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With $T$ defined by $T(\mathbf{x})=A \mathbf{x},$ find a vector $\mathbf{x}$ whose image under $T$ is $\mathbf{b},$ and determine whether $\mathbf{x}$ is unique.$A=\left[\begin{array}{rrr}{1} & {0} & {-2} \\ {-2} & {1} & {6} \\ {3} & {-2} & {-5}\end{array}\right], \mathbf{b}=\left[\begin{array}{r}{-1} \\ {7} \\ {-3}\end{array}\right]$

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Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 8

Introduction to Linear Transformations

Introduction to Matrices

Missouri State University

Campbell University

Baylor University

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In this example, we have a linear transformation. T that maps are three back to our three. It's going to be defined by t of X equals eight times X, where here are matrix. A is defined now for this transformation. Let's also let B B the vector set equal to negative 17 Negative three. Now the questions are for what? X in our three is t of X equal to be so let's determine what the solution X is in this equation now. T of X recall is equal to eight times X, so the question could be rephrased As for what X is a of X equal to be. But as soon as we write our question this form, we can get the augmented matrix and begin row reducing to solve for X. So let's first take the Matrix A. It's one negative to three 01 negative, too, and negative to six negative five then augments with the Vector B, which is given to be negative 17 and negative three. So now that we form the Matrix, the next step is to begin row reduction. So I'll copy Row one and our objective is to obtain the role reduced echelon form. So for the pivot one obtained here, we need to eliminate all entries above and below. So in this case, we need to eliminate to negative two and three. So let's multiply Row one by to add it to row two and we will get 012 and a five. Next, if we multiply row one by three or negative three, rather at two. Row three will obtain zero negative too. Ah, positive one and a two. So now we have a pivot here with zeros below. For this pivot, we have zeroes zero above, but not below. That means on the next step, we're going to be needing to eliminate that value. Negative too. So let me copy Rose one and two. Next. The first row is 10 negative to negative one. And our new row two is your A 1 to 5. Now, to eliminate the negative to multiply row, to buy positive too. And add to row three will obtain zero zero ah, five and looks like there's a tiny mistake here. This should have been a zero. And that means this entry here should be equal to value of 10. So now let's check our pivots at this stage here in here, we have to pivot positions with zeros above and below. And our next pivot position will be this highlighted. Five. We need to make the 501 So for the next row operation divide Row three by five we'll obtain 0012 and let me copy the rest. So we have 0125 and 10 negative to negative one. Now, on the next step, from this pivot position we see here, we need to eliminate the entry to and negative to found above. So this time let's start by copying row three, which is your 012 Then multiply row three by negative to add the results to row two, we'll obtain 01 zero and then a one for the next operation. Multiply Row three by positive to this time and added to row one will attain 100 three. Now, at this stage, we have the identity matrix here augmented with values on the right, and this tells us that X one must be equal to three. X two is one and x three is too So here's our conclusion. Then if we set a vector X to be equal to 312 then the image of X under T is equal to the vector B, and this solves this problem.

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