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(a) Give the elementary matrix that converts41(6) Give ine elementary matrix that converts96:(c) Give the elomentary malrix that conworts...

Question

(a) Give the elementary matrix that converts41(6) Give ine elementary matrix that converts96:(c) Give the elomentary malrix that conworts

(a) Give the elementary matrix that converts 41 (6) Give ine elementary matrix that converts 96: (c) Give the elomentary malrix that conworts



Answers

Evaluate the determinant of the given matrix $A$ by using (a) Definition $3.1 .8,$ (b) elementary row operations to reduce $A$ to an upper triangular matrix, and (c) the Cofactor Expansion Theorem. $$A=\left[\begin{array}{rr} 6 & 6 \\ -2 & 1 \end{array}\right]$$.

And this problem were given a matrix a B C D, which is equal to three identity matrix 1001 And we're multiplying that by, um, this matrix right here. Nine negative seven negative five and four. And so because this has two rows and two columns, our average our final matrix is going to have she rose in two columns. And so A is going to be equal to a matrix with four entries, two on the top and two on the bottom. And so I'm going to begin by writing out R A B c d. Because we know that that is equal to 1001 And so to multiply the matrices, we're going to start with this first top left country right here and is going to be equal to nine times one plus negative seven times zero. We got that by taking the first row in the first column of each respective matrix. And so then we're gonna do the same thing for the next entry. We're gonna take nine times zero this time Organized the second column 9 10 0 plus negative seven times one and then we're going to repeat this process for the next two entries, The sun is going to be the bottom row on the first column. So negative five times one plus four times zero. And then lastly, we have the bottom row in the second column Negative five times zero plus four times one. And so we can see that the zeros all cancelled out and they're all equal to zero. Which means we're only left with the entries. I have, ah, one better multiplied by one. So nine negative seven negative five and four which you can see is the same thing that we multiplied the identity matrix by. So this property shows us that if you have a matrix multiplied by the identity matrix, it's going to be equal to that original matrix.

Mhm. All right. They want to use technology to solve this problem. So I have a graphing calculator handy and we are going to find the determinant of a four by four matrix. So we're gonna hit second matrix. We're gonna go over to edit so we can get that right arrow twice. Or an edit. A enter wanted to be a four by four matrix if one. Mhm. Negative 13 seven. Yeah. Oh yeah. Yeah. Mhm. So once we get uh the entries in there we want to hit second mode we want to quit this screen and now we're going to calculate. So what we're gonna do is we're gonna hit second matrix and we're gonna go over to the maths we're at the right arrow. There's a determinant. Find the determinant of the matrix that we want. We gotta tell the calculator hey. Matrix A just did all that work for you. Mr calculator. Find the determinant. And so we get that are determined as 2886. Yeah.

In this problem would give in this matrix, and our job is to show the elementary matrices. Then get this matrix in tow Row echelon form. Okay, so this one is gonna be pretty long, so just bear with me. So let's start off with the first row. If you notice there's a one in this position and so the first rule is actually done for us. So it's focused on the second row. According to write row, I feel on form. This first pot should be a zero. Let's focus on doing that. And the way to do that, we have a one right here. We can utilize that to change the second row. So essentially we're gonna be doing is adding negative two times, two first row, so negative to kind of the first row is negative two native for negative six and negative eight. And so we're gonna be adding these values to the second rope. So where is that going to look like the first row stays the same. The third roast is the same. But the second row, we're gonna be adding these values down here and so we end up getting is zero negative one negative too. 93. Okay, so the next step is trying to get this spot to be a one, and that is fairly simple. We just need to multiply it by negative one. And so if we do that again, the first road stays the same. Third row stays the same, but the second row, all signs are going to switch. And now, according to Rochel, on form, we have 01 and then random digits. And so the second row is complete, and now our job is to solve the third row. Okay, So we're gonna try to do is get this position to be zero. And we can utilize this one right here to do that. And so we can add Native three times in the first row. And if we do that, let's see what we get. So the first word stays the same. Second row stays the same. The third row. We're gonna be adding negative three times the first row, which is going to be negative. Three native, six negative nine native truck. So we add these values to the third row and we end up with zero negative to negative four and negative six. Okay, now, the next goal should be getting this spot to be a zero. And we can utilize this one right here to do that. And so we can add two times the second row and what that ends up being first throw stays the same. Second row stays the same, and third row ends up being 0000 And this is actually fine. This is a valid row, actual on form. So this is going to be massing the form one. Any random numbers here, 01 random number is and zeros all across.

So in this problem, we're gonna be taking this matrix and reducing it to row actual inform and then showing the elementary matrices that we used to get there. So the first thing we would notice with this problem is that there is a one in this position. It's the first row is actually done for us. So we need to focus on is turning this first position in the second row to be a zero. The easiest way that you do that is utilized this one right here. And so if we add native to time of the first row, this will get it done. So if we go ahead and sell this out, the first horse stays the same. Third row stays the same in the second row, is going to be adding negative too times each value of the first row. And so we end up with zero negative one negative to negative three. The next thing we want to work on is getting this night of one to be a one. Is this what to do that is more provide this row by negative one so the first word stays the same third row stays the same and each value of the second row is gonna flip signs. And now we're done with the second row now working on the third row, we want to get this value to be a zero. We can utilize this one right here to do that. Yes, we can add negative three times the first rub. If we do that, we end up with first row staying the same second, roasting the same on the third row, adding negative three times each value of the first round. So we end up with zero negative, too negative, four and negative six. Then we want to get this value to be a zero. And so to do that, we can utilize this one right here from the second row. Since we add two times the second rope and when we do that, we get first. So sane. Second row, the same the third row. When you add negative two times a second row, you end up just getting zero. So this is our row echelon form. And now we want to keep track of the road elementary matrices that we used. So essentially we have to apply it to the identity matrix each move that we used. And so the identity matrix is as follows. Since the max of the row and column is there, the minimum of the Roman column is the row. It is three. The identity matrix is going to be the three by three identity matrix, and each of the elementary matrices is just applying each of the moves that we made onto the identity matrix. And so here are answers. The first elementary majors will represent by E one is using this first move right here, adding negative two times the first row. And so we're gonna apply that to the identity matrix. And so the first row in the third row say the same. But the second row is adding negative two times the first row and so it's a negative 200 That is the 1st 1 The second elementary Matrix is taking this one, multiplying the second row by negative one. And so we apply that to the identity matrix and that is going to be first row, saying the same third row, staying the same and multiplying the second row by negative one third me tricks relook here. The third step that we use is adding the third row times negative three of the first row and so we end up keeping the second row, keeping the first row and adding negative three times the first road to the third Rob to end up with negative 30 can want. And for our final elementary matrix we're adding two times the second row to the third row, so two times the second row we end up with this first row stays the same. Second most is the same and never adding times to the second row to the third row and so we end up with 02 and one and these other four element.


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