in-exercises-19-22-determine-the-values-of-h-such-that-the-matrix-is-the-augmented-matrix-of-a-consi

Question

In Exercises $19-22$ , determine the value(s) of $h$ such that the matrix is the augmented matrix of a consistent linear system.
$$
\left[\begin{array}{lll}{1} & {h} & {4} \ {3} & {6} & {8}\end{array}ight]
$$

Samantha Lincroft · Accepted Answer

So in this problem were asked to find the value of each that would make the system consistent. And so they do this. It&#x27;s really more about finding what values would make it inconsistent. And so the first thing that we&#x27;re going to do to do that is started as though we were going to solve this system. So we&#x27;re gonna so you can fill out about zero that three to get a zero in the left hand corner. And when we do that, we get this new matrix here. And what would make this inconsistent is if this right hand lower term eight minus four h was equal to zero. And so, in order the check for that, we&#x27;re going to say that H minus four h cannot be equal to zero. This will give us everything that&#x27;s consistent. And so when we saw that we get that four h equals eight and that H equals two, or in this case, H cannot possibly be, too. In order for the system to be consistent. If H is anything I&#x27;ve ever been to, we won&#x27;t get a zero there, and it will be consistent

Samantha Lincroft · Answer

So in this problem were asked to determine for what values that h this system is consistent. So they do this. We&#x27;re gonna start by going about this as though we were trying to solve this system of equations anointing to determine if it&#x27;s consistent. And when we do that, we see that we have in this bottom row zero H plus 12 and zero, which is basically saying that H plus 12 times some variable is equal to zero. And what that means is this kind of two possibilities here. If H is negative 12 then we basically have zeros all in the bottom row. And this would be consistent cause it would just be a line. And if h is 12 it would also be consistent. So no matter what, this system will be consistent.

Alina Krajewski · Answer

and this problem were asked to find the values h dummy some consistence. For now, it&#x27;s pretend as if H is a number and attempt to solve the system. Let&#x27;s look at X one and used their operations to turn this negative to Lex one into easy year old. To do that, we can add two times were one she wrote to to get a new road to zero for close to H zero. So what this for close to a judge for presents is our x two. And what this neuro to is telling us is that four plus two h X two is equal to zero. So this four plus two h is just a number that Times X two is equal to zero. So from this equation, we can see that X two is equal to zero. It&#x27;s because of that our four plus two h can be equal to any number, since any number times zero is equal to zero. So the answer to this problem is that the system is consistence for all values of each

Alina Krajewski · Answer

so here were asked to find the value of eight. Shot. Makes this matrix of a consistently in your system. For now, let&#x27;s pretend as if H is a number and attempt to solve the system. Let&#x27;s look at X one and attempt to turn this negative six export into easier to do that. We can add three times for a one she wrote to, and that&#x27;s going to give us a new row. Two off zero zero five plus three h So this new road to tells us that for the system to be consistent, zero must equal five plus three h so we can algebra ll it break We solve for H vice attracted five from both sides and then divided by three and back. Is this h People&#x27;s negative five over three?

Luis Amaro · Answer

hello, but it was well tomorrow. Today will be finding the value of age such that the matrix is there augmented makers because on your system before against that, I want to bring it back in tow. Deborah Interview To write this matrix in the form off system, so plan your equations, you will get two X plus three y equal age and for X plus X Y equals seven. I remember back in algebra assistant Lanier system basically means is that if you were to solve will be questions, the end result will be that zero go zero Or in other words, the two equations are equivalent to each other. So if so right now we&#x27;re to find age such that this pope equations are equivalent and in the sense of linear algebra when sobbing and solving values, the matrix will be using your robe reduction war. So first do is that we set wrote to the values wrote to Equalling, wrote to minus two minus run in the sense were to try to get all the values, the bottom l equals zero. So as you see here, if you were to use roll reduction for you&#x27;ll get 007 minus two h. But what we&#x27;re trying to do is to just do get the bottom one equals zero, the bottom right in one sequence. So so you&#x27;ll have so where you do it. Set the expression seven minus two h equals zero on which you once you solve it, you get age equals 7/2. So where you plug in, it&#x27;s something over twitch. So someone over to for age, if you would solve it then you get the results of 00 Therefore, age is equal to 7/2.

Arden Mccarthy · Answer

for this problem, we&#x27;re going to solve the set of linear equations by using gassy elimination, which means we want to simplify our matrix into reduced echelon form. So our first step for doing that is to write out The Matrix that corresponds to the set of equations up were given to us. So the first equation reads. Zero x one plus x two plus three x three minus two x four is equal to zero. The 2nd 1 reads to x one plus x to minus four x three plus three x or is equal to zero. The 3rd 1 reads to x one plus three x two plus two x three minus x four equals zero and the last equation that were given reads negative four x one minus three x two plus found X three minus four of four is equal to zero. The first step that we&#x27;re gonna do is we&#x27;re going to cancel out this term to make it a zero. We can do that by setting our two equal to our two minus r three, and that is going to give us the following matrix. First row is gonna say the same. The second row is going to become zero negative, too negative, six or zero and then the other two rows are going to stay the same. Okay, Our next up is going to be to make the second row, actually a row of all zeros and we can see that we could do that pretty easily by looking at how the first row is pretty much just a scaler of the second World. So two kids love this road. They call Ciro&#x27;s ruins that are too. People Thio two are one plus are too. When we do that, we get the following major. So again the first roast is the same. The second row is gonna cancel out completely, given the substitution into the equation for our two that we wrote. And again the third and fourth rows are going to stay the same. Okay, next up, we want to make this entry or negative for injury a zero. And when we do that, this will be the only entry in the column. That&#x27;s not zero, which is good for the form that we want to get into. So in order to eliminate that, we&#x27;re gonna set our four equal to to our three. Those are for once we do that, we get following matrix. So again, these rose rows one through three are gonna say it seems we&#x27;re not doing anything to them. And row for is going to become zero three non negative six and zero by the equation given next up, we want to cancel out this entry right here. That&#x27;s gonna be our next step. So in order to do that, we just have to do some subtraction and we have to say, or three is equal to are three minus are for and that is gonna give us the following matrix. The first row is gonna stay in the same that there was going to stay the same or Robles heroes. Our third row is is going to become to zero negative seven, 50 bug subtraction. The question on the other page and our last row is gonna say the same. So 039 negative. Six and zero. Oh, it so next up, what we want to do. ISS We want to cancel out most of this road because, like a difficult operation that we already did, we can see that Row one is pretty much just a scaler multiple of row three, so that should be pretty easy to cancel out. So what we&#x27;re gonna do is that our four people to our four minus three are one. Well, we do that. We&#x27;re going to get then used to roast, stay the same, and this last row is going to become a row of all zeros. So now what we want to do is we want to put our all zero rose at the very bottom of the Matrix. So one of them is already in at the bottom and the other one right here, we can just switch with the road below it. So that&#x27;s gonna be our two is gonna be ableto the current are three, and our three is gonna become equal to the current are too. And this gives us the following matrix. Okay, Next up, we want to switch these two rows with each other because as per rules of reduced echelon form, we want to make sure that the left most leading entry is at the top of the matrix. So are one is going to war too, and our two is going to become a recurrent or one that&#x27;s gonna give us this following matrix. Then again, we have are all zeros on the very bottom. Okay, so now our next step is we want to divide are one by two. So that this leading entry is a one which matches the form that we want to put our Matrix into. So are what is he? Will one have times are one, and that is going to give us the following matrix. That&#x27;s gonna give us 10 negative seven halfs. But house zero our second row&#x27;s gonna say the same. So one 013 to 0 and then our rose three and roast. We&#x27;re also gonna say the same. Okay, so now we&#x27;re ready to solve for our variables. So we&#x27;re gonna label this X will in x two x three explore so we can see the columns are correspondent What variables. And we can also see that ex one next to these variables. Herbal pivot values. So that means that our free variables rx three the next four. So what that means is we want to solve for X one and X two in terms of x three, next four. So we&#x27;re gonna start with the first row. So are one reads That&#x27;s one plus zero x two minus seven halfs x three. Let&#x27;s buy a house. Explore is equal to zero X. You cancels out because it&#x27;s coefficient zero and we&#x27;re left with x one minus 7/2 x three. Let&#x27;s buy half explores equal to zero and we want to solve for X one in terms of our free variables. So to do that, we just put the other free variables on the right hand side and we get that X one is equal to seven. House X three minus five halfs, that&#x27;s for okay. Now we&#x27;re going to go to our second row and that reads zero x one plus x two plus three x three Bonus two x four is equal to zero and from there not cancels. We&#x27;re gonna sell for X two in terms of x three and explore so x two equals negative three x three close to export. So we sell for variables. And if you want to write that in a vector form, we can do that by writing ex one. Next to is equal two next three times the following vector. We have a seven. How? Spector at a negative three. Er sorry. Seven house and negative. Three is sailors. And then we&#x27;re going to add that to explore times following vector, uh, native five house and two. And we just got these numbers from these entries here. So the X three terms here and here We plugged those in right here and our export values. We got them from here and here. I put them in here.

Problem

In Exercises $19-22$ , determine the value(s) of …

Problem 19 Medium Difficulty

In Exercises $19-22$ , determine the value(s) of $h$ such that the matrix is the augmented matrix of a consistent linear system.
$$
\left[\begin{array}{lll}{1} & {h} & {4} \\ {3} & {6} & {8}\end{array}\right]
$$

Answer

matrix is consistent as long as $h \neq 2$

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

Linear Algebra and Its Applications

Kayleah T.

Maria G.

Kristen K.

Michael J.

Absolute Value - Example 1

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matrix is consistent as long as $h \neq 2$

Linear Algebra and Its Applications

Kayleah T.

Maria G.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Kayleah T.

Maria G.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications