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In Exercises 19 and $20,$ choose $h$ and $k$ such that the system has (a) no solution, (b) a unique solution, and (c) many solutions. Give separate answers for each part.$\begin{aligned} x_{1}+h x_{2} &=2 \\ 4 x_{1}+8 x_{2} &=k \end{aligned}$

No Solution if $h=2, k \neq 8$Unique Solution if $h \neq 2, k \neq 8$Infinite Solutions if $h=2, k=8$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

Baylor University

Idaho State University

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in this video, we have a system of two linear equations and two unknowns x one next to, and we're going to analyze the consistency and a number of solutions when we change their values of H and K. The first thing to do here is to make an augmented matrix out of this system. So from the X one column will have 14 and the x two column. We have h eight. We're augmenting with the value of two K. Now we just need to put this in echelon form, since our goal is not to solve the entire system, and that means that can be done one step. Once we eliminate that for, I will say that our row operation here is going to be row to replaced with itself. And if we take negative four times row one added to row to, then we'll wipe out this quantity here. So first copy Row one, it's one H and to then multiplied by negative four. Adding to Row two gives us zero first. Then it's H minus or excuse me eight minus four age, and then it will be K minus eight. 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 eight minus four h must be equal to zero and K minus eight is not zero. So this situation happens when let's put an end here. So this happens when H is too and que is not equal to eight will 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 and we know from the uniqueness and consistency here, um that a unique solution will happen specifically when this entry is a pivot. And how do we make that a pivot? We just have to insist. Eight to minus four H is not equal to zero. Well, that would be negative, for H is not equal to negative eight. And this implies H cannot be equal to a positive two. So when this is satisfied, will have a unique solution. Now, what about the last possibility? We know that for any linear system of equations, there is either no solution, a unique solution or infinitely many solutions. Here we can say that there are in fine, nightly, many solutions when let's say what happens in that case we'll have in Fallujah May suit is infinitely many solutions precisely when there is a free variable. So how do we make it free? Variable? We would just have to insist that this quantity goes to zero and at the same time recall that this must also be zero other else. Otherwise, there'd be no solutions. We would be inconsistent. So let's write down. We have infinitely many solutions when eight minus four h equals zero and K minus 80 So this would imply that H is too and que is eight. So these values will give us infinitely many solutions. If we insist h is not too will have a unique solution. And last, if h is to make a is not eight, there would be no solutions. And these are all analyzed based on where the pivots would be

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