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Find all $\mathbf{x}$ in $\mathbb{R}^{4}$ that are mapped into the zero vector by the transformation $\mathbf{x} \mapsto A \mathbf{x}$ for the given matrix $A .$$A=\left[\begin{array}{rrrr}{1} & {3} & {9} & {2} \\ {1} & {0} & {3} & {-4} \\ {0} & {1} & {2} & {3} \\ {-2} & {3} & {0} & {5}\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

McMaster University

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So we want to find all vectors x such that a X maps to the zero vector. So we want to reproduce. This augmented me tricks of a and zero vector. So first we know we want to get rid of these two. So first and third row are going to remain the same. So let's get rid of the one in the second equation. So I'm gonna take negative one times the first equation and add it to the second. So we get a zero here. Negative three. Ah, negative. Nine plus three is negative. Six, we get another negative six and a zero. Now, I'm gonna take twice the first equation and add it to the fourth to get a zero here. Two times three is six plus three is nine. Two times nine is 18 plus 0 18 Ah, two times two is four plus five is nine. And, uh, anything would do a zero. Here is zero. Ah, So next I want a pivot in the second row. Second call a position. So in this position here. So in order to do that, I'm going to swap the second and third equations. So 139 to 0. Second becomes 0123 Then we have zero minus three, minus six months. Six and zero, nine 18 9 So now we have one where we want it. And we can use this one to get rid of, um, everything below it. So let's do that. So first column remains the same, and so does the second, their first and second row. So now I'm gonna take three times the second equation and add it to the third. So we get a 00 another zero, three and zero. Now, I'm gonna take negative nine times the first equation at it to the third. So we get a zero Where the nine waas Negative. 18 plus 18. A zero. Um, we get a negative. 27 plus nine is negative. 18 and zero. So now, um, we want to go to the next pivot position, which would be right here, but we can't make this a one, so we have to move on to the next s. So we have this negative, Uh, this three here. So let's just take the third equation and divided by three so we can get that our next pit. So everything else remains the same. So dividing everything By three, we still have zeros and our three becomes a one, which is what we want. We want our pivot in the third row and last Rover means the same. So our last step, we need this guy to be a zero. So I'm just gonna take third equation multiplied by 18 and add So all that's gonna do is switch that to a zero. So let's try to squeeze. Listen here if we can. So Last Room's always zeros. So now it's right out on the equations that we have. So we have x one plus three x two plus nine x three plus two x four is equal to zero. Second equation gives us x two plus two x three plus three x four equal zero third equation gives us x four equal zero and the last equation tells us nothing. So we can see that since we didn't have that pivot in the third column, we have x three is ah, free variable, so x Ford's equal to zero. We know that. So this first equation becomes next. One plus three x two plus nine x three Because for getting rid of these terms. Second equation becomes x two plus two x three So we know X three is free so everything's gonna be written in terms of X 30. So the second equation um, we could get an expression for X two, so x two is equal to negative thio x three and let's plug this into the first equations. So we have x one plus three x two plus nine x three equal zero. So now we can get an expression for X one. Uh so first let's simplify x one minus six x three plus nine x three Equal zero negative six plus nine is three so x one is equal to negative three x three So let's see what we have next one next to x three x four is equal to x one is equal to negative three x three x two is negative two x three Ah x three is just x three because that's what we're ready in terms of and x 40 So we can just rewrite this as a vector times, uh, X three to be a little bit more clean. So we have negatives. Very negative, too. One and zero. So this is our final answer

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