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Enter the matrix2 46 5 7 1 4 9 2 6 34 1 2 5 2 5 6 2 4 5 1 9To prevent typing errors you may copy and paste following Maple command for entering A.A:= Matrix([[2 , ...

Question

Enter the matrix2 46 5 7 1 4 9 2 6 34 1 2 5 2 5 6 2 4 5 1 9To prevent typing errors you may copy and paste following Maple command for entering A.A:= Matrix([[2 , 2, 2, 5,2, 4,4,8,3, 1, 3], [7 , 4, 8,6,5,7, 1, 4, 4,9,4], [8,9,9, 2, 6, 3, 3,3, 8,7, 6], [2, 6, 3,6,9, 8,7, 6, 3,9, 8], [3,2,4,1,2,5, 6,9, 9, 7, 8], [1,4,2,5,8,5,4, 9,2, 4,3], [2,8,6,2,4,7,1,7,1,3,7], [5,2,5,1, 9,8,3,7, 9, 1, 2],[3,9, 8,8, 1,4,4, 6,5,9,2], [9,2,9,8,7, 7, 5,1, 9, 8, 4J]);[Note that Maple does not display the full matri

Enter the matrix 2 4 6 5 7 1 4 9 2 6 3 4 1 2 5 2 5 6 2 4 5 1 9 To prevent typing errors you may copy and paste following Maple command for entering A. A:= Matrix([[2 , 2, 2, 5,2, 4,4,8,3, 1, 3], [7 , 4, 8,6,5,7, 1, 4, 4,9,4], [8,9,9, 2, 6, 3, 3,3, 8,7, 6], [2, 6, 3,6,9, 8,7, 6, 3,9, 8], [3,2,4,1,2,5, 6,9, 9, 7, 8], [1,4,2,5,8,5,4, 9,2, 4,3], [2,8,6,2,4,7,1,7,1,3,7], [5,2,5,1, 9,8,3,7, 9, 1, 2],[3,9, 8,8, 1,4,4, 6,5,9,2], [9,2,9,8,7, 7, 5,1, 9, 8, 4J]); [Note that Maple does not display the full matrix when it has more than 10 rows or 10 columns However; you can still work with the matrix using Maple commands:] Use Maple to create the vector b that is column 2 from A and the matrix C that is made from columns and 3 to 11 of A (in the same order as the columns of A): Now solve the matrix equation Cx=b and enter the 5th component of the unique vector solution for x in the box below: (Your answer should be an exact fraction, not decimal: ) (19205*_tO[1 , 1JV112637



Answers

Find a single elementary row operation that will create a 1 in the upper left corner of the given augmented matrix and will not create any fractions in its first row. (a) $\left[\begin{array}{rrrr}2 & 4 & -6 & 8 \\ 7 & 1 & 4 & 3 \\ -5 & 4 & 2 & 7\end{array}\right] \quad$ (b) $\left[\begin{array}{rrrr}7 & -4 & -2 & 2 \\ 3 & -1 & 8 & 1 \\ -6 & 3 & -1 & 4\end{array}\right]$

Okay for part A. We wanna work for part A and part B. We want to make the top last # one. And usually the idea or what we would do is just divide by divide the first row by the top left number to make to make it one. But that sometimes makes the other numbers in the first row a fraction or something that's not an integer. And if we did do divide by In this case -3. The other numbers in the first row aren't going to be introduced. So we'll do it another way. And the other ways to make chain or other role operations are adding other rose to the first row or just switching Rose. So in part A let's try adding two copies of Road to the row one cell change no one two no one those two copies a road to. Okay now let's see what we get. And I added two rows of our two I'm two copies of our two Road to two row 1 because adding two times 2 2 -3, get this one. Okay, so let's see what our matrix is going to be. Okay, so we had negative three plus two copies of two which is gonna be one, negative one plus two copies of negative three. Which is going to be negative seven oh two plus three. I mean two copies of three, we're just going to be eight 4-plus 2 copies of two. It's going to be eight And rode to and road three are just going to be the same. So two negative three, three, two and oh yeah, zero two negative three and one. So that's part A let's go to or B for the matrix were probably noticed that the first number in the third row is one so we can just switch row one and row three, so We'll switch row on in road three and what do we get? So the first row is going to be Row three in the Original Matrix. So one for Native three and three then Road 2s is not going to change. So to negative nine, three and two then rode three was switched with Rogue one. So our 3rd row and our new matrix is going to be row one in our original matrix. So zero negative one 85 and zero oh. And that gets

Okay, So the idea here is how to show a system of linear equations on represented as matrix forms. If we have a system of linear equations, a 11 times first variable, plus another constant times here, second variable and so on until your last very lost. Like if you have three. Those three males, like the X y Z, is you would commonly see them and then what it's equal to right? And so these a a and B are just sort of Constance. Okay. And then in the next equation and so on and so forth. OK? And so the process here is then, that we shouldn't break it up into a matrix for the constants attached to each variable. Right? So we would get a 11 And then he went to a way to anyone and independent on how many you have here. Yeah, this committee. And okay. And next to that. Okay, So this is our n by N. Matrix. Then we'll have our variable matrix, right? And this is, of course, been end by one matrix. But that means we you can see if they can be multiplied right, because there's an elements in the road with an end in the comb. So we go through the multiply, you'll get a 11 times x one a 12 times x to you with a one end times X and right That gives us our first equation. And so then that's gonna be equal to are be one here, okay? And then, of course, is also a And I want matrix Rex. If we were to multi parties together, then we're gonna get and by one matrix, and that results from here. And so this is how you would represent our system of linear equations with they a matrix and X matrix and a be matrix. Okay, so in part B has this looking at how to show that we can find the solution using our ex matrix be equal to the inverse are a matrix Times B. And so we ended up we ended here, you know, this would be represented as sort of a matrix Times X matrix Sequels are bi matrix. Right? As we ended with this, a X equals B, but we're multiply both sides by the inverse it multiply both sides but universe, then the inverse times, eh? that's would be equal to the identity, right? So that causes the identity Times X being equal to the inverse times B. But it any times X is just equal to X. Okay, so that's how we get that we can find X from the inverse times be

For this problem. We have been given four matrices A, B, C and D, and told to use a computer or a graphing calculator to do some matrix arithmetic. So in this case, I am using the desk most graphing application on my computer. And as you can see, I have already entered in A, B, C and D, and we're going to do some multiplication with two of these matrices A and C Now, before I use our application here, let's take a look at what we should expect to see. It's always good to make sure you understand the results, so the results makes sense when you get them. Make sure you haven't made me type something in wrong. So here where we have two matrices A and C A is a four by five matrix. Hey, four rows and five columns See is a five by five and we're gonna be looking at two, uh, multiplication problems a times C and C times A. Well. When we do our multiplication, I have a four by five times a five by five. The columns of the first matrix need to match the number of rows in the second So these two numbers together they need to match which they dio and our resulting matrix will be those outer numbers. So when I multiply a times C, I'm gonna end up with a four by five matrix. Now, what if I do see times a Well, that's a five by five times a four by five. These numbers do not match. Not like they did on the first one. That means I cannot do this multiplication. I can do a times C, but when I try to multiply C times A, I should get an error. So let's see how this actually works. Okay, First of all, let's do a times c so a look. Sorry about that. A times C. I don't know why that keeps coming up a times C and I do get a four by five matrix. Right. Um I'm gonna multiplying across the rows and a down the columns of See, Here are my results. What if I put in those C times? A. I get an error message and the error says I cannot multiply matrices with incompatible dimensions. And that's what we saw before those rows and columns. Numbers need to match Otherwise we cannot do the multiplication. So a times C is a valid multiplication. See, times A is not so no thes do not equal each other.

For this problem. We have been given four matrices A, B, C and D, and we've been told to use either a computer application or a graphing calculator to do some matrix arithmetic for this problem. I'm using the desk most graphing application on my computer. And as you can see, I've already entered in all four matrices A, B, C and D. Now, before we actually start work using this application to do the multiplication, let's take a look to see what we expect to have a za result. Okay, so I'm using two of our four matrices here. C and D, C and D are both the same size matrix five by five. Okay, because they are both square matrices, exactly the same size. We can actually do both of the desired multiplication. See Time's D and D times. See now, just remember when we do multiplication See is a five by five. He is a five by five. So the number of rows number columns excuse me in the first matrix have to match the number of rows in the second a za long as they match, we can do the multiplication and the numbers that are left will tell us the size of our resulting matrix so we can multiply these together and we will end up with a five by five. And we actually end up with the same sighs matrix on the other side. Because again, five by five times, five by five thes inner numbers match those outer numbers, tell us what to get. So maybe these will be equal to each other. We're at least gonna have the same size matrix as our result in either case. So let's take a look and see what we actually get is a result. First, let's find see times d. Okay, so here it is. It is a five by five, just like we expected. But what happens when we multiply de times? See? Oh, now again, it is a five by five. The size is the same, but the similarities and there the elements do not match from one to the other. So even though both multiplication zones, uh, work are valid, both of them give us the same size matrix C, D and D C are not equal


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