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

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

Algebra

Chapter 2

Matrix Algebra

Section 3

Characterizations of Invertible Matrices

Introduction to Matrices

Missouri State University

Baylor University

Lectures

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In Exercises 33 and $34, T…

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in this example, we have a transformation. T that's indicated here. And possibly the most important thing we know about the transformation is that it's linear. So now that we know that it's a linear transformation that motivates us to write the transformation in a better notation, we'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'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'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'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's not the primary thing we'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'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'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's cheating to stop here. We needed to describe this transformation, so it's now calculate a inverse so a inverse will be won over the determinant. But that would be one over negative one. So it'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'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'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'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'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

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