in-exercises-19-and-20-choose-h-and-k-such-that-the-system-has-a-no-solution-b-a-unique-solution-and

Question

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.
$\begin{aligned} x_{1}+h x_{2} &=2 \ 4 x_{1}+8 x_{2} &=k \end{aligned}$

Michael Jacobsen · Accepted Answer

in this video, we have a system of two linear equations and two unknowns x one next to, and we&#x27;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&#x27;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&#x27;ll wipe out this quantity here. So first copy Row one, it&#x27;s one H and to then multiplied by negative four. Adding to Row two gives us zero first. Then it&#x27;s H minus or excuse me eight minus four age, and then it will be K minus eight. Now let&#x27;s analyze what we have here. First, let&#x27;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&#x27;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&#x27;s say what happens in that case we&#x27;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&#x27;d be no solutions. We would be inconsistent. So let&#x27;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

Arden Mccarthy · Answer

this question asks us to solve the given land your system, using any methods. So we&#x27;re going to use the Gaussian elimination method, which means we need to put our Matrix and to reduce national on form and then solve for the variables from there. So the equations that were given we can use to make a matrix, which from the first equation the first row is gonna read. X one plus three x two plus zero x three Since there&#x27;s no x three term plus four x four equal zero, the next one is going to give us x one plus four x two plus two x three is equal to zero then our third equation gives us negative two x two minus two x three minus x four equals zero. After that, we have two x one minus four x two plus x three plus Explores equal to zero and finally we have Excellent, but it&#x27;s two x two, but his ex Terry Plus at four is equal to zero. So now we work on this matrix to put it in, reduce special informed. So the first step we&#x27;re gonna do is we&#x27;re gonna switch Row one and road for So are one of those are for and our four equals. Our current are one. We do that we get this matrix. Okay, once we have that matrix, what we&#x27;re going to do is we&#x27;re going to try to knock out this term up here to make it a zero. So in order to do that, we need to take the second row and said it equal to our tumor in his 1/2 are one. So once we do that, we worked that out and we get this matrix so the first row stays the same. And then the second row changes because of the previous row operation we just did on it and all the other roser going states name cause we&#x27;re not doing any row operations on them. Okay, Now our next up is we want Thio make this entry is zero. And so, in order to do that, we take the fourth row and we subtract 1/2 our one from it. When we do that, we get this nature. So first rose, the same second room&#x27;s the same. Third row is gonna be the same. And then our fourth row is going to become zero guys. Negative 1/2 seven house and zero. And then our last road stays the same. Okay, so now what we want to do is we want to make this entry a zero, and we&#x27;re knocking all of zeroes out and this column or not, knocking all the non zero entries out in this column and making them zero. So we&#x27;re left with this as the only non zero element in the column, because it&#x27;s gonna be a pivot. So in order to knock out that one right there, we have to take the fifth row. So are five and subtract 1/2 are one from it. When we do that, you got this matrix. First Row stays the same as just a second. That&#x27;s just third and the fourth. And in the fifth room becomes 00 Negative three. House. What? Huh? And zero. So now we&#x27;ve successfully knocked out all the zeros, the non zero elements in that column except the first pivotal one, which is the two. And our next step is going to be to work on the second column. So what we want to do is we want to take our or three term right here, and we want to make this a zero. So to do that, we say or three is equal to 1/3 or three. Uh, 1/3 or two. Plus our three. Let me do that. We get this matrix, our third row is going to become 00 Negative. Three halfs. Negative. 76 zero, and then the other roads, we&#x27;re gonna stay the same. Okay? So as you probably guessed, the next thing that we want to do is get rid of this five and make it a zero entry. So to do that, we want to take the fourth row. It&#x27;s a tract from that 56 of the second row. When we do that, we get this matrix here. We&#x27;ll see that are five got knocked out. Not our fifth roses. Okay, so the next thing that we&#x27;re gonna do is we&#x27;re going to switch are three with our four, and we&#x27;re just doing that. Thio help the steps for getting into recession form. So the pivots fall nicely with each other in that staggering format. Okay, so now after that, what we want to dio is we want to make this element zero So in order to do that, we take or fourth row and subtract 67 or three from it. Once we do that, we get the following matrix, and then our fourth row becomes this. We see that we&#x27;ve knocked out that determine its zero now. Okay, Now our next step is to make this a zero. So in order to do that, we take our fifth row. So are five. And from our five, we subtract 67 are three. Once we do that, we get this new Matrix. But this is zero. And then negative. 95 over 12. And then our last road becomes 000 and then negative 20 over seven zero. All right, so now what we want to do is we want to get rid of this tournament. We want to make it a zero. So in order to do that, we take or five. We said that equal to or five one is 12 over 19 are for still, once we do that, we had this matrix, which is to native four warned 10 063 House Fate of 1/2 0 00 Negative. 7/4. 47. 12 zero 000 Negative. 95 over 21 0 And then finally, our row of all zeros. Now what we want to dio is we want to make sure that each of these pivot points are all equal to one because that&#x27;s a requirement bird. There do special informants in threes and gassy and elimination. Uh, well, put it and reduce social inform before we do any any more solving. So we can do these all the same stuff we can say our one is equal to 1/2 are one are two is equal to 1/6 are too are three is equal to negative 4/7 are three and or four is equal to negative 21 95th are for It&#x27;s me do that. We get this new matrix and then we start calling for our variables x one x two x three The next four So we&#x27;ll start with this road right here and work our way up. So this first row So we&#x27;re selling for explore the streets Explore is equal to zero. We moved to the third row appear which reads x three waas negative 47 over 21 times X four is equal to zero. Since we know from here that exportable zero, you know, that cancels, which means that X three has to equal zero. Okay, Still, for X two, we do exactly the same thing. X two plus 1/4 x three minus 1 12 x four equals zero. We know that X four is equal to zero. No exterior is equal to zero. So these will cancel, Which leaves us with x two is equal to zero And for our last one x one we see x one minus two x two plus 1/2 x three was 1/2 We&#x27;re running out of room a little bit, so all right, a little smaller one minus two x two 1st 1/2 x three. 1st 1/2 that&#x27;s four equals zero. And since we have X for X three index to all equal zero, he&#x27;s all cancel. Which leaves us with X one equals zero. So that is gonna be our solution. It&#x27;s the trivial solution. And we have that x one x two x three and four is equal to the zero doctor. So all these equal zero

Julie Silva · Answer

So we&#x27;re giving the equation three x squared minus each expose four K is equal to zero and we&#x27;re being asked to solve for H and K, and we&#x27;re being told to use exercise 68 72 to help us. So the first thing they tell us is that the some of the solutions is negative. 12. Well, what we found in number 72 is that the summer solutions would be equal to negative, be over a So in this particular case, are some is negative 12. So we know that negative 12 will be equal to negative B. So we have minus negative, each divided by a, which is three. So in this case, we can solve for H. These two negatives in the numerator will become positive. So we&#x27;ll have negative 12 is equal to H over three. And then we just need to multiply both sides of our equation by three and three times Negative 12 is negative. 36. Okay, so now we&#x27;ve solved for H now in 68 they told us that the product of our solutions will be equal to see over a well are the product they told us is 20. See, in this case is four K and A is three. So now we just have to solve for K So I&#x27;m gonna multiply both sides by three. So we get 60 is equal to four k and I will divide both sides of our equation by four and 60 divided by four is 15 and now we&#x27;ve solved for H and K.

Robert Daugherty · Answer

16 3 dot to Problem number 23 were asked to take a look at this linear system and find all values of the constant K that would guarantee me a unique solution for this system. So unique solution means that the determinant is non zero. So the determinate of the system is going to be the determinant. So right, the majors a coefficient, So one okay. And then K four. So to have a unique solution, this cannot be equal to zero. So that means that when you take the determinant here, so one times for minus K times K cannot be zero. That means that four minus k squared not equal to zero. So the only time that that would ever be equal to zero as when K is equal to plus or minus two. So OK, cannot be plus or minus two. So that&#x27;s the condition. So every other value of K would give me a unique solution. So all values all real number for K, with the exception of plus or minus two to get a unique solution for this system,

Ashley Jordon · Answer

we&#x27;re trying to find when our system has two solutions, meaning it touches twice. So I see if K equals negative, too. We have two solutions. That means if K equals positive, too, we also have two solutions. That&#x27;s because K is being squared. Same thing would happen if Kay was three. We have two solutions, and if Kay was negative three, we would also have to solutions.

Ashley Jordon · Answer

for this problem. What we&#x27;re changing here is the K which is the radius of the circle, and we&#x27;re trying to find what values of K are going to give us no solutions to the system. So right away, I noticed. Right now there&#x27;s no solution, these two graphs or not intersecting. But once I get to to you would intersect and then once again at three. But numbers larger than three is no solution. The number is less than two would be no solution. Let&#x27;s look at it on the negative end if Kay was negative. Same thing of K is less than negative, too. There&#x27;s no solution. And if K is less than negative three, there&#x27;s also no solution.

Problem

In Exercises 19 and $20,$ choose $h$ and $k$ such…

Problem 19 Hard Difficulty

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.
$\begin{aligned} x_{1}+h x_{2} &=2 \\ 4 x_{1}+8 x_{2} &=k \end{aligned}$

Answer

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

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Grace H.

Caleb E.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Linear Algebra and Its Applications

Grace H.

Caleb E.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Grace H.

Caleb E.

Maria G.

Michael J.

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

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications