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In Exercises $7-12$ , describe all solutions of $A \mathbf{x}=0$ in parametric vector form, where $A$ is row equivalent to the given matrix. $$\left[\begin{array}{rrrrrr}{1} & {-4} & {-2} & {0} & {3} & {-5} \\ {0} & {0} & {1} & {0} & {0} & {-1} \\ {0} & {0} & {0} & {0} & {1} & {-4} \\ {0} & {0} & {0} & {0} & {0} & {0}\end{array}\right]$$

$\mathbf{x}=\left[\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}} \\ {x_{4}} \\ {x_{5}} \\ {x_{6}}\end{array}\right]=x_{2}\left[\begin{array}{c}{-4} \\ {1} \\ {0} \\ {0} \\ {0} \\ {0}\end{array}\right]+x_{4}\left[\begin{array}{c}{0} \\ {0} \\ {0} \\ {1} \\ {0} \\ {0}\end{array}\right]+x_{6}\left[\begin{array}{c}{-5} \\ {0} \\ {1} \\ {0} \\ {4} \\ {1}\end{array}\right]$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 5

Solution Sets of Linear Systems

Introduction to Matrices

Campbell University

McMaster University

University of Michigan - Ann Arbor

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in this example, we have a homogeneous system of equations written in this matrix form eight times X equals zero, and we've been told that a, which is just the coefficient matrix, it hasn't been augmented with the zero Vector. It's real equivalent to this very large four by six matrix. So our goal here is now to determine the solution set to that matrix equation. Well, let's analyze the pivots to get started in column one we have zeros below, so we're in good shape. Then we have to go down and write to the next non zero entry to find her next pivot, which is here, and that leads us to our first problem. We need a zero in this location because our goal is to take a and write it and roll reduced echelon form. So let's start by copying a and I'm going to take in first the fourth row, which is six entries, all zero very large matrix. Then copy Row three, which is 00 00 one negative four. Now we're ready for a row operation, but I'm going to first copy Row three as well, which is 001 00 negative one. Okay, so now visualize multiplying row. To buy a positive, too, and adding it to row one, we'll obtain one negative. 4003 and negative seven. Now with our pivots were all set on this column. This column has been corrected. If we go down one and rights to the next non zero entry, we find that this is the next pivot position. And tragically, there is another problem right here. So we need to eliminate the three. It's a lot of copying, but it's good practice. So first, let's copy down the first or the last three rows saw Go with the last rose. 000000 Row three is all zeros except at 14 in the last two positions. One negative four rather then in Row three, we have 001 00 negative one and we're ready to make our role operation, which will alter row one so visualize multiplying row three by a negative three and adding it to row one. The result of that operation is will have one negative four zero zero. Then a new entry here become zero. We're multiplying this by negative three so that's positive. 12. Subtract seven and we get a value of five here. Hopefully, we're about there, but let's look at the pivots. Once again, this is a pivot column with zeros below and calm. Three's a pivot column. Zeros are above and below, and Colin five, is also a pivot column with zeros above and below. If we try to find the next pivot column, we would have to go down and go to the next non zero entry. But there is none. So that tells us we found all of our pivot columns just for the sake of organization. Let's indicate the free variables as well, So that would be X two, since it's a non pivot column x four and x six. So I'm just going to record that here. So I recall that these are the free variables in the system. So now the next step, which is a good idea to take until your practiced at this is convert this coefficient matrix into actual equations, where the right hand sides are all zero to correspond to. This is your vector here, so no one is going to say that X two. Excuse me. X one minus four x two plus five x six equals zero. Row two tells us that x three minus X six equals zero. Then Row three tells us that X five minus four x six is zero, and we do not obtain any interesting information and row, for it would just simply be the equation. Zero equals zero. So now we're ready to begin describing the solution. Set first to get Parametric Vector form, Let's organize and get their step by step. Let's say what the size of the Vector X has to be. Well, A itself is the coefficient matrix, and it has six different columns. So it'll be a victor in our six containing x one x two x three x four x five and x six. Didn't quite make that large enough. Now lets say what X one through X six must be equal to. So to do that, visualize going back to these equations and first solve for X one, you'd have to add four x two and subtract five x six from both sides to do that. Now go on down the line. But let's also go back to recalling what the free variables are Whenever we have a free variable, we just simply copy it in. So x two is equal to x two and it goes there. X three week unsolved by adding X six doble sides so x six is going to go here for X three, but then X forest free. So I'm going to copy and x four in then for x five. We have an equation for X five. If we solve here, we see the X five is equal to four x six. So that will go here. And then finally, x six is a free variable. So I'm going to just copy x six. I wrote the variables in this column white like fashion just for the sake of organization. And now this takes us to the last step. We have three free variables and so we're copy them in turn x two x four x six Separated by addition. Then we need three constant vectors. So all right, them in like this and are only stipulation on x two x four and x six is that they are numbers in our the set of real numbers. Now to determine the constant vector that's associated with X two. We just go to the X to hear, column and pull coefficients. Those coefficients are for one and then four more zeros. Four more zeros because X two never appeared again. Now go to the x four column or X four vector in this column for X four. We only have X for appearing once, and that was in this position. That means we start with three zeros until we get to the X four or four throw. Pull the coefficient, which is a one and two more zeros, since there is no entries below here. So exports taking care of last. We go to x six and just pull the coefficients for all X six variables. They'll be negative. 501 zero, four and one. So this is our Parametric vector form for the solution set

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