negative 1, and 1.

01:11 Then we have the constant of 5, 5, and 4.

01:17 Okay, so now we want to get this in row echelon form, where everything here on the diagonal is equal to 1, sorry, everything beneath the diagonal is equal to 0, which as you can kind of see we're already a good step with the way there.

01:40 So we want to get this 2 here equal to 0.

01:45 So if we take, right, we got our new row 2 is equal to, we take two copies of row 2.

01:52 So this becomes 4 and add row 1, this negative 4 plus 4 is going to be 0.

01:58 So we can do that.

02:00 Okay, so we've got row 2 is equal to 2 row 2.

02:07 Plus row one right now it's right out row two with two times row two which is gonna be four negative two two negative two and ten so now we can add row one to that right so we get negative four plus one zero zero and five so now we can add these up and this is going to give us our new row two value zero zero negative 2 plus 1 is negative 1 2 negative 2 and 15 right now we have our new row 2 which we can erase our old row 2 and replace it with our new row 2 which is 0 negative 1 2 negative 2 and 15 so now we've got everything beneath the diagonal equal to 0 right the 0 this is 0 this is 0 this is 0 this is 0.

03:10 Now we want to have everything on the diagonal equal to 1.

03:14 Okay, so the way we're going to do that is by taking whatever value is on the diagonal for that row and multiplying the entire row by the reciprocal of that value.

03:26 Okay.

03:27 So if we had negative 1 ,000 in the diagonal, we would multiply the entire row by negative 1 over 1 ,000.

03:36 Here we have negative 4, so we're going to take.

03:40 Our new roll 1 is going to be equal to 1 over negative 4 times roll 1, right? negative 4 times negative 4, or negative 4 divided by negative 4, that's going to be 1.

03:56 Okay.

03:59 1 times negative 1 4th is going to be negative 1 over 4.

04:06 Anything times 0 is 0.

04:09 And we have this 5, right? so then we've got multiplying that times negative 1 .4, we get negative 5 over 4.

04:21 So now we can erase our roll 1.

04:24 Erase this equation here.

04:26 Now we want to get this value here, this negative 1, to be 1.

04:32 Okay.

04:33 Now we know that if we take negative 1 times negative 1, we're going to get positive 1.

04:39 And multiplying anything times negative 1 is the same as just flipping the sign, right? like that.

04:46 So, what we can do is instead of doing a whole bunch of complicated reciprocals, let's just change the sign on all these, right? that becomes 1, that becomes negative 2, there comes positive 2, and it becomes negative 15.

05:02 So now we've got it in raw echelon form, right where everything on the diagonal equals 0, everything beneath the diagonal, equals 0 and everything on the diagonal is equal 1.

05:13 Sorry about that.

05:14 Now we want to convert it back into, a system of equations.

05:20 So that way it's a little easier what we're working with.

05:24 It's a little easier to see what we're working with...

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