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Point) Give a 3 x 3 elementary matrix E which Will carry out the rOw operation R, + RxTest that Eactually works for carrying out this roW operation by computing the...

Question

Point) Give a 3 x 3 elementary matrix E which Will carry out the rOw operation R, + RxTest that Eactually works for carrying out this roW operation by computing the product EA for the matrix~2 ~3 ~1A =55 ~2EA =

point) Give a 3 x 3 elementary matrix E which Will carry out the rOw operation R, + Rx Test that Eactually works for carrying out this roW operation by computing the product EA for the matrix ~2 ~3 ~1 A = 55 ~2 EA =



Answers

Find the determinant of the matrix. Expand by cofactors on the row or column that appears to make the computations easiest. $$\left[\begin{array}{rrr}-1 & 2 & -5 \\0 & 3 & -4 \\0 & 0 & 3\end{array}\right]$$

Okay. Problem number nine. We have a matrix and were asked to, um, show a row operation three halves times ra one plus row three. And that's going to replace row three. And so we're going to keep row one the same with elements to negative six and four. And we're going to keep row to the same with elements 12 and negative three. But we're going to change row three. So let me you do this in a couple of steps. First of all, let's figure out what three halves of our one is. Three halves of row one just off to the side. When you multiply two by three halves, you get three. When you multiply negative six by three halves, you get negative. Nine. And when you multiply four by three hats, you get six. Okay, so we're gonna add that to row three. So the first elements negative three plus three, we get zero the middle elements one plus negative nine. We get negative eight and the third elements negative. Two plus six. We get four. So there's Arnie Matrix having done the desired row operation

Okay, so for this for just demonstrating elementary row operations on a practice matrix, and for this one, they want us to take twice the second row and add it to our third row. So what that tells us is that in our answer, our first row is not changing. Our second row is not changing, and just the answers to our third row are so we do have to know what twice the second row is. So this 15 negative to one I have to double. And that doubles to two 10. Negative four and positive, too. What I do with those answers is now I have to add them to my third rope, so I'd have to plus a negative too, which is zero 10 plus two, which is 12 negative for plus negative, too, which is a negative six and a two plus a five, which is seven. And then these numbers that we figured out when we doubled don't actually show. But all in the Matrix we would on Lee have what the answer is. Once we've combined those answers

Okay, We're trying to find a trace. The Matrix in this case is a three by three matrix. So I'm going to write. The matrix is a is one tu minus one three to minus two seven five in minus E. If you have actually seen my other video, that was a two by two. This is very simple and easy to trace of. Okay, is equal to make it quick. Fred Lynn. Here you get all the diagonal components and you add them together. So in this case, we're gonna add one plus two plus minus three. So the answer for this is zero. Thank you.

When you want to find the determinant of a free life we matrix, we always tend to wanna work with the road or column, which has the most ever I shall. We're gonna work with the first word on when you find the determinant of a three by three matrix Schefter, consider in the form or Petroplus minus press minus, plus minus and plus minus part. So it keeps alternating between those do so just looking at this first way we're gonna use the minors off the first word to calculate that. Determine up the whole matrix. So, looking at the minor for one, we would eliminate the first way on the first column, leaving us with just a determinant off one minus tay minus one and five on. Then we wouldn't want the mine off minus four, which changes to a plus for because we got two negatives. So it would be laugh, uh, by this, free by this to three on five and then because, as we stated earlier in the video, we want to work with the roads which contains zero. We don't have to at the third element because obviously it's just more proud prize ever so we can just work with this, um, and then work out the determinant off these. We have one times five, which is five. Take away minus two times minus one, which is just two of minus two plus four minus three times five. That's minus 15 minus minus two. Test features plus six, which is equal to minus free plus or times minus nine and then is equal to minus three. Thus, I mean minus Sorry. We've got negative nine minus 36 which is equal to minus 33.


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