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Let $T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ be a linear transformation that maps $\mathbf{u}=\left[\begin{array}{l}{5} \\ {2}\end{array}\right]$ into $\left[\begin{array}{l}{2} \\ {1}\end{array}\right]$ and maps $\mathbf{v}=\left[\begin{array}{r}{1} \\ {3}\end{array}\right]$ into $\left[\begin{array}{r}{-1} \\ {3}\end{array}\right] .$ Use the fact that $T$ is linear to find the images under $T$ of $3 \mathbf{u}, 2 \mathbf{v},$ and $3 \mathbf{u}+2 \mathbf{v} .$

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Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 8

Introduction to Linear Transformations

Introduction to Matrices

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for this example were given a transformation t but were not given very much information about it. What we do know is that the vector you maps 2 to 1 and the vector V maps to negative 13 So given this information, this is what we want to compute. Let's first find the image of tea of three. You under T. Well, this is difficult because we do not have three you to tell us where it will map to. However, we do know that this transformation t is linear. So what that allows us to do is write this as tee times three times t of you. Then, when we're told team maps you into 21 this becomes three times the vector to one, resulting in 63 Now what's interesting about this particular example is they did not actually need to tell us what five to is. Nor do we really need to know what 13 is. So keep that in mind that that information specifying U and V is helpful in a sense, but mostly a trick in the question. Next, let's evaluate the image of two times V so t of two times V by the same strategy, using linearity of t. This is equal to tee times. Be so this will be well. First, we know that V is mapped into negative 13 by the transformation T. So we'll make that substitution here and have two times negative 13 resulting in altogether negative to six. So now if we can do t three u and T of two V, let's determine next the image of three u plus two v under this transformation. In other words, let's calculate tea at three. You plus two V. Really, This is a problem about getting used to linearity. And let's break this down step by step, then first, when we have scaler or vector addition on the inside that losses to write TF three you plus T at two V, then the vector the scaler Multiplication here, since Tia's Linear allows us to write three times to give you plus two times TSV However, we have just calculated the images of three you and to V to obtain 63 and negative to six. So by substitution this becomes 63 plus the vector negative to six. So the image of three u plus two v under this transformation T is altogether +49

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