Solve using elimination. If the system is linearly dependent, state the general solution in term

Solve using elimination. If the system is linearly...

Key Concepts

Linearly Dependent Systems

A system is said to be linearly dependent when one or more equations in the system can be expressed as a linear combination of the others. This characteristic typically results in infinitely many solutions, as not all equations provide independent constraints on the unknowns.

General Solution and Parametrization

When a system is underdetermined or linearly dependent, the solution can often be expressed in a general form with one or more free parameters. This parametrization involves assigning one variable as a parameter and expressing other variables in terms of that parameter, thereby describing the entire family of solutions.

System of Linear Equations

A system of linear equations consists of multiple equations with the same set of unknowns. In solving such systems, one seeks values for the unknowns that satisfy all equations simultaneously. This forms the basis for many algebraic methods used in linear algebra.

Elimination Method

The elimination method involves combining equations to eliminate one or more variables, thereby reducing the system to one with fewer unknowns. This step-by-step process simplifies the solution process by systematically canceling out variables until the remaining ones can be easily solved.

Transcript

00:01 In this problem, we are given the system of equations shown and asked to provide the solution.

00:06 Now, our first step in this might be to make the first equation have a coefficient of 1 in front of its x, in order to eliminate the other xes.

00:17 And as you can see, in the second equation, we have a negative 1 in front of the x, and so a simple way we might make it have a coefficient of 1 is to multiply times negative 1.

00:29 Then we're going to switch that with the first equation, in order to give the first equation, and x of the coefficient of 1.

00:34 And when we do this, our first equation is going to be x minus 2y plus 3 z is equal to 3.

00:44 Our second equation then is going to be the original first equation, and the third equation stays the same.

01:04 Now what we're going to do is use the x in the first equation to eliminate the xes in the other two equations.

01:10 And so we're going to take the first equation, multiply it times negative 3, add it to the second equation.

01:20 And then we're also going to take the first equation multiply times negative 3 and add it to the third equation.

01:29 And when we do this, our new equations are the following.

01:32 The first equation stays the same.

01:39 The first equation becomes negative 3x plus 3x is 0x, negative 2y times negative 3x, negative 2y, add that to the negative 4y.

01:47 Of the second equation, we get 2y.

01:50 Then we get negative 3 times negative, or negative 3x times 3 z is negative 9 z.

01:55 Add that to the 5 z of the second equation we get minus 4 z.

02:01 Then we get negative 3 times 3 is negative 9.

02:04 Add that to the 5 of the second equation, and we get negative 4.

02:12 And for the third equation, we get negative 3x plus 3x, that's 0x.

02:16 Then we get negative 3 times negative 2y, that's 6y.

02:21 Add that to the third equation we get 6y minus 2y, that's 4y.

02:26 Then we get negative 3 times 3 z, negative 9 z, add that to the third equation, negative 9 z plus c, that's negative 8.

02:38 Then we take negative 3 times 3, that's negative 9, add that to the 1 of the 3rd equation, we get negative 8.

02:48 Okay, and now our next step might be to make the y in the second equation have a coefficient of 1, if it's possible, and it seems like it's going to be simple.

02:59 And here, sure enough, if we divide this whole equation by 2, because every coefficient in this equation is visible by 2, we will give the y coefficient of 1.

03:10 So if we're going to do that, and when we do that, our new equations are the following.

03:23 Our new second equation is y minus 2c equals negative 2, and our new third equation is actually just the same.

03:37 Okay.

03:38 Now we're going to take the second equation, multiply it times negative 4, and add it to the third equation to eliminate the coefficient in front of the y, giving it, giving us a zero in front of the y, which is what we want.

03:52 And so when we rewrite all of these equations, the first equation stays the same.

04:03 The second equation also stays the same...

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