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Find the general solutions of the systems whose augmented matrices are given in Exercises $7-14$ .$\left[\begin{array}{rrrrrr}{1} & {-3} & {0} & {-1} & {0} & {-2} \\ {0} & {1} & {0} & {0} & {-4} & {1} \\ {0} & {0} & {0} & {1} & {9} & {4} \\ {0} & {0} & {0} & {0} & {0} & {0}\end{array}\right]$

$x_{1}=3 x_{5}+5$$x_{2}=4 x_{5}+1$$x_{3}=t_{1}$$x_{4}=4-9 x_{5}$$x_{5}=t_{2}$$t_{1}$ and $t_{2}$ are free variables.

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

Campbell University

Baylor University

University of Michigan - Ann Arbor

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in this video, we're going to find the general solution to the system of equations whose augmented matrix A is displayed here. First, we need to rate reduce this into a row reduced echelon form. And so let's look to the pivots. The first pivot is here and we have zeros below, so we're all set on that position. Next, we're going to pivot off this position here, which means we need to make this quantity equal to zero. So our operations going to be too replace row one with the quantity three times. Row two added to row one. When we do that, just imagine multiplying this entire row by three, then adding the result of this row and it will kill this coefficient here. So after taking those operations are matrix takes this new form that we see displayed here. Next we have a pivot. Here was euros above and below after our last row operation. Didn't this pivot position there zeros above and below And the next pivot position is found by taking this going down one row and go right to the next non zero entry. So this is our next pivot position to work with we need to eliminate any quantity that's not zero above and below. So this quantity must be eliminated. The operation will take is replace Row one with itself, plus row three, and that will give us a zero Here will have in this position a negative three and a five here. So let's go to that next row operation after taking that step. This is what our Matrix looks like and from our last pivot position would go down one row. Then there is no more non zero entries, which says we have three pivots altogether. Now we're ready to describe our solution. Set. Let's be first called this x one x two x three next four x five and recall that we were augmenting. And so this was going to be the vector B on the right hand side. So imagine that each row of this matrix is really an equation and we'll have four. Our first equation. That X one minus three x five is equal to five. Then x two. Let me call the X one then x two. We have a variable here and that tells us x two minus four x five is equal to one. Now go to the third row. Next, the third row is going to use a variable x four, and that means we're skipping over X three, since there's no pivot in that column. But that means X three is free. That means it could be any real number whatsoever. So now on the third row, we have the X four plus nine. X five is equal to four and next on the X five row or Excuse Me column, we have no particular very or pivot there, so we know that X five is also free. Next, we need to solve these equations for the basic variables in terms of the free variables. So from what we said here we have that x one is going to be three x five plus five x two is equal to four x five plus one x four is equal to negative nine x five plus four and finally we have the X three and x five are free variables. So this is the solution set to this system of equations

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