row-reduce-the-matrices-in-exercises-3-and-4-to-reduced-echelon-form-circle-the-pivot-positions-in-t

Question

Row reduce the matrices in Exercises 3 and 4 to reduced echelon form. Circle the pivot positions in the final matrix and in the original matrix, and list the pivot columns.
$\left[\begin{array}{cccc}{1} & {2} & {3} & {4} \ {4} & {5} & {6} & {7} \ {6} & {7} & {8} & {9}\end{array}ight]$

Carson Merrill · Accepted Answer

So we&#x27;re given the following matrix. We have 1234 5678 44567 Actually, and then 6789 And we&#x27;re wanting to get into reduced echelon form. So first thing we&#x27;re gonna do is try to focus on this first column. We focus on the first column. We acknowledge that we have a one here, so that&#x27;s good. Eso this is going to be row equivalent to one zero zero. That&#x27;s how we want to do it. And how are we going to do that? Well, we&#x27;re going to use replacement, so we see that we&#x27;re going to replace our to, um, minus four times are one, and we&#x27;re gonna place are three by subtracting six are one. Okay, because that&#x27;s how we get that. And then this will be left with 234 down here. We will have, um, two times for eight, so we&#x27;ll have a negative three. Two times, uh, six is 12. So I have a negative five here. Then we&#x27;ll end up getting three times negative for which is negative. 12. It would be a negative six, and then we have three times negative six, which is negative. 18. So that&#x27;ll make this a negative 10, then. Lastly, we have four times Negative. Four. That&#x27;s negative. 16. So we&#x27;ll have a negative nine here, and then we get negative. 24 plus nine. So that&#x27;s going to give us a pause. A negative 15. Okay, so then we want to go down to the next row, and we see that this is the next thing that we&#x27;re focusing on. That negative three. So to get rid of that, to make it a one, we&#x27;re going to skill this. We&#x27;re going to scale. Row three wrote to. Actually, we&#x27;re going to scale it by a negative one third. So we do. Negative one third times are two. And then just for the sake of making things simpler, we&#x27;re going to, um, scale row three by 1/5. Okay. It&#x27;ll just make it simpler, but we don&#x27;t have to do it, so we&#x27;ll get 100 234 one, 23 And this will be one too. Three. All right, Now, um, we can we see what we do is we can just cancel the top in the bottom room or that the second in the third row. This will be through replacement. We will dio are three minus are too. And then we, um we can also replace the top row are one minus two times are too, since this is a to and this is just a one, So using this row replacement will end up getting 10 00 This will be a one. This will be a zero, and we&#x27;ll see. We&#x27;ll get zeros across the board. So now this is what we have. Um And then we would end up getting zero here. This would be and negative one, and this would be a negative six plus force. And negative too. So this is about as reduced A We can make it on doing so. We now wanna look at pivot positions that&#x27;s going to be the first non zero number in a row. So we see that there are two pivot positions. Um, and these pivot positions didn&#x27;t change, so this would be a position, and this would be a position. They are the same. And, um, the pivot columns would be, uh, column one and calm to that would be the pivot positions or the pivot columns.

Arden Mccarthy · Answer

this question is asking us to solve the given system of linear equations based on the reduced matrix out we have been given. So well, go ahead and first label our columns with the variables that the representing. So this is X one. This is X two, this is X three. So for the 1st 1 we can write out the equation with the coefficients for X one x two x three has given to us by this matrix. So the coefficient of X one is one. So we can just leave that out since it&#x27;s a one. Or we could put that in if we want. And then the coefficient of X to an X three are both zero. So we can basically just ignore these and write X one equals negative three. Because these cancel out one times X one is just one X one. So we get that X one is equal to negative three. Uh, this same thing we&#x27;re gonna bring you the same thing. For rose to one rose three. We have zero x one plus one. Next to zero x three is equal to zero. We simplify this, we get X two is equal to zero. Kathy&#x27;s cancel out and one times X two is just x two. So lastly, will Sof uh, this final road here? So we have zero x one plus zero x two was one x three is equal to seven. We simplify that we get that extra is equal to seven so we can write our final answer in Dr notation. If we want, we can say this x factor of these three components x one x two x three is equal to negative three zero seven. Okay, so we&#x27;re going to do the same thing for this. We&#x27;ll label our variables, x one x two x three x four and also, uh, just tow. Remind ourselves this lost column is what the equations in this matrix are equal to. So for the first row B right x one zero x two plus x three plus zero x three was negative. Seven x four is equal to a and what we want to do before we solve. This is looking or matrix and see if any of these variables our free variables and what that means is that one of these variables x one x two of three or X four is not a leading injury. So here we see that this is a leading entry. This is a leading entry and this is weeding entry, which means that this is not and none of the entries in the X four column are reading injuries. So that means, uh, export is our free variable. And we want to solve all of our equations that we&#x27;re gonna write down. We&#x27;re gonna want to solve for X one, x two and x three in terms of x four. So we&#x27;ll go ahead and do that. So he&#x27;s cancel because these are both multiplied by zeros. We&#x27;re left with X one minus seven. X four is equal to a And like we said before, scores are free variable. So we want to solve for X one in terms of export. So x one is equal to eight plus seven x for Okay, so now we&#x27;ll go ahead and look at our second row. We have zero x one plus x +20 x three plus three of four is equal to two. He&#x27;s cancel. We have x two plus three x four is equal to two and we solve for x two in terms of our free variable of exports, so actually was equal to two minus three x four. So then our last row, we&#x27;re gonna want to solve for X three. So we have zero x one. The zero x two was x three plus explore is equal to negative five. He&#x27;s cancel. We solve for X three in terms of explore, So x three is equal to negative five. Linus Export. Now what we&#x27;re going to want to do is do what we did in the first problem and write that in vector notation. So we have the general expect, er that we&#x27;re selling for which has components excellent X two and X three. And what we&#x27;re going to do is we&#x27;re going to May one. Dr. That&#x27;s just these numbers that have no other variables involved. So we see our X one has a next to has it too. Exterior has the negative five. We&#x27;re going to add that to explore times it&#x27;s coefficient, so we can see that the coefficient for X one for X four, And, uh, this overall equation for X one is seven. So we&#x27;re gonna write a seven here, uh, for X to the coefficient of X four is negative three. And for X three, it is negative one. And if this is looking a little bit confusing to you, we&#x27;ll just go over here and know that we can rewrite this as we can. Reword right each of these lines the x one x two x three as x one It wa ce eight waas x four times seven, which is seven. Explore and we can see that that is exactly what we got here. We can do the same thing out for Ex tube just for practice. So x two is equal tune too. Shot number here. Plus that&#x27;s four times negative three, which is negative three x four, which is the exact same thing that we have here. And lastly, X three is equal to negative five plus negative one x four, which is also the same thing that we see here, okay, onto the next one. So we&#x27;ll do what we&#x27;ve been doing right down the different variables in these systems of equations. We have excellent Xbox three x four, index five and we then want to figure out what our free variables are, and we can see that This is a leading entry. This is leading entry. This is leading country. So that leaves next to an X five to be our free variables because they have no leading injuries. So we&#x27;ll solve this just for myself. The other ones, uh, we&#x27;ll start with the first column right here. You see that? This is X one minus six, x two plus three x five is equal to negative too. And remember, we don&#x27;t have an extra year and explore because these are both coefficients of zero, which cancels he&#x27;s out. So then we want to solve for X one in terms of our free variables, which our ex too and x five So we have X one is equal to negative two plus six x two was er sorry no plus minus three x five and then we&#x27;ll do the same thing for the next row where we have, uh zero x one plus zero x two plus x three closed zero x four plus or expire is equal to seven. We want to solve in terms of our free variables which we have his x five and we get the, uh X three is equal to seven minus for X five. Um, and now we have our last row that we want to D&#x27;oh! And so this one is going to be solving for X four. So we are going to I&#x27;m gonna open up a other patriot. Quick. Um, so we&#x27;re going to have X four five x five is equal to eight, and then we want to solve it in terms of our free variable, which is that&#x27;s five, and we have explore is equal to eight minus five ex spies. So now we have the variable. So we want to solve for which is excellent x three, next four over leading injuries. We&#x27;ll write down what we got for those. So we got X one is equal toe. Uh, negative two plus six x Q minus three x five. Well, go ahead and write that down for extremely. We know that that equals seven minus four x five. Then for last variable, which is explore, we have a minus five x five. If you want to put it in vector notation like we did with the other ones, we write x one. It&#x27;s three that&#x27;s for is equal to thinking of 27 a Waas X two, which is one of our free Barry Bulls time. Some doctor Waas That&#x27;s five times some vector. So now let&#x27;s go ahead and fill that in for X two. We see that X one has a coefficient of six. Since there&#x27;s no X two terms in either extra explore, we know that these have a coefficient of zero because that&#x27;ll cancel it out. And for X five, we can see the coefficients for X one. It&#x27;s negative. Three for X two. It&#x27;s negative for and for explore its negative five. Okay, so now we&#x27;ll go ahead and go into the last one. So for this, before we even start solving for variables, we can kind of do a little shortcut and just take a look at this bottom row. So what this bottom row is saying is that zero x one. I&#x27;ll label the columns again for the variables we have x one x two, the next three. It&#x27;s saying zero x one plus zero x two zero Expiry is equal to one. So because he&#x27;s a zeros zero times, any number is gonna be zero. So we have zero plus zero plus zero is equal to negative one. Er sorry is equal to one. And some of all zeros is going to Syria. We have zero equals one, which is obviously not true. So for this matrix because we have this which is invalid, we say that this matrix has no solution.

James Kiss · Answer

you want to use their operations, Get this into producer Ashland form. So let&#x27;s get a leading one in Rhode won by just taking the opposite. Uh, Roh wants, like most buying by negative one and putting that back in row one. So row one will not be positive. One negative to negative one Positive, too. In rows two and three are the same. Now is your folks and getting zeros below the leading one in row one. And so we could, um, put in road to row two minus three times World one. So now wrote you would become from above three minus two or three times. One would be zero negative. Two minus three times negative too, which is four and one minus three times negative one, which is also for and four minus. Three times two is negative. Two rose, one in three are the same. Okay, so now we gotta zero. There are leading one. Let&#x27;s go. Just go for getting a zero in E um, Row three column one. So we can do this by taking road three from above. It&#x27;s attracting to 10 01 and putting that road three so he gets to minus two times one, which is zero negative for minus two times negative two, which is zero, uh, negative two minus two times negative one, which is zero and four minus two times two, which is zero. Rose one and chew remain the same. No, we can get a leading one in road to if we take just 1/4 of road to you. Put that in road to then that becomes 011 negative, 1/2 rose. Three in one are the same. We can get a zero off the leading one in wrote to If we take row one above and add two times wrote to to it and put that back in Row one. So it becomes one plus two times zero, which is one negative two plus 2 10 p. One a zero negative one plus two times one is one in two plus two times negative to would be to minus one or one, and that&#x27;s as far as we&#x27;re going to be able to go. Final answer

Cyrielle Lorio · Answer

Okay, So what we have here isn&#x27;t lead by five. Medics were in very asked if I this row in excellent form into the terminus trial. So if they&#x27;re going to have your lessons, a maintenance is in role. Excellent form even satisfy some following conditions. First, the first element in the first row should be an an zero second edition and zero leaving entry off the next room should be to the right of the leading into the big stroke. So it should resemble like this like a masterpiece. And finally Rose wiggle zero elements, if any, are below those having in an zero. So we can do these paper for me. A series of elementary row operations, including Re Skilling and Roll, adding a multiple of Juan drove to the other room and swapping that gross. So for this example, what we&#x27;re going to go first is to cancel the leading coefficient of art by performing are too minus 1/2 off our one. By doing that, we now have to one three for whom? Zero. They got the 1/2 1/2 negative one and to so pretty one but its event. Next, you want to cancel the leading coefficient of artery. And we do this by subject being are one, So all right, three. With that, we now have to one three for no zero negative. Want half 1/2 negative. One who and zero do negatives do one and fight. Next. We&#x27;re going to swap. Are too well are three now. We have so one three for Oh, zero. No negative too. One five zero negative. 1/2 one Have they get the one and who? For our next that which is our final step. We want to cancel the living coefficient of r three. So I&#x27;m just gonna delete this. I hope you re still for living the solution. Okay, so this will be our fort and finance that we want to cancel the leading coefficient of arteries, so it satisfies for condition that we have mentioned a while ago. So we do this by ad being one Ford off our do who are pretty. We now have no one to be four who zero to get them to one. But he and 000 negative. 3/4 and 13. 04 So this added spice a condition that we have mentioned for a row. Excellent for Therefore, this is the by not form off our medics. So we&#x27;re also ask so that they mean their ground and my destination round gets the maximum number of linearly independent vectors in the Matrix. So this is a cold of the number of non zero rose in this row. Insolent meetings. If they&#x27;re to look with our example, we have 12 and day does with non zero element, which is throw one roto and three. Therefore, the round off our made drinks is three.

James Kiss · Answer

use row operations. Teoh, um transform this matrix to produce rural Ashland form. So first, let&#x27;s get a leading one in Row one. What I&#x27;m gonna do on that this is one approach. There are many ways to do this many different steps you could take, but I&#x27;m just gonna change, throw one in a row to around and replace that, uh right our new matrix. So no one would be one negative 11 negative. Four on three. Negative to negative three. Negative one. And we&#x27;re leaving around three alone. Then we should have a goal of getting zeros below leading one in row one. So what if we take road to and subtract three times? Row one and right? That wrote to and we&#x27;ll take Road three minus two times, Row one, and write that in row three. So row one excuse me stays untouched. Road to, uh, becomes three minus three times one. Give myself from space up here. Negative two Minus three times negative. One negative. Three times. Three times one 1,000,000. That&#x27;s negative. Three minus three times one and then negative. One minus three times a negative for stand for row three. Let&#x27;s just do one of the time. I don&#x27;t want to get too messy. So here&#x27;s Row one and Row Brie three minus three times. One is zero negative. Two plus three is one negative. Three minus three is negative. Six. Negative one. Most 12 is 11. No, let&#x27;s go row three miles. Two times roll one. Put this in Row three. So roll one and road to will remain the same and we take growth. Reince attraction to times were a one, so it would be to minus two times one three minus two times negative. One in five minus two times one in 14 minus two times negative for tu minus two times one A zero three minus, um two times negative one is three plus two or five five minus two times one is three and 14 plus eight It&#x27;s 22. OK, now let&#x27;s get zeros Blow leading one and wrote to So what if we do Road three minus five times Road to and put that in row three. So rose one into will be lost alone in a row. Three would be, um, zero minus five times zero, which is zero five minus pictures. Me, Yeah. Five minus five times one is zero three minus five times six would be 33 and 22 minus, um five times 55 is negative. 33. No, let&#x27;s get a leading role in Row three. We can take Ah, 30 third of Row three and put that in Row three. So rose one and two remain. The 33rd of 00 30 30 00 33rd of 33 is one of 33rd of negative. 33 is negative one. No, let&#x27;s get zeros above the leading one in Row three. So in order to do that, we could take a road to and add six times row three to it and put that road to so row one. I was left alone in Row three. There&#x27;s less alone road to would be it again. Row two times 6 10 Throw 30 times 6 10 0 is zero zero oxidant. One plus 6 10 0 is one negative. Six plus six times one is zero and 11 plus negative. 65. No, let&#x27;s get, um, but if we now do roll one minus road three and put that in row one. So those two and three are untouched. Throw one Minus row three one minus zero is one negative. One minus zero is negative. One one minus one is zero and negative for minus negative. One is negative. Four plus one or to give three. And I think we only need one more step here. What if we and Row one and two together and put that in row one? You So one plus zero is one negative. All +10000 Negative. Three plus five is too. There is our final. The answer. We&#x27;ve got our three by three identity matrix on the left and it reduces to 25 negative one, right?

Thomas Emment · Answer

I can&#x27;t give this augmented matrix on. We&#x27;d like to get into reduced wrote actual on form. So for some time, if we replace road one on road will be switched around. Row one on rate. So right. Why? With row to since Roe ones pivot very to pivot, the 1st 1 is already at one. So just to say that&#x27;s doing some difficult maths this early in the morning will change that round 21 minus 64 Okay, so now it was the We need this one to be our pivot essay. We&#x27;re gonna have to do row, uh, to minus two lots off road, one of the worst we&#x27;re gonna be replacing. Where it&#x27;s you with that. So no one&#x27;s gonna unaffected one minus two Teoh minus three, but would have to minus two zeros. We wanted one plus four, Which is gonna be five, uh, minus six, minus four with U minus 10 and then four plus six, which should give us 10. I always see what appeared to be once. We&#x27;re gonna have to do 1/5 off road to be replacing right to with that. So we have one minus two to minus 301 minus two. That Okay now? But I pivots when you have a zero. So clearly we&#x27;re gonna have to have row one plus two lots off road to. That was awful tasting red one. With that eso we&#x27;ll have It wasn&#x27;t effective. Zero minus to add two zeros we wanted to add minus four will be minus two on, uh, minus three. Add four is good. Minus one on the bottom row is gonna beat. Unaffected on this should be our, um, matrix in reducing X one for we&#x27;ve got our pivots going to the bright both of up its I want and zero but it&#x27;s

Problem

Row reduce the matrices in Exercises 3 and 4 to r…

Problem 3 Easy Difficulty

Row reduce the matrices in Exercises 3 and 4 to reduced echelon form. Circle the pivot positions in the final matrix and in the original matrix, and list the pivot columns.
$\left[\begin{array}{cccc}{1} & {2} & {3} & {4} \\ {4} & {5} & {6} & {7} \\ {6} & {7} & {8} & {9}\end{array}\right]$

Answer

$\left[\begin{array}{cccc}{1} & {0} & {-1} & {-2} \\ {0} & {1} & {2} & {3} \\ {0} & {0} & {0} & {0}\end{array}\right]$ Pivots are in row 1 $\operatorname{col} 1$ and row $2,$ col 2

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

Linear Algebra and Its Applications

Grace H.

Heather Z.

Kayleah T.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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$\left[\begin{array}{cccc}{1} & {0} & {-1} & {-2} \\ {0} & {1} & {2} & {3} \\ {0} & {0} & {0} & {0}\end{array}\right]$ Pivots are in row 1 $\operatorname{col} 1$ and row $2,$ col 2

Linear Algebra and Its Applications

Grace H.

Heather Z.

Kayleah T.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Heather Z.

Kayleah T.

Caleb E.

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

Linear Algebra and Its Applications