suppose-that-all-the-entries-in-a-are-integers-and-det-a1-explain-why-all-the-entries-in-a-1-are-int

Name: suppose that all the entries in a are integers and det a1 explain why all the entries in a 1 are int
Uploaded: 2019-08-23
Duration: 2 min 29 s
Channel: Numerade
Description: Suppose that all the entries in $A$ are integers and det $A=1 .$ Explain why all the entries in $A^{-1}$ are integers.

Question

Suppose that all the entries in $A$ are integers and det $A=1 .$ Explain why all the entries in $A^{-1}$ are integers.

Viswesh Uppalapati · Accepted Answer

in this video, we&#x27;re gonna be going over how to solve problem number 18 from section 2.3. Wishes on Cramer&#x27;s rule in determines. So the question because a prompt that, um and says that the Matrix has and has all entries to be integers so like whole numbers. Um, and the determinant of a is one. And it s says to explain why all the entries in the inverse of a have to be introduced as well. So Thio do that. It&#x27;s pretty simple. We just consider this formula that was given to us and they were made, Which is that that the inverse inverse of matrix is one over the determinant of the Matrix times, the advocate of the Matrix. So you scale every entry in the as yet of the Matrix by the determinant. So, um, if we have like, let&#x27;s say, three by three matrix with nine entries, The advocate of the Matrix is no the cool factor of every entry transposed in a matrix. So if you want to find a co factor for that, uh, for the 1st 0 for example, you basically block or like cross out the row and column that contains that entry and take the German of the remaining two by two. And you repeat that for every single element and then transpose that and put it in the matrix in the same spot and transpose it. And they&#x27;re actually alternating signs of plus minus minus, plus minus plus minus. Plus. When you take the determinants, you should have a plus or minus sign based on what entry it is corresponding to it. So that&#x27;s how you find an adjective, a matrix. So basically, this just involves taking the determinant of a two by two, which is basically, uh, the diagonals multiplied by each other and take their difference. So when you multiply indigenous together or subtract, add them, they&#x27;ll still be integers. You&#x27;ll end up with integers. Therefore, the inverse is going to be scaled by one over one because the determinant of a is given to be one times the advocate of a. So the inverse is just the as you go today and since we know that the advocate of a contains only integers, because taking a determinant will give us a whole number. If we use all integers, then we also know that every entry and the advocate of a or the inverse of

Chris Trentman · Answer

were given the Vergis ease of a parallelogram and were asked to find the area of this parallelogram. The verdicts is our 0052 64 and 11 6. So we see that parallelogram already has a vertex at the origin. And we see that this parallelogram he&#x27;s determined by the columns of the Matrix A which has entries one of our vertex seizes five to and the other Vertex its opposite from that is 64 These are the columns we have that the absolute value of the determinant today is Thea absolute value of 20 minus 12 which is eight. So this is eight and so this is the area of the parallel.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be solving problem number 22 from the book from section through going through which the baseline determinants in the cream result and the area of parallel grounds. Um, so you were given four coordinates its label them a bcd and rasa. Find the area the parallelogram generated by these four coordinates. First, we need to find two factors that are part of that parallelogram. So we can do so by just, um So to find the vector a B, for example, we just take be just five negative, too. And subtract the accident like component from a car. Um, subtracting X and Y component of beef by a So you get the vector to be five minus zero, which is five common. I give to minus negative, too, which is negative. Two plus 20 So that&#x27;s the first specter. And we need one more vector here. Ah, which is different from the first specter. So or perpendicular to so, um, way use the same one of the points has to be the same. So we use a as a common point. Um, so a C would be just the recover one minus zero common negative too, which is native three comma, one minus negative. Two is one plus two with three. So those are the two points We just put them in a, um in a two by two matrix. So five zero negative 33 And the area is that&#x27;s called us A. So the area is the determinant of of this matrix a year on and take the absolute value of it. So it&#x27;s gonna be five times three minus negative three times zero, which is absolutely of 15 because made of three times zero is just zero, possibly 15 is 15.

Chris Trentman · Answer

were given the Vergis ease of a parallelogram, and we were asked to find the area of this parallelogram. The Vergis is our negative. 20 03 13 and negative 10 So, first of all, we noticed that none of the vergis ease air the origin, so I wanted to shift parallelogram to the origin. We can do this by subtracting. One of the vergis is from each of the other. Vergis is So for example, we could do this by subtracting the Vertex negative 10 So we end up with a parallelogram with new vergis is negative two minus negative. One is negative. One. It&#x27;s going to be negative. 10 and then 03 minus negative. 10 says 13 and 13 minus negative one. Well, that&#x27;s me too. Three. And of course, the origin. 00 So visualizing the new parallelogram, we see that we have a point. Vertex it negative 10 as well as the vertex at 13 and at to three. So parallelogram as the same area as the original and moreover, is determined by the columns of the Matrix A in this as the column vectors which are the Vergis, is negative 10 and 23 So it follows that the area is equal to the absolute value of the determinant of this matrix, which is the absolute value of negative three, which is simply three.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be selling problem number 20 from section 3.3 of the book, which is based on determinants and ah Cramer&#x27;s rule. So, um, it gives us four points or or coordinates and then it the question asked us to find the area the parallelogram generated by these coordinates to do so, we just need to find two vectors of that parallelogram and take their determinant. So it&#x27;s easiest to label these first a B, c d. And we need to find two vectors. And since a a 00 let&#x27;s make it easier to start with 00 So the vector a B would be, um, the coordinate B minus A. It just gives us ah, negative to come up for because negative to minus zero is negative. And for many zeroes Ah, for And the vector um, the vector a c would be ah, for native five because it&#x27;s again subtraction by zero is the same. Um, and we put these in in a parallelogram are in a two by two determinant like this because they&#x27;re vectors are two sides of a matrix are not a matrix. A parallelogram on the on the coordinate plan, and And we take the cross product here, which is, um, negative. Negative. Two times negative. Five minus four times four times four. Yeah, which is just, uh, 10 minus 16. And you take the absolute value. The determinant to find, um, the area so area pickles, absolutely. Of the chairman of A, which is call the absolute value of 10 minus six years, which is equal to six.

Joshua Eastwood · Answer

So this question says a one and our are integers And then it asked us to show that this this a one minus a one to the ire. So anyone are in the end, divided by one minus R has to also be an integer So if we imagine first of all, that we take a one out of the top get one minus are to the end divided by one minus R All right, so we know this is an integer, um one minus are to the end. One&#x27;s an imager. Are is an integer. Okay, so it&#x27;s some number that&#x27;s an integer and then it&#x27;s raised to and is always a natural number. So if you take any agent reason a natural number, you&#x27;re gonna get an integer and then one minus an integer is also an integer, So that&#x27;s an integer. And if one mine is an integer is always an integer, then that&#x27;s also manager. Hey, so you get an instant. Your time&#x27;s an integer divided by integer, which will give you an integer. Now that might seem like a bunch of just random talk, so just think about it this way. If I have an integer, and then I&#x27;m multiplying it by an integer each time. All of these numbers in my Siri&#x27;s, whether it&#x27;s plus or minus, are going to be integers. Do they have to be because two time to be, for example, is gonna give you six? There&#x27;s no decimal. There&#x27;s no fraction. So if you multiply and injured by an integer, you&#x27;re going to get another integer. And if you add up a series of integers, they&#x27;re gonna stay is interesting because there&#x27;s no place where you can introduce anything else, all right, so hopefully that makes sense.

Problem

Suppose that all the entries in $A$ are integers …

Problem 18 Easy Difficulty

Suppose that all the entries in $A$ are integers and det $A=1 .$ Explain why all the entries in $A^{-1}$ are integers.

Answer

since each entry of adjA is an integer, entries in the inverse of $A$ are also integers.

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

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Caleb E.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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since each entry of adjA is an integer, entries in the inverse of $A$ are also integers.

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Caleb E.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Anna Marie V.

Caleb E.

Michael J.

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

Linear Algebra and Its Applications