mathbfm-in-exercises-37-40-let-t-be-the-linear-transformation-whose-standard-matrix-is-given-in-exer

Question

$[\mathbf{M}]$ In Exercises $37-40,$ let $T$ be the linear transformation whose
standard matrix is given. In Exercises 37 and $38,$ decide if $T$ is a one-to-one mapping. In Exercises 39 and $40,$ decide if $T$ maps $\mathbb{R}^{5}$ onto $\mathbb{R}^{5} .$ Justify your answers. 
$$
\left[\begin{array}{rrrr}{-5} & {10} & {-5} & {4} \ {8} & {3} & {-4} & {7} \ {4} & {-9} & {5} & {-3} \ {-3} & {-2} & {5} & {4}\end{array}ight]
$$

Barsha Rana · Accepted Answer

we have to find if the given transformation, um is 1 to 1, but not, um in order to find out the kind of cancer we have to use the road reduction that we know already and then looking at the rank and looking at the number off pivot columns pivot positions in each column. We can decide if it is wonder one or not. So let&#x27;s start by, You know, we will keep the first. The role always fixed. Thank you. We will try to the malls on the streets here that understanding a screen. So if the first ro is fixed, I&#x27;m going to fix it at the same bad deuce. What? I&#x27;m going to change. But I am going to do the following task first. Odd to what will be my new second row? Well, it will be the previous second room. Added bit eight. Bye, five times album. Okay, so how How much will that be? Let&#x27;s see. See, first IHS, we have to multiply eight by five with the entire first. True. So what will be it if I multiply by eight by five, this will become minus eight, right? This will remain zero This will become minus eight on this will become, um, four into eight by five. So it will be 32 boys. Thanks. Correct. So I&#x27;m showing it for one. And then I will expect that you would be able to follow for the the rest of the rules 100. So now our took That means what? Previously my art to first position was eight. So eight plus minus eight. How much we like it. We will get here. Cedo. Same way three plus zero. I&#x27;m sorry. It&#x27;s not zero. Still, so 10 will then become If I write 10 over here, See, it will be 16th. Correct. I have a 16 over here. So my first well was 33 plus 16. It will give us 19. Then I have again minus eight plus minus four. So it will be minus 12 and then seven. Plus, Did you do by five? So seven less 32 divided by five. It will be 67 because 75 sir. 35 35 plus 32 is 67. So we get 60 seven. Bye. Fine, right. Same way now. The next action that I&#x27;m going to do is are three are three will be the older three. Plus are one is going to be changed. So whatever numerator off art that will come in the denominated divided very five. And whatever is the coefficient off are three that will go in there New Morita. So four by five, Hard one, Correct. So again, if I clean up the screen little bit I&#x27;m sorry. Then we will see what will be those numbers. And as I again say that for one particular But I&#x27;m showing all the steps I hope you would be able to figure out for the remaining glands. So now if I multiplied with four by five this number four by five I will change it to, uh, minus. I&#x27;m sorry when you take up minus for if I multiply this number with four by five, I will change it toe plus off eight. If I change this number, multiply this number with four by five I will change it to minus four and four by five. So 16 by fine Now what am I doing? I am adding this toe The previous told you so. Let us add element by element four plus minus four. It will yield zero again. Minus nine plus eight. It will yield minus one plus five minus four. It will yield plus one and then minus three plus 16 by five. So it will build one bite. Fine. Next the same way. If you see on four, he&#x27;s going Toby at Fort plus the coefficient off. Three. Bye. Coefficient off first. So playing well, it one. So definitely the first term will be zero. The second term would be minus eight. The third term would be eight and the last time would be eight by five. Now, if you see these two rules 00 if you C minus one left, minus seat minus 11195 minus 88855 It&#x27;s like there is an eight right throughout. So why can&#x27;t we subtract the fourth rule from the first one after we multiplied the first one? Right? So let us try that. I am not writing the first two rows. Imagine There, there, in the third one, we will have this and my new four through will be our for prime, which will be eight times third row minus only for truth. So that will be zero my insider. Whatever multiply you get zero the eighth time for a third row. So it will minus it minus the next troop. So it will be zero. It will be zero. It will be zero. So now, at this point, if we take a look at the system, this is giving of a metrics that somehow looks like this minus five 10 minus 54 0 19 to a minus 12 67.5 zero 19 minus 12. 67 by five. Zero minus 115 I mean, one by five. Seattle minus one one one by five and then zero zero zero Syria. Now, what is so particular about it that it is the moment if it position food. Hello? Okay, look. Awesome, Trump. This means that this system off recreation have a free baby. This is means nontrivial solution on a dark to me. We dad? Mm, Yes, please. A little nice Mobile insulation. Mhm. It&#x27;s more than one position means the consternation. P confirmation. Qin&#x27;s the transformation t don&#x27;t be. Come on, transformation. So that&#x27;s a That&#x27;s a visit

Ashley Boni · Answer

So we have our matrix A and we want to determine if the mapping is 1 to 1. So if we take our matrix A into Matt Lab, which it suggests we use, we type the first row semi colon, second row semicolon, third row, semi colon and fourth row. And then if we take, um r R e f which is the command for a reduced Rochel on form of A and give it a name, I just called her are for I guess, reduced. Um, it gives us an output of, uh, this matrix 1070 01 negative 90 0001 and the last row of all zeros. So, ah, by theorem 11 a mapping is onto if the equation a X equals zero has only the trivial solution. So that means that ex would have to be the zero vector. Ah, But since there is no pivot in the third call, we know that we have a nontrivial solution. So Ah, the solution two our X equals zero is nontrivial. Therefore, um, the mapping. I guess tea is not want one

Ashley Boni · Answer

s so we have the linear transformation with the matrix A given. And we want to figure out if the mapping ah t is onto. So we need to look and check that there is a pivot in every row. So if we take the Matrix A and put it into Matt Lab, which it suggests we use, um we just input a equals and we type all the entries in the first row and then put a semi colon, second row, semi colon and do the same. So we&#x27;ve input all five rows four and then row five and then close our brackets. And then if you use the command, our equals r r E f of a. This tells us to, um er sells my love to take a kick the deuce row astral form and signed that matrix Thio makes your car anyway, So if you do this, you get an output of one 0005 So we have our first pivot 010 zero, minus four. But her poo for a second row third row is 001 00 So we have our third pivot. Ah, fourth row is 000 11 So we have our fourth and the last row is all zeros. So since there&#x27;s no pivot in our last row, this means that tea is not onto, since the columns don&#x27;t span or five.

Ashley Boni · Answer

So we have the formula for the standard matrix of a transformation t and that is given by t of E one TV too, and t of a three. So these are the columns off our matrix. And you wanted me to any three artists the standard basics vectors with the one in the, um, 1st 2nd or third position. So all we need to dio is plugging. The vectors were given because we&#x27;re already given t of a 20 to 33. So TV one is 13 T v two is four negative seven and TV three is negative five.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be answering problem number six from section 1.7, which from the book with Based on Later Independence, we&#x27;re given this. Ah, augmented They&#x27;re not I&#x27;m into Matron&#x27;s. But just the 1st 3 columns of this matrix is the matrix for the problem to the problem to start. And the promise says if to justify answer and to determine whether, um, this set of our this matrix as linearly independent for a matrix to be literally independent or for its columns to be literally independent, uh, the Matrix can own the matrix equation. X equals zero can only have the solution, the trivial solution, which is just this, in this case, just a column of zeros based on depending on how many variables it has. So every single variable has to be zero, which is a travel solution. So X equals zero is a matriculation that we learned section one point for and, uh, hey, here is just the given matrix. The Matrix were given, and zero Is this Colin Dobie Koncak cabinet to the end of the matrix. Ah, and this is how you set up in X equals zero matrix and you perform row operations elementary row operations till you get the staircase of zeros and one&#x27;s in, Um, the diagonal of the matrix. Um, this is the Matrix. This is the same Matrix Elementary, uh, that this is the same matrix as the previous matrix that we have here as elementary operations results in the same matrix, but with different values. So here, essentially, we&#x27;re just holding a system of equations. Since in this matrix, we have, ah, diagonal ones. And the right side is still zero after the elementary row operations element hero operations include subtracting arose one road from another or dividing are scaling it with ah factor, um, or switching rose. So so elementary row operation results in the same matrix, but with different values. So essentially, we&#x27;re just solving the same size is Miss essentially Alexa Living system. Since we have ones in these diagonals. And let&#x27;s just say we can label each of the columns to be a variable x y z, um, here, the one here shows us that Z equals zero and one y plus our why minus four Z equals zero. So why minus are so y equals zero sends you zero and X plus three z equals zero and says zero X is also zero. So we know that this right column is just indicative of, uh, the solution set. So we know that here the, um sorry. One sec. So we know that here, um, the solution column Vector X is just 000 where each each entry represents a different variables to the 1st 1 is X the second one&#x27;s a why and what their own see. So which is exactly like the trivial solution. Therefore, this matrix are the columns of these. This matrix are literally independent. I don&#x27;t show a lot of the math behind, um, performing elementary row operations because I won did this beforehand before the video started. And it like it&#x27;s a very long process so it wouldn&#x27;t fit in one slide. So if it&#x27;s a matrix larger than, like a three by three or three by four, uh, it would be best in your times interest toe to use like a matrix calculator to get it in, uh, like basic isil on former reduce the echelon form so that it makes the problem easier if your teachers

Ashley Boni · Answer

so we won&#x27;t show that he is a linear transformation by finding a matrix, um, such that when we multiply it by vector X with four components, we get this as an output. So this transformation is mapping a vector in our four to vector or well, just the number in are a real number because this is just a number. If we plug in ex one next three x for so we know we need a matrix to multiply x one x two x today, that&#x27;s four. So that means we need to have four different columns one to multiply each of these, uh, vectors. But it&#x27;s being mapped to just are so we only need one room so we can stop. So we need four different components that multiply at claw on x two x three x for and we need it to give us to x one plus three x three minus or explore. So the coefficient for X one is two coefficient. Rex to zero coefficient for X three is three and the coefficient we need on export is negative. For so ignoring these lines. This is our matrix that we need to multiply

Problem

$[\mathbf{M}]$ In Exercises $37-40,$ let $T$ be t…

Problem 37 Easy Difficulty

Answer

by the above result, the transformation $T$ is not one-to-one.

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Heather Z.

Absolute Value - Example 1

Absolute Value - Example 2

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by the above result, the transformation $T$ is not one-to-one.

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Heather Z.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Catherine R.

Anna Marie V.

Heather Z.

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

Linear Algebra and Its Applications