Find the solution to the homogeneous system 2 x1+x2-2 x3=0,2 x1-x2-2 x3=0. Then solve the inhomo

Find the solution to the homogeneous system 2 x1+x2-2 x3=0,2...

Key Concepts

Superposition Principle

The superposition principle states that for linear systems, any linear combination of solutions to the homogeneous system is also a solution, and the solution to an inhomogeneous system can be written as the sum of a particular solution and any solution to the homogeneous system. This principle is fundamental in understanding the structure and behavior of solutions in linear algebra.

Homogeneous Linear Systems

A homogeneous linear system is one in which all the constant terms are zero. This system always has at least the trivial solution (all variables equal to zero), and its set of solutions forms a vector subspace. Analyzing the homogeneous system helps to understand the structure of the solution space, including the dimension and the free parameters that describe the null space of the system’s coefficient matrix.

Inhomogeneous Linear Systems

An inhomogeneous linear system has nonzero constant terms on the right-hand side. When such a system is consistent, its general solution can be expressed as the sum of a particular solution and the general solution of the corresponding homogeneous system. This means that the solution set is an affine subspace, which is essentially a translation of a vector subspace.

Parametrization and Free Variables

Parametrization involves expressing the solutions of a system in terms of free parameters when the system has infinitely many solutions. Free variables arise when there are fewer independent equations than unknowns. In both homogeneous and consistent inhomogeneous systems, these free parameters characterize the degree of freedom in the solution set.

Transcript

00:01 Okay, so we want to go ahead and solve this linear system that's non -homogeneous.

00:06 And we know that for non -homogeneous equations, we have the superposition principle, which says that our general solution is just the sum of our homogenous and our particular solutions.

00:16 So we can go ahead and solve the homogenous equation, which is just this system here.

00:28 So we're writing this in the form we like to do.

00:30 We'll get the following system.

00:37 And we'll write at h for homogenous and be clear that this is not this actual system that we're trying to solve.

00:44 It's just the first step.

00:45 So we'll have d plus 2x2h.

00:50 Sorry, it should be d minus 1.

00:52 D minus 1, x2h equals minus 2x1.

00:57 So then we'll go ahead and perform that d minus 1 operation on the top equation to get our substitution.

01:06 To work.

01:08 So we'll have this and we'll obviously substitute d minus 1 x2 for a negative x1h and i will make the positive with the negative side in front.

01:22 So expanding our, if we expand this and then have a positive 2, we'll end up this x1h is 0, and we'll end up with our final polynomial as follows.

01:40 And we know if we solve this characteristic equation, we get x is, lambda 0, lambda negative 1, which means that our equation is just c1 plus c2, e to minus t.

01:57 The e to 0 t becomes just a constant.

02:02 So then this is our x1 homogenous.

02:06 Finding our x2 homogenous is simple as well, because we know the relationship between them already, that is just equal to d plus 2 x1h.

02:16 And we know that derivative of a constant just 0 and adding 2 of it is just 2c1.

02:24 And then derivative is negative 1 and then positive 2.

02:31 So that's just a positive 1, c2, e to the negative t.

02:35 So these are our homogenous solutions.

02:39 And now we want to find our particular solutions.

02:42 So we're going to go ahead and do that.

02:45 And we'll take a good look of, we'll rewrite these two equations right here, except including our non -humogenous terms.

03:00 So then we get the following, that d plus 2 x1h minus x2h is equal to t, and d minus 1, x2h, is just negative 2 x1...

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