In Exercises 19 and 20, choose h and k such that the system has (a) no solution, (b) a unique so

In Exercises 19 and 20, choose h and k such that the system...

Key Concepts

Systems of Linear Equations

A system of linear equations is a set of equations with multiple variables that are solved simultaneously. The goal is to find variable values that satisfy all of the equations. The analysis of such systems involves checking conditions for the existence and uniqueness of solutions, and it forms a fundamental component of linear algebra.

Determinant and Unique Solutions

For a system of linear equations represented in matrix form, the determinant of the coefficient matrix determines whether a unique solution exists. If the determinant is nonzero, the system has a unique solution because the matrix is invertible. In problems where parameters appear, the values of these parameters affect the determinant, which in turn determines the uniqueness of the solution.

Consistency of Systems

A system is said to be consistent if there exists at least one solution; otherwise, it is inconsistent. Consistency involves ensuring that the equations do not conflict with each other. When parameter choices lead to conflicting equations (often signaled by a zero determinant without proportional right-hand sides), the system is inconsistent and has no solution.

Dependent Equations and Many Solutions

If a system has a zero determinant and the equations are multiples of each other (i.e., they are dependent), the system has infinitely many solutions. This is because the equations describe the same geometric object (such as a line in two dimensions), meaning that there is not a single point of intersection but rather an entire overlapping set of solutions.

Transcript

00:01 In this video, we have a system of two linear equations in two unknowns, x1 and x2, and we're going to analyze the consistency and number of solutions when we change the values of h and k.

00:14 The first thing to do here is to make an augmented matrix out of this system.

00:19 So from the x1 column, we'll have 1 .4.

00:22 In the x2 column, we have h8.

00:25 We're augmenting with the value of 2k.

00:31 Now we just need to put this in echelon form, since our goal is not to solve the entire system.

00:36 And that means that can be done in one step.

00:38 Once we eliminate that four, we'll say that our row operation here is going to be row two replaced with itself.

00:47 And if we take negative four times row one, added to row two, then we'll wipe out this quantity here.

00:54 So first copy row one.

00:55 It's one, h, and two, then multiply it by negative.

01:01 To 4 adding to row 2 gives us 0 first then it's h minus or excuse me 8 minus 4h and then it'll be k minus 8 now let's analyze what we have here first let's determine this situation when there will be no solutions so there are no solutions when well that happens specifically when there is a pivot here that would mean 8 minus 4h must be equal to 0 and k minus 8 is not 0 so this situation happens when let's put an and here so this happens when h is 2 and k is not equal to 8 we'll have a pivot here and that means there is no solutions now what else could happen instead of no solutions we could have a situation where there is a unique solution.

02:14 And we know from the uniqueness and consistency theorem that a unique solution will happen specifically when this entry is a pivot.

02:23 And how do we make that a pivot? we just have to insist 8 minus 4h is not equal to 0...

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