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Question 5 By finding the pivots, sorting basis for the column space ofRepresent each dependent column as a combination of the basis columns. Find matrix A with thi...

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Question 5 By finding the pivots, sorting basis for the column space ofRepresent each dependent column as a combination of the basis columns. Find matrix A with this echelon form U, but a different column space. (Note: A could be any matrix which have reduces

Question 5 By finding the pivots, sorting basis for the column space of Represent each dependent column as a combination of the basis columns. Find matrix A with this echelon form U, but a different column space. (Note: A could be any matrix which have reduces



Answers

use elementary row operations to reduce the given matrix to row-echelon form, and hence determine the rank of each matrix. $$\left[\begin{array}{cc} 2 & -1 \\ 3 & 2 \\ 2 & 5 \end{array}\right]$$.

Okay, so what? We have you It's a three by four matrix were in were asked if I betray login form and to determine its around. So if we are doing a view over lessons I made the existing aero engine for if it satisfies the following condition first, the first element on the first row should be in a zero second each non zero. Leading entry on the next rose should be the right off the leading entry in the previous row. So you should look like this resembling, as 30 eights entered Rose with all zero elements of any are below those having a non zero. So we can do this by performing a series of operations, including the scaling a role, adding in multiple the one road to the other room and swapping for Rose. So for this example, what we want to do first, it's the cancel, the leading coefficient off far too. So we do this vapor forming. I do minus 1/2 off our one. By doing that, we now have to negative one 30 for zero negative, 3/2 negative one house one one negative, five and zero like next. We want to cancel the leading coefficient of R three by performing artery mine us 1/2 off our went. So now we have to negative one 34 zero negative 3/2 negative 1/2 1 and then zero Negative nine over to negative 3/2 and then three. Next they want to swap. Arto we had are three. We're doing that. We now have negative 134 zero negative nine over to negative rehabs. Three, then zero negative. Three haps negatively. 1/2 1 And for our final step, we want to cancel again the leading coefficient of party. So it satisfies the conditions that we have mentioned it When a group, I'm just gonna believe his first. So we want to cancel the leading coefficient of R three by performing artery minus 1/3 off our therefore Already, final answer is too negative. 134 zero negative. Nine Hafs Negative three haps city and 000 You know, So this already satisfies the criteria is that we have mentioned a while ago. So this is the final r e m off the matrix. So we want to determine a swelled around and by definition, wrong is the maximum number of linearly independent vectors in The Matrix. So this is equal to the number of non zero rules in its raw Angela and me tricks. So we are the local example. We have one Turow's with non zero element, which is rolled one, and therefore the run off this matrix is the

Okay, So what we have here is a three by two matrix were in very ask defying Israel Regiment form into the 30 minutes trump. So if we are to review our lessons and matrix isn't it is in its room. Excellent. For if it satisfies the following conditions first, the first element in the first row should be a non zero second kitchen and zero leading entry. The next rules should be to the right of the leading and in the previous year. So it should look like this. Like a stir keys entered Rose with old zero elements, if any and below rules having a none zero. So we can do this by performing a series of operations, including re skating. A room adding a multiple of one road to the other room are swapping that rose. So for this example, we have a three by three matrix and the first role off the first cold. Um, zero. But this we expect that the first non zero element of the medics would be on the second. No. So what do you want to do? First is to swap. I do Well, are they? By doing that, we now have zero one three zero the five and 01 for next. We want to cancel the leading coefficient in our entry By performing are three minus 1/3. Are you by doing the amputation on your own? We now have zero one three zero 30 five and 0073 So this is now the road accident form off the matrix and were also asked to determine its trunk. So, by definition, the wrong is the maximum number of we nearly independent back parts in the Matrix. So this is 1/4 the number of non zero rose in its row. Excellent metrics. So we're to look on our example. We have three rose with none zero element, which is throw one road poop and pee. Therefore, our runk is

So we're given this matrix and the task is to take it into row echelon form. So for a two by two matrix such as this, the row echelon form is gonna look something like this. We have a one right here. We have any number right here. 01 So are cast is to show the elementary matrices that get this matrix to equal that matrix. So the first up that we should do is trying to get this first spot in the first row. T equal one. So how would we do that? Well, we can add negative too of the second row. So by using this one, multiplying it by negative two and then adding into the first row, we can achieve that one in the first spot. So if we do that, it would look something like this negative Two times one is native to we added to three to get one negative two time night of two is four. We added to five. Yet nine second row says the same. The second thing we can work on is trying to get this digit to be a zero, and we can do that by utilizing this one right here. So if we were to add native one of the first row, then we kept what we desire. It's the first row, stays the same, stays one and nine. The second row. We add negative one in the first row. So negative one of the first row is negative one and negative nine. So if we add that we get zero and negative 11 now our final test, she said this to be a one and because this is a zero, we can actually do something very simple. We can simply multiply this row by the inverse of this number, which is negative. 1/11. We do that. We get our answer. 19 zero and one so we can go ahead and check with the row. Echelon form is which is one any number zero and one. No one can see

So we're given this matrix and our task is to produce it to row Ashlan form and then keep track of the elementary matrices that we used to get there. And so, for the first step, we want to change this into a one. Yes, we're gonna utilize this one to do that and so we can add negative two times the second row. When we do that, the second row is gonna say the same. And the first road, we're gonna motor pie each value of the second world by negative two and added to the first row. Since we end up with one and nine now because there's a one here, we're done with the first row now working on the second row. We want to make this a zero and we can use this one to do it. And so we add negative one times the first row we did. That first course is the same. The second row is going to change. 20 Nate 11. The final step would be to get this to be a one. We didn't do that by multiplying by negative 1/11. We do that. First rule stays the same. Let me get zero one to this is in row echelon form. And now what we want to do is tracked each step that we took and apply it to the identity matrix to get the elementary matrices. And so because there are two rows in this matrix, our identity matrix is going to be a two by two in the two by two identity matrix is as follows. To know let's find the elementary matrices so well represented by E One is the first elementary matrix. So we go up here, we take the first step. We apply it to the identity matrix negative two times a second row and added to the first row to the second roads in a sit same in the first row. It's going to add on you get to the second room to end up with that, the second elementary matrix. We apply the second step, adding native one time in the first room in the second room. It's the first person of the same, and then we apply a negative one time of the first road and added to the second to be a native one one in the final one So I'm in. Tree Matrix is applying the third step, which is multiplying second row by negative 1 11 We do that. The first rule of say the same. The second room changes Teoh. Negative 1 11 There we have 23


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