in-exercises-1-10-assume-that-t-is-a-linear-transformation-find-the-standard-matrix-of-t-t-mathbbr3-

Question

In Exercises $1-10$ , assume that $T$ is a linear transformation. Find the standard matrix of $T$ .
$T : \mathbb{R}^{3} ightarrow \mathbb{R}^{2}, \quad T\left(\mathbf{e}_{1}ight)=(1,3), \quad T\left(\mathbf{e}_{2}ight)=(4,-7), \quad$ and $T\left(\mathbf{e}_{3}ight)=(-5,4),$ where $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$ are the columns of the $3 	imes 3$ identity matrix.

Ashley Boni · Accepted 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.

Michael Jacobsen · Answer

in this example, we have a transformation t that&#x27;s going to be linear, and what it does is it takes the first column of the two by two identity matrix, and it maps it to this element and are too. Likewise, the second element of the identity matrix is mapped to negative 5 to 00 of our two. Our goal, then, is to find a matrix. A such that t of X could be expressed as a Times X If we confined such a matrix A. Then we can use The Matrix to evaluate T at any Vector X, not just these two vectors. So the first step is to note that the standard Matrix A if we have these two elements, is given as follows. The Matrix say will be t evaluated one for the first column and t evaluate at E two for the second call. So for this situation, we know that t one is 3131 TV two is negative. Five to 00 And so this matrix A is the standard matrix for the transformation. T

Ashley Boni · Answer

So we want to find the Matrix Tea for a mapping from our TUTTO are too. That leaves the one unchanged and maps to took you two plus two equals. So where do we start? We know that the Matrix has the form t of even one as the first column and t of e to as the second calm. So we want to be want to remain the same. So in other words, we want t of you want to equal one waas zero e too. And we want teeth of e too the equal. You too. Plus three you won. So in order to write this, then matrix form TV once the first column. So it&#x27;s one yuan plus zero e too. And TV too is equal to 31 plus one e two and that is our matrix t.

Henry Carnick · Answer

so in order to solve this problem, but we first need to do is one, uh, find t of 100 TF 010 and t of zero 01 The first observation I would like to make is if we look at the vectors 0 to 3 and the fact two times 011 We end up with zero 01 If you play the transmission to both sides like so the first thing you&#x27;ll note is ah, transfers. T is linear, which means I can distribute the function across terms on the left hand side. So what, we end up getting on The next line is t of 0 to 3 minus tee off. Two times 01 One is then, of course, equal t of zero 01 But another opposition make Is that because tea is linear, I think take this to and move it outside of the function and have it look like so And so what? This gives us this que 0 to 3 minus two times t of 011 which equals t of zero euro one. So knowing they given a information about But we know about what happens, we evaluate ta zero trophy and teaser 11 We now can solve 40 001 directly. And what would end up getting there&#x27;s, uh, is ah six minus five four minus two times three minus one minus one, which is then equal to six minus 54 plus minus six to to which is, of course, equal to zero minus three six, which is equal to t 001 Another observation I would like to make is that if we look at 011 minus zero euro one, we end up with 010 And if we apply the transformation of both sides again, we could make the observation that we can distribute our our transformation t because it&#x27;s linear. So we end up with t of 011 minus t of 001 is equal to t of 010 Our previous work tells us what see officers Air one is we already know what to t of 011 was before. Ah, so now we can solve this directly, and what we end up with is going to be three minus one last one minus zero Niner three six, which has got equal three to minus seven, which equals T of 010 So the last observation I would like to make is that if you look at one from the too calm a zero, that&#x27;s a fact. Two times zero 10 we end up with 100 playing the transformation of both sides. We make the same observations that&#x27;s before that t is linear. Push means I can distribute t uh, across the terms. So we end up with TF 120 minus T f two times 010 which equals t of 100 But note the fact that again, to use a linear so I can pull out the to, which means we end up with tee off 120 minus two times t of 010 which equals t of 100 And because of the our previous work, we can evaluate this directly. So we end up with is two column ice 11 minus two times 32 column minus seven, which of course, is equal to two compromised 11 plus 96 minus or 14 which, of course, is equal to minus four comma minus five 15 which is equal to T of 10 euro. So using all the information, we can now construct the matric representation off our linear transformation and it&#x27;s going to be minus four minus five 15 three to money seven zero line of three in six.

Michael Jacobsen · Answer

in this example, we have a transformation. T mapping are two into our to, and what it satisfies is that the image of the one will be the one minus two B two, and the image of E two is e to. Given this information, our Cole here&#x27;s to determine if we map X under the transformation. T What The Matrix A must be, in other words, were looking for the standard matrix a off the transformation t But by formula, this matrix A can be written as TV one for the first column and t of e two for the second column. So it&#x27;s determined what these values would be in this situation. First, we know that T V one equals one minus two, e two, but the one is the first column of the two by 28 any matrix? So it is a 10 My has two times e to you can think of e to as being of calling vector containing zeros. But we put a one in the two slot, so it is a 01 Then if we do the scaler, multiplication and the vector addition will have ah one and negative too altogether. Next we&#x27;re told that TV too is equal T too. But you two is 01 as before. So now we&#x27;re raid the states with the standard matrix A of t will be first go to TV one. We found that that isn&#x27;t one negative too. Then for TV to we have a 01 and this is our standard matrix A for the transformation T.

Michael Jacobsen · Answer

in this example, we have a transformation. T mapping are two into our to and what T is going to dio is take points in the plane and rotate them by an angle of three pi over two in the counter clockwise direction. Our goal here is to find the standard matrix A of T so that he can be written when evaluating X as eight times X. Her goal is to find the standard matrix A. We know that we can calculate t one and th e to. Then our work is going to be complete where e one e two are the first and second columns of the two by two identity matrix. But we have a great formula for when we&#x27;re ever were rotating points by an angle in the counter clockwise direction. The First Inc. The first element is going to be co sign at three pi over two, and the next element will be signed at three Pie divide by two. And if we change the angle, we just have to change what goes in here. Likewise, for the second set of points we have that t at e two is going to map into negative sign at three pie Divide by two for the first century and co sign of three pi over two for the second entry. So let&#x27;s evaluate each one of these components or will have all together. In the first case that co signed three Pi over two is a zero and signed three. Pi over two is negative. One for the second case Sign a three Piper to is negative one. So with the extra negative, we get a positive one here and co sign is zero at three pie or two. This tells us our standard matrix A can be written as TV one TV, too. But now it&#x27;s equal to the falling matrix. We put in zero negative one and 10 and this causes the appropriate rotation by this ankle.

Problem

In Exercises $1-10$ , assume that $T$ is a linear…

Problem 2 Easy Difficulty

Answer

$\left[\begin{array}{ccc}{1} & {4} & {-5} \\ {3} & {-7} & {4}\end{array}\right]$

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

Linear Algebra and Its Applications

Grace H.

Maria G.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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$\left[\begin{array}{ccc}{1} & {4} & {-5} \\ {3} & {-7} & {4}\end{array}\right]$

Linear Algebra and Its Applications

Grace H.

Maria G.

Kristen K.

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.

Maria G.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications