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Find the general solutions of the systems whose augmented matrices are given in Exercises $7-14$ .$\left[\begin{array}{cccc}{1} & {-2} & {-1} & {3} \\ {3} & {-6} & {-2} & {2}\end{array}\right]$

$x_{1}=2 x_{2}-4$$x_{2}$ is a free variable$x_{3}=-7$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

Jason S.

August 30, 2021

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Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

01:32

In mathematics, the absolu…

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in this question. We want to solve the win your system by using gassy and Jordan elimination, which means that we want to simplify our system of equations and matrix form to reduce Groeschel. Inform. So we start off by writing but the original matrix for a problem, So we have to x one plus two x two plus two x three is equal to zero negative to x one plus five x two plus two x three is equal to one and eight x one plus one Ex two plus four x three is equal to negative one. So the first step that is the most obvious choice to do is to cancel out the negative to in this road to find our leading entry for the left. Most column. So we get we want to do are two is equal to our one plus are too, which gives us the following matrix and then the third row doesn't change. Okay, The next step that we want to do is eliminate this eight so we can have a row of her column of zeros underneath this leading entry. So we do. Our three is equal to four or one minus R three. The simple by that and we get the following matrix and then we can eliminate our last road by doing our three equals R three plus negative are to basically just attracting the two rows. And from now we get this matrix. Okay, so our next step iss thio scale the following rose so we can get a leading entry of one. So are one is equal to 1/2 are born and our two is equal to 1/7 or two. And we're going to continue this on another page and this gives us the following matrix. And then again, the last road doesn't change. Okay, now what we want to do since again remember, we want this matrix in reduced Rochel inform is we want to make sure that the new miracle injuries above you pivot points is equal to zero. So what that means is we want to change this entry to zero. So what we can do is weaken D'oh! Our long is equal to our one, Linus are too. This will give us one minus zero, which is so 11 minus one, which is 01 minus 4/7. Just three Sevens and zero minus 1/7 which is equal to negative 17 and then the rest of the rose doing the same. Okay, and now we're ready to solve the system. The equation. So first step that we want to do is see what our free variables are so x one has a pivot, so that's not a free variable X to also has a pivot. So that's also not a free variable x three. That column does not have a pivot point, so we know that X three is our free variable. So we want to solve for X one and x two in terms of x three. So let's start with X one. So the first row x 10 x two plus 3/7 Next three is equal to negative 17 we solve for X one, x one equals negative 1/7 minus 3/7 x three and there we have our excellent terms. Anthrax, too. We're gonna go ahead and look at the second row, which says x two plus four. Seventh X three is equal to 1/7 saw board in terms of our free variable x three, we get X two is equal to 17 minus 47 Thanks three. And now, if we want that in better form, we have X more Next to is equal to negative 17 and 1/7. Because those are he's terms right here. Right? X one is negative. 1/7 which is what we have on this line. A toe and x two is has a positive 1/7 setups. This spot of one here. And then they both come next three terms we have plus X three and then the coefficient of X three for both of them. For X one, you see that it's negative 3/7. And for X two, it is negative 47

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