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In Exercises $17-20,$ show that $T$ is a linear transformation by finding a matrix that implements the mapping. Note that $x_{1}, x_{2}, \ldots$ are not vectors but are entries in vectors. $$T\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{1}-5 x_{2}+4 x_{3}, x_{2}-6 x_{3}\right)$$

$A=\left[\begin{array}{ccc}{1} & {-5} & {4} \\ {0} & {1} & {-6}\end{array}\right]$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 9

The Matrix of a Linear Transformation

Introduction to Matrices

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University of Michigan - Ann Arbor

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in this example, we have a transformation. T that maps the three dimensional space are three into the two dimensional space are too, and the way the transformation is implemented is it takes x one x two x three and maps it to this for century. And then the second entry is X two, minus six x three. Let's re express what this notation could be. Also written as if we let x be the vector containing x one x two and x three. Then this becomes t of X equals. Let's also change the notation here and write this as a column vector will be x one minus five x two plus four x three for the first entry, and the second entry is next to minus six x three. Now let's see what we can do with this particular vector. Here we have three variables to consider. X one x two x three. Let me say that this is equal to x one times a vector plus x two times a vector plus x three times 1/3 vector. So what we're doing here in this step in green is we're reversing the process of vector additions and scaler multiplication To do that for X one, go to the column here that contains X one and pull coefficients will have a 10 then for X to its negative 51 And for X three coefficients are four and negative six. Now that we've taken this step, we also note that a linear combination in this form can also be written as a matrix times a vector. The vector would be the vector containing x one, x two and X three and the matrix is formed by just pulling each one of these vectors. So it's 10 negative. 51 and four negative six. So if we let a be 10 negative 51 four negative six. The same matrix we found then t of X, where we started with is going to be equal to eight times X where this is our matrix. A. This proves that t is linear since we have a standard matrix A and since we can write t of X equals a X where the Vector X was completely arbitrary

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