in-exercises-33-and-34-t-is-a-linear-transformation-from-mathbbr2-into-mathbbr2-show-that-t-is-inver

Question

In Exercises 33 and $34, T$ is a linear transformation from $\mathbb{R}^{2}$ into $\mathbb{R}^{2} .$ Show that $T$ is invertible and find a formula for $T^{-1} .$
$$
T\left(x_{1}, x_{2}\right)=\left(-5 x_{1}+9 x_{2}, 4 x_{1}-7 x_{2}\right)
$$

Michael Jacobsen · Accepted Answer

in this example, we have a transformation. T that&#x27;s indicated here. And possibly the most important thing we know about the transformation is that it&#x27;s linear. So now that we know that it&#x27;s a linear transformation that motivates us to write the transformation in a better notation, we&#x27;ll call it T of X, where X is a vector containing X one, and X two is equal to a matrix eight times X, where the Matrix A is going to be the standard matrix. We need to determine what that matrix A is going to be. So it&#x27;s also rewrite the transformation. T as t of x one x two Using columns is equal to negative five x one plus nine x two Again. Since we&#x27;re using columns, the comma will correspond to a new row, which contains four x one minus seven x two. In this notation, it follows that our standard matrix A of the transformation T is going to be obtained as follows. Go to column one and pull coefficients. It&#x27;s negative five and four. Then pull the coefficients from Collins column to which are nine and negative. Seven. So this is our Matrix, and now that we have a transformation. T written nasty of X equals a X with this matrix A. We know for sure that this is a linear transformation. But now we might wonder, is this transformation convertible? One way to check is to calculate, determined of a the determined of A is formed by first multiplying the main diagonal five times negative seven which response in a positive 35. Then put in a minus sign and multiply the off diagonal nine times four, which results in at 36. We get all together a value of negative one and this shows a inverse exists. But that&#x27;s not the primary thing we&#x27;re interested in because a inverse exists and a is the standard matrix of t. The thing were really interested in is the fact that now t is an in vertebral transformation. So we started with t being linear. We found a standard matrix. We know that this matrix is in vertebral. And so now we&#x27;ve made a very strong conclusion about the transformation. T t has an inverse. Now that begets the question. If t is in vertebral, what is the inverse mapping recall? This is the original mapping, and we&#x27;re going to find a formula for its inverse. It turns out the best thing to do here is go back to this type of format. T of X was written as a Times X. We wanted redo this with tea in verse. It turns out that T inverse of X will be equal to a inverse Times X. Well, it&#x27;s cheating to stop here. We needed to describe this transformation, so it&#x27;s now calculate a inverse so a inverse will be won over the determinant. But that would be one over negative one. So it&#x27;s the negative of a particular matrix where we go again to the main Dagnall, which was negative. Five. Negative seven and interchange them. So it&#x27;s negative. Seven. Negative five. Now go back to the matrix a again, the off diagnose and purple, and we need to multiply it by a negative one. So it becomes negative four and negative nine. So then, if we multiply this matrix by negative one, we obtain 74 95 So we now know that T inverse of X is equal to 74 95 times a vector X where x is x one x two If we stopped here. That still would not be the best thing to do because our transformation t was given in this notation. Not in this notation. So let&#x27;s take a few more steps so we can write the result in the same notation, which is always considered the proper thing to do. First we&#x27;ll multiply The Matrix Times the vector we obtain seven x one plus nine next to in the first row and four x one plus five x two in the second row. Now we&#x27;re ready for a conclusion. T inverse of X, written as X one comma x two horizontally, Much like in this notation is equal to seven times x one plus nine times x two for the first entry or first row and then for X one plus five x two for the second entry. So if this is our transformation T, it will be in vertebral and its inverse is provided here

Runpeng Li · Answer

no. In this problem work even a linear transformation that is tea with two variables X one x two and his ex wife&#x27;s, too is map into you six x one minus 862 and negative five x one last six X two. All right, So, in order to I didn&#x27;t find whether the transformation is in vertebral, We first need to separate this vector into the product Off matrix. And, uh, the unknowns, which is a vector as a vector, has two variables. So in this case, it will be six activate make five and seven matrix, and the and No. One vector is ex one next to so you can check. This is the correct way to separate this two separate this vector here. So let&#x27;s just do the mortification. Um, we have six times x one minus eight times x two, which is a first entry here, and we also have negative five times six times es Juan plus seven context, too. That is exactly the second enter here. So that&#x27;s correct. Yeah, So right now we need to figure out whether this finger transformations in vertical So by our our textbook told us in order to check the only thing that we need to charities to check weather of this mission Esther has determined into zero or to say to check whether this is this matrix, um, is incredible. If there&#x27;s matrix convertible, then this union transformation he seemed Berko. So the way to check whether it&#x27;s matrix easier burden always to check the determinant. So in this case, at six, activate 57 All right, And that would be determinant will be six times seven minus negative, five times negative eight, which will be passed 40. So that&#x27;s for you too. Minus one to you. So this is too, which is none. Zero. So we conclude that tea is convertible. Okay, so in the next is to find out the inverse off t. So again, by our textbook, we know, um, ask, which is the inverse off T is a inverse Times X. Where here. The a B. The basics here. Six and six. Next eight and naked 57 So we need to find a inverse to do that on the way in the way that I would I would recommend it to find that I would recommend to do to find out the universe is to using the galaxy in divination. Here we have six connective aid. Next, five, seven and 1001 Remember, we need to turn the left part off this matrix to the identity matrix. And then the right hand side of the right right part off this matrix will be our first. So Well, first use, uh, we add the first row by second angle. So six last negative five. It&#x27;s one negative. Eight plus seven negative. One 1010 plus 11 and make it 570 one. Okay. And by now, we we times the first roll by five. So that will be by connected five. Five, life naked. Five seven 01 and Oh, sorry. Um, just a excuse me on this will be just should be wish time. Um, we usedto the personal times by seven. To cancel to cancel the, uh Is this entry here? So that should be, um seven elective. Seven, 77 Next. 5701 And we add the first rule by the second roll. So this will be two zero seven hate and elected. 5701 All right, so right now we can just ah, uh, thing we need to do is to is to cancel too. Cancel out this entry here. Kids out. Thank you. Five. Here. So do so. In order to do that, we multiply the first roll by You have I have. So that would be, uh, five zoo on five. Half times seven. That will be 35 half. And I have times eight will be 20 and connective five will still be there. Seven 01 So we can add the second row by the first roles. We have zero here and seven and 35 over too. And 21 for the first roll. I still have 50 35 over two and 20. Now we divide the first roll by fight, so we have 10 And, um, that would be seven over two and four and divide a second rule by one by seven. So that will be one. Um, and this will be you live over too, And this would be three. Okay, so I were in verse is 7/2 or and I&#x27;ve had three heard in war. Our members asked, will be, uh, the inverse of a times bearable, which is, uh, ex lax to vector. That is seven or times. Uh, I&#x27;ve have three times. Ex one next to okay. And to do the matrix pontification, we have seven x one class or axe too. And five have X one last three x two and this is our members off. This is in verse. Oh, tea and

Ashley Boni · Answer

So we&#x27;re given that t of X is equal to the specter, which depends on X one and X two. And we want to find a next wanna next to such that t of X is equal to the vector. Negative 149 So first, let&#x27;s write this in the form of a matrix A times X So we want some matrix times x one x two, two equal the specter here. So we need thio. Multiply X one by one annex to buy minus two in the first opponent For the second, we have negative X warm. So minus one plus three Next to and for the last component, we have three x one minus two extra. So this is our vector. Our army tricks, eh? Going you to do with on. We want a X equal negative 149 So we need to roll. Reduce the augmented matrix of a and we need to put the vector we want on the right hand side So minus one for nine. So to produce. First, we want, uh, these two values to be zero. So I&#x27;m gonna take the first row and add it to the second row So I get a zero here. Um, negative two plus three gives us a one and negative one plus four gives us a three first broken stay the same. And the last row. I needed tohave a zero in place of the three. So I&#x27;m gonna multiply the first year of times negative three and add it to third. So I get a zero here. Negative two times negative. Three or six minus two. Gives us four negative. One times negative. Three is three plus nine is 12. So now I have a one in this position, which I&#x27;m happy with. So now I want to make the entries above and below zero. So I&#x27;m going to use that second rule to do that. So let&#x27;s see. Our second row can remain unchanged. We&#x27;re done with that. So now I&#x27;m gonna take second row, multiplied by two and added to the first room. So are one stays the same? We&#x27;ll get a zero here and three times to a six minus one. Gives us a five ah zero here and then we have negative four times. Want it&#x27;s negative for post for zero negative. Four times there is negative. 12 post wall was the girl. So this tells us that X one equals five x two equal city and the third row doesn&#x27;t tell us anything. So our vector that we wanted to find Waas 53

Rakvi . · Answer

We just want to find a and was in this problem. That is easy. Quartile. 1123 No determinant of is is three times one minus two times one, which is three minus two, which is one. Okay, so to obtain inwards, you just take the matrix obtained from co factors and divide by determinant, which is one in this case. We just take the matrix obtained from the co factors. So you will get 31 minus one and mines. So that&#x27;s your angels.

Rakvi . · Answer

recalculating words in this problem that is, or two by two matrix whose entries are as follows. This is minus for minus one. And so So, as I said, we want to find inwards, Sophie. So let us first get played your German and off you literally. Nando is minus four times to which is minus it minus two times minus one, which is minus two and minus eight. Plus two is minus six. Yeah, and now in worse is just one upon determinant trophy times a matrix which is obtained from co factors off the original matrix. So in particular for two by two matrices You have that you just entertains thes two entries and take negative all thes two other entries, so you&#x27;ll get to minus four and here you will get one and minus. Ah, so this becomes two by minus six. Store divided by minus six is minus one, divided by three and minus one, divided by six minus two divided by minus six is just 1/3 and minus one, divided by minus six. Who just toward so that&#x27;s your in worse. And you can quickly check that if you multiply a everyday in Worse, we will get I didn&#x27;t

Robert Daugherty · Answer

section three nights to Problems 65 computer exploration. They give us a four before matrix a show that this matrix is in vertical and then solve the system X equal b where B is the transpose of the row matrix that you see there. So, first of all, it&#x27;s just find that it is in vertical. So that means if I look at this matrix, which you see here, all I need is its determinant, and that should be non zero. So the determinant is negative. 20. This is in a vertical matrix. So I&#x27;ve shown that the determinant of this matrix A is negative 20. That means this is an in vertical matrix. Now, in order to solve the system they want, we need to be able to write X equal a inverse times This column matric 371 negative forces go back to our calculator. And so I need to figure out. So what is the inverse? So how do I invert this matrix? There&#x27;s the inverse, but I could just right, um, made the inverse of this matrix times and then let&#x27;s just create the column matrix. So I need the for by one column matrix of four rows and one column, and that&#x27;s gonna be filled with 371 negative 437 one negative four. And then that should give me my final solution. So if I do this then there ago and I see the solution that I&#x27;m looking for and let&#x27;s just write that down. So in this case, where you gonna solve this? X is going to be the column matrix off 19 10th negative five when Half and 12 5th So that would be the solution matrix for this particular system.

Problem

In Exercises 33 and $34, T$ is a linear transform…

Problem 33 Hard Difficulty

In Exercises 33 and $34, T$ is a linear transformation from $\mathbb{R}^{2}$ into $\mathbb{R}^{2} .$ Show that $T$ is invertible and find a formula for $T^{-1} .$
$$
T\left(x_{1}, x_{2}\right)=\left(-5 x_{1}+9 x_{2}, 4 x_{1}-7 x_{2}\right)
$$

Answer

$T^{-1}\left(x_{1}, x_{2}\right)=\left(7 x_{1}+9 x_{2}, 4 x_{1}+5 x_{2}\right)$

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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$T^{-1}\left(x_{1}, x_{2}\right)=\left(7 x_{1}+9 x_{2}, 4 x_{1}+5 x_{2}\right)$

Linear Algebra and Its Applications

Grace H.

Catherine R.

Caleb E.

Maria G.

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.

Caleb E.

Maria G.

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

Linear Algebra and Its Applications