test-cramers-rule-for-a-random-4-times-4-matrix-a-and-a-random-4-times-1-vector-b-compute-each-entry

Question

Test Cramer's rule for a random $4 	imes 4$ matrix $A$ and a random $4 	imes 1$ vector b. Compute each entry in the solution of $A \mathbf{x}=\mathbf{b},$ and compare these entries with the entries in $A^{-1} \mathbf{b}$ . Write the command (or keystrokes) for your matrix program that uses Cramer's rule to produce the second entry of $\mathbf{x}$ .

James Chok · Accepted Answer

Okay. So in this example, we want to see how these four transformations will affect the determines off A when we use random former for matrices. So I created a love another program here, where we will generate random matrices depending on this shape. Yeah, so if you want to use it on five by five and 656 matrix Just changed history. Five comma five and six. Common six. Then what this will do is that this will find the ratio between eight transfers on A. So it&#x27;s going to wipe transfers and eight to the ratio between negative, eh? And eh, The ratio between two. A two and ratio of 10 8 to a and then everything worry about what&#x27;s called us. But that&#x27;s what it does. So what you find out is that the determined a transpose should be equal to two determinant off, eh? Determined off negative. A is equal to negative one times by and determines all, eh? So here I&#x27;m just gonna make a random a general n by N matrix. And then this one will be true to the end determinants. Oh, hey, this is 10 to the end tournaments away and in this case because this is for for so many years, for this will just be one to the end. So it&#x27;s gonna be equal to determine off A is going to be too to the form, so you&#x27;ll get 16 and this is tentative. Fall and you&#x27;ll you&#x27;ll see this when you perform this program.

Nez Nikoo · Answer

okay. We want to find the value of X knowing that we were given matrix a times some X metrics A&#x27;s a three by three. So you know that this guy&#x27;s three by three is equal to B, and we&#x27;ve given value be and they&#x27;re asking us to use the Cramer rule to do this. Okay, so the way we&#x27;re going to do it is we know X one. We have to find a determined off a. Okay, I&#x27;m not going to go through detailed calculation, all finding determine of a because you should know that by now. So, please, just if you do not know that, go back to previous videos and, um, find out how exactly find the determinant of it. Three by three matrix. So x one you cool there is for X. They&#x27;re going to be three values. It&#x27;s going to be x one x two on the x three. And if you know how to find one, you just matter of substituting a crack column for on matrix A. So on the bottom, we&#x27;re gonna have determine it. A, which is a three by three matrix have already calculated that that is minus 32 using the Cramer rule. What you do is you write basically matrix again and you substitute be for specific columns for X one. You could be on the first column, so I&#x27;m gonna right here. Five, 72 That&#x27;s a B. Then you write the rest of the A, as is 351 Okay, And then you find a determinant off the top, and you already know the bottom minus 32. Again. All you do have to find a determine it is. You write the first column 572 That&#x27;s how I do it. You can different ways. You can find a determine it. So that&#x27;s up to you. And then you just do them. Basically this product product product you, Adam subtract from 333 again. Please make sure, you know, and you&#x27;re comfortable with how you find a determined and over three by three matrix. When you do all of that, you&#x27;re gonna get a minus eight here minus minus eight over minus 32. Gives you 1/4. So that&#x27;s your X one x two. You do the same thing that we do for X one. We know the bottom is determined of a So that&#x27;s minus 32. Now what you do on top is you substitute be for the second row. So what you do is you&#x27;re right for two to 35 point, Actually, I&#x27;m gonna go different color. So you know, when I substitute what color? So this one, I&#x27;m just gonna physical like this. This is the one that I substitute. Now I&#x27;m going to write it here. Five. Seven to that&#x27;s so much better to see. So now that&#x27;s the substitution that you do being there. Cramer&#x27;s rule again becomes a minus 32. And when you find a determinant on the top, you get minus two. So the answer becomes 1/16. Okay, we got one more to go x three. And if you want to pause it, do it yourself. Pause the video. Make sure you know how to do this one. Because if you any do it yourself, you get the right answer. That&#x27;s when you really know you understand it. So posit. Come back. I&#x27;m going to continue now. We&#x27;re just going to substitute for the third row and I&#x27;m gonna change my color. So you see it? Five, seven to and When you find a determinant on the top, you get 21. I mean minus 42 over minus 32. And when you simplify it, there&#x27;s 21/16. Okay, guys, here&#x27;s the answers. One over for his X one, which is this guy. This is X two. This is X three. Thank you so much.

Nez Nikoo · Answer

okay. We want to find the value of X. We know what the matrix of a is is a three by three and bees a three by one. And we want to use Cramer&#x27;s rule because that&#x27;s what the question is asking us to do. So in order to find X one on, what we need to do is we need to find the determinant of a. So I put here Determine of a determine of a is a three by three matrix and you can look at it and go through it yourself. It&#x27;s You should know how to do that by now. And that basically is to 71. I&#x27;ve done that on a piece of paper, so I&#x27;m just gonna write it here t to 71. Now the Cramer&#x27;s rule says the matrix A. When you want to find the first value you&#x27;ve plugging to be, which is right here into the first call. Okay, so be is three minus one four. Then write the A as iwas three, four, five in six, minus seven and nine. Now you have to find a determinant on the top, which is the same but again by three. By three. And if I&#x27;m just gonna show you how you start the determinant you don&#x27;t need I&#x27;m not going to write all the details here because you need to know. You practice yourself. I give you the answer is three minus one for here. You repeat the first column, you repeat the second column and you do your magic off diagonal multiplication and adding and subtraction the other way around. So when you do all of that, the top becomes 30. And we know where the bottom is to 71 truth. So now we have X one. That&#x27;s X one. We&#x27;re gonna find X two. We do the same thing. Cramer&#x27;s rule on X two We have to not substitute be for second column. So it becomes three minus 14 for second column. We write to 71 on the bottom because that&#x27;s determinant of a. And then you just repeat the outer columns from Matrix a 5 to 2 six minus seven and nine. When you find a determinant on the top is 59 over to 71. Okay, last item is X three. I&#x27;m gonna write the X three right here. Start X three, right here equal to now for X three, you need to substitute The third column would be okay. So I like to kind of right to be first when I whatever I substitute. So it&#x27;s minus three minus 14 So I kind of verify that first knowing that that&#x27;s the substitution be. Then I go put the rest of the a whatever is therefore to to 345 Now I like to substitute by column since we substituting my column, I write it that way also. So it&#x27;s less chance of error in the bottom 2 71 Okay, so the X three again You right here. 5 to 2, 345 Whatever. You feel the most comfortable mitt to find a determinant, you do it that way. I like it this way that I go on, we&#x27;ll fight this but this at these three. And then I multiplied these three. Subtract, subtract, multiply, subtract. When you do all of that, you&#x27;re going to get a 81 over to 71. So those are the three answers. This is our X one x two on x three. Okay. Thank you.

Nez Nikoo · Answer

okay. We&#x27;re trying to find the value of acts when the Matrix A is known and B is known. So the formula is a times X equal to be Now we know a and we know B But we want to find what one of the values for expertise formal to be basically rally. We know A is to okay minus two and four x is unknown, So I&#x27;m gonna know that I&#x27;m gonna put x one x two. How do I know that is to row and column? I explain it right after I r a to B and B zero in three. So what happened is you have it two by two. And you know, the answer is A to buy one. So the only way that can happen is you need to have another two. Why one matrix in here? And we&#x27;re trying to find, um, to buy one matrix here. And so this is going to be the former. And the way it works is we trying to find this true Cramer rule? What is Cramer&#x27;s rule? Cramer&#x27;s rule tells us that each element X one is equal to determinant of this new matrix over determined a determinant of a is two times for minus times for minus minus two times eight. So we know that&#x27;s 24 now on top. What you have to do is you have to substitute for the first column B. Okay, so what you do is you substitute to be for the first column. So if I do that is going to be zero minus three and write the 2nd 184 Find it, Determine up top. I know that&#x27;s going to be 24 because zero times four minus minus three times eight is going to be 24 over 24. So the first value is one to find the second value using the creamy rule you write the determine of and the bottom, which we know is 24 on the top. You write the first column, as is, which is a two minus two. And now you substitute being here, which is a zero, any three. So when you do that, um, what you&#x27;re gonna get is his minus three. I&#x27;m sorry to zero, uh, 20 minus two and minus tree. So you&#x27;re going to get minus six. That&#x27;s zero over there. Zero times minus 20 So minus 6/24. So you get one over for, and that&#x27;s x two. So you find the to and in order to write the X properly used, X is going to be like the first values one and minus four. So if you plug in the X one next to in this equation, this becomes a valid. That&#x27;s how you find the answer. Thank you.

Nez Nikoo · Answer

okay. We want to find a value of X. We have determined A and B so basically is a we know a times X, we do not know and be So you basically write a two by two is three, five, six and two times it to buy one that we do not know we want to find which is expert So I&#x27;m gonna put here x one x two equal to be and we want to find it using Cramer method First you have to do is find determine off a but should be easy by now. So determinant of a is basically six minus 30. So that&#x27;s minus 20 four. Okay, now finding X one just use it. Cramer&#x27;s method You put determinant of a on the bottom, which is minus 24 and the top What you need to do the first column. You do a substitution. You could be as the first column. So we basically moved this be on X one as the first column. So is going to be four nine here and a five and two. I solved this later. I want to show you ex to what you&#x27;re gonna do is better to have these organized rated. You go again, minus 24 here and in here. We&#x27;re gonna write three, six, and then again, substitute four and nine, which is B in this second call. Come back to x one. We&#x27;re gonna have a eight minus 45 over minus 24. So that gives you a minus 37 over minus 24. So the answer is 37 over 24. I don&#x27;t know why this 37 kind of gets ugly, but no, but it correct my seven guys. Sorry. Right here. Go back. Right 37. And this one here, we&#x27;re gonna have a minus 24 and three times nine is 27 minus 24. So the X two is going to be minus three over 24. So that is minus 1/8. Because you can simplify. So the answer for X IHS 37 over 24 and second Valley is 1/8. That&#x27;s it. All right. Thank you.

Problem

If your version of MATLAB has the flops command, …

Problem 34 Easy Difficulty

Answer

And, the command for producing $2^{\text {nd }}$ entry of the solution is already in the programming.

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

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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And, the command for producing $2^{\text {nd }}$ entry of the solution is already in the programming.

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Alayna H.

Caleb E.

Kristen K.

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

Linear Algebra and Its Applications