find-the-general-solutions-of-the-systems-whose-augmented-matrices-are-given-in-exercises-7-14-lef-6

Question

Find the general solutions of the systems whose augmented matrices are given in Exercises $7-14$ .
$\left[\begin{array}{cccccc}{1} & {2} & {-5} & {-6} & {0} & {-5} \ {0} & {1} & {-6} & {-3} & {0} & {2} \ {0} & {0} & {0} & {0} & {1} & {0} \ {0} & {0} & {0} & {0} & {0} & {0}\end{array}ight]$

Arden Mccarthy · Accepted Answer

this question is asking us to determine whether the homogenous system given tow us has a nontrivial solution and they want us to do that without doing any math out. So what&#x27;s important is to know the difference between trivial and non tribunal so trivial is basically a solution with all zeros. So what that means is that given any variables x one x two x tree all the way to extend all of these values when we solve our system of equations are gonna be with a serious well of X one equals zero extreme equals zero all the way. Thio accent equal zero nontrivial solution means that one of these values so either x one next to all the way to extend one or more of these values is not equal to zero. So now that we know the definition, we can take a look at our Russian. So the first thing to know is that the last question. So let&#x27;s look at the last equation that only involves the extra variable. So here we have a number. We can call that number, eh? Times x three is equal to zero. The solving for X three. We get extra is equal to zero divided by a and zero divided by any numbers in vehicle A zero. So that means that X three is equal to zero. We can now do this for X two and solve for X, too. So we have some. We have some number times x two minus some number times x three is equal to zero. But since we already know that X three is equal to zero, we know that this equal zero, which means that in order for this to be true X to house to also equal zero. So now we go ahead and look at X one. So we have some number times x one plus some number Tom text two plus some number times x three is equal to zero. We know that X three is equal to zero. The next to is equal to zero said these both cancel out. You&#x27;re both equal to zero and we&#x27;re left with X one equals zero, which means that X one has to be equals zero. So that means that our solution for X one x you and X three is just equal to the zero vector. So every single very blows equal to zero. And if we check our definition from before of trivial versus non attributable, we can see that all of these variables how swollen extremities three. Our bull are their own zero solution. So that means that the homogeneous system that was given to us on Lee has trivial solutions.

Michael Jacobsen · Answer

in this video, we&#x27;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&#x27;s look to the pivots. The first pivot is here and we have zeros below, so we&#x27;re all set on that position. Next, we&#x27;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&#x27;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&#x27;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&#x27;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&#x27;re ready to describe our solution. Set. Let&#x27;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&#x27;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&#x27;re skipping over X three, since there&#x27;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

Arden Mccarthy · Answer

for this question, you want to use gassy and Jordan elimination to solve built in your system of equation. So what that means is we want to take the Matrix that we&#x27;re going to write for the equations they&#x27;re given and put it in Reduced row echelon formed. So the first step that we&#x27;re going to do is write out The Matrix to solve for the variables given the equations given, So our first equation read zero X one minus two x two plus three x three is equal to one three x one plus six x two minus three x three is equal to negative two and six x one plus six x two plus three x three is equal to five. So our first step and putting this in reduced Rochel inform is we want to switch these two rows so we&#x27;re gonna set our one equal to our three and our three equal to our current are blond and what this is gonna give us is following matrix so we can see that road&#x27;s been switched. The second row stays the same because we haven&#x27;t done anything to it and the first row is become third row So our next step is we want to eliminate this entry. We want to make this entry zero so we could make sure that this pivot value is the only non zero entry in the road. So to do that, we&#x27;re going to set our two equal to our two minus 1/2 or one and that gives us the following future. First row stays the same. Second row becomes 03 negative. About half negative, not half just for plugging and the values into this equation here. And our third room also stays the same because we didn&#x27;t do anything to it. Next thing we want to dio is we want Thio eliminate this entry. So this leading entry of three is the only non zero elements in its column. So to do that, we set our three equal to or three waas. 2/3 are, too. When we do that, we get the following majors First row states to say second row stays the same and this third row become zero 00 and negative too. Again, from plugging in these values for our three and these feelings from or to into this equation that we wrote so our next up is we want to make this entry one. And once we do that, we can start eliminating the entries above it pretty easily, which will help us get it into producer special inform. So in order to make that entry born, we can see that we need to set our three equal to negative 1/2 times are three not gives us this matrix. First row stays the same. Second row, stays the same, and the third row become 0001 Okay, so now to knock out this entry, So to make it a zero, we want to set our two equal to are too. Was 9/2 our three. Because what that will do is that&#x27;ll give us that this entry is equal to negative nine halfs, plus not half times one so not half plus negative. The house is equal to zero. So that gives us this matrix. Okay, so now our next up is we want to eliminate this country. So in order to do that, we do a very similar thing. Toe what we did right here. Well, you want to set? I&#x27;m gonna write and start reading over here. We want to set are one people Thio are one minus five or three. Okay, so we&#x27;re going to do that. And that is gonna give us the following majors. First row stays the same. Sorry. First row doesn&#x27;t say the same. And word we just knocked out that five entry. So that is going to become 6630 second rose days as it was before. And the third row also stays as it was before. Okay, so next up, we want to make this entry a one so we can start eliminating the non zero entries above it. In order to do that, we said our too well to 1/3 are too. And that gives us the fall of matrix. First row stays the same. Second row, become 01 negative, 3/2 0 and the last row stays the same. So now we want to eliminate this entry. So this is the only non zero entry and its road. So in order to do that, we have to set our one equal to our one. Minus six are, too. Once we do that, we get the following major 60 one. They&#x27;re sorry. Six zero, 12 zero Kira. And the second worst is same. And the last roses say so. Now we want to make this entry one. So we want to divide the entire road by six, though, are one he was 16 or one, which gives us the following matrix. Okay, so now our matrix is in reduced. Groeschel inform, and we want to find a solution for it. But what we can notice really quickly is this last room right here. Reeves X one plus x two. But we&#x27;re sorry. Uh, it breathes zero x one, but zero x two was zero x three is equal to one. All of these canceling to zero. Which means that we have an impression that reads zero is equal to one. This is obviously not true. Which means that there is no solution for this system up equations.

Arden Mccarthy · Answer

in this question. We want to solve this assume of linear equations galaxy an elimination, which means that we want to put our matrix in reduced echelon form. So our first step is to create our matrix out of the when your system given to us. So for the 1st 1 the first equation it&#x27;s zero a minus to be plus three C is equal toe one for the second equation. It is three a plus six B minus three C is equal to negative too. And for the last one, it&#x27;s six a plus 16 plus three C is equal to five. And here we have our columns representing the different variables A, B and C. So our first step is to put this matrix in reduced echelon formed. So you first step that we&#x27;re gonna d&#x27;oh is we&#x27;re gonna try to cancel out this value would make it a zero. So how do we do that? We can say that our three is equal to our three minus two are too. And this will give us the following matrix so the first road doesn&#x27;t change. Second row doesn&#x27;t change. And for third room, we have our three so six, minus two or two. So minus two times freeze. Run a six. So six minus six is Europe. And then if we do this, if we plug in the numbers for the same equation right here, we get that third row is equal to zero negative six now and and died. Okay, so now it&#x27;s pretty clear that this entry right here, the three is going to be our leading entry for the first column. So to make it in reduce special inform, we want to make sure that this is the first call up in the Matrix. So we&#x27;re gonna switch are one or two, which is basically just saying our one is equal to the current or two and are too, is going to be equal to the current are one right? So that that&#x27;s just switching. These two rose around. So we do that and we get following matrix 36 Negative. Three. Negative, too. Zero negative. 231 And then the third row stays the same. Okay, so now what we want to do is we want to figure out the leading injury for thes second row or the second column and we want to figure out if it has one. So we&#x27;re gonna want to do something with the entries. Negative too. And negative six. So let&#x27;s try to cancel out the negative six in the third row. So to do that, we see that we need to set our three equal to negative three or two and add it to our three. So what that will do is that will give us the following row of zero, and then negative three are too is equal to six and then plus negative six is all people to zero. So we were able to cancel out that term and clinging in the numbers into, uh, the rest of the numbers into the equation that we just wrote. We find that the remainder of the row is zero and six and us before the other two rows stay the same because we don&#x27;t perform any row operations on them. Okay, so now if we want to finish this by putting it in reduced social inform, we need to make sure that all the leading entries which we see are three negative too. And six are equal to one. So for our one, that means we need to multiply it by 1/3. There are two and he&#x27;s multiply the road by negative 1/2. And for our three, we need to multiply it by 16 So we do that and we get this matrix we get one two date of one negative. 2/3 were just dividing the entries in the first grow by three. And for the second born get 01 negative three halfs negative 1/2. And for this third row, I ran out of room down here, so I&#x27;m just going to rewrite the whole matrix on this page. So and just, uh, for the third row were doing, or three it were dividing the entries in the third row by six. So we know that, uh, this entry is going to be equal to one and the rest of the entries Rami equal to zero. So, keeping that in mind, we&#x27;ll write our final major, except we got, which is 12 negative. Want negative 2/3. This is just coffee ing over. What was on the other page. 01 negative three halfs. Negative 1/2. And then for the third road 000 one. So now that it&#x27;s soon reduced ocean on form. We can take a look here before we even start trying to solve. I mean, if we write this equation out, we see that we have zero x one plus zero x two zero X tree is equal to one. These are all equal to zero because zero times anything is zero means zero is equal to one, which is obviously not true. So that means that there is no solution.

Ryan Conley · Answer

all right, so just all of this problem. You take this mattress e and turned into reduced row echelon form. So reduced for OS will inform is when you have a major C like going we have there, it&#x27;s gonna be a three by four. So that means reduce or echelon form means that you have diagonal wine of ones from the top corner. And then you have everything else in every single row. Lo the warns is zero, and then hopefully we can also get all of these other space is to be zeros as well. And if we have three free variables will be able to do this. But if we have variables that are not free than some of these upper corner zeros will actually be numbers. But that&#x27;s all right. So then what will be left in this last row will be our answers to our systems of equations. So to get this major see up here So first of all, for this system of equations, all you need to understand is that this room is the first variable Will call it X one. This room is the second variable will call it x two and then this last one will call x three and then this last last row is gonna would be is representing what is on the other side of the equal sign. So this top run, for example, is saying Negative three X one minus four x two plus two X three equals zero and that&#x27;s what this top liners representing. So now we need to translate this into reduced, really, Rashwan for which is this sort of form All right, here. So to do that, we&#x27;re gonna be able to do what&#x27;s called Roe operations, which basically means that you can add they&#x27;re subtract anyone row from any other row, for example, wrote to you can subtract for a one from 02 or row one from row three. You can also multiply that row by a constant. So you could, for instance, what we&#x27;re gonna do. This one is we&#x27;re going to so we know we&#x27;re going to start with the first rail. We know what we want to get this nine negative nine in this negative six to be zeros. So we have a negative three. So negative nine negative dine ship. Oh, I actually wrote this down row my apologies. This negative three should actually a possibility. So negative. Nine plus three times three, which I&#x27;m gonna represent by row one. He&#x27;s going to give us a zero there and then right on this one to plus negative six plus two times Rohan, which is a three. Well, give us serum. So now we&#x27;re going to go through and we&#x27;re going apply these rail operations toe every single column in the Metris E. So we&#x27;re not performing any real operations on our top row, so we&#x27;re just That&#x27;s gonna remain the same. So we&#x27;ll go ahead and fill that in three for to zero. Now, this we already decided already. Know that&#x27;s gonna be a zero. We&#x27;ll move on to the next com. So 12 plus three times negative for is 12 minus 12. She&#x27;s gonna be zero and negative. Six plus two times three is also gonna be zero, and that&#x27;s all zeros. That all speech. Then on the final row, you know that&#x27;s gonna be a zero eight plus two times negative for is eight minus eight, which is zero. This one&#x27;s gonna give you negative four plus four, which is zero and is here, so Now we&#x27;re into this for so immediately. You can see that there&#x27;s no way that we&#x27;re gonna get that diagonal row of ones. But that&#x27;s OK. So since sees both went to zero, these two bottom to rose went to zero. That means we know we&#x27;re gonna have to free variables within our system, meaning that one very one variable will have an equation that defines its value. And that equation will be based on the two values of the other two variables, which will be able to be any any value in the book, any number ever. So to get that one equation that will define our system of equations for us, we need to take one more step to get it closer to reduce Romash one for, and that is to get a one in this top form here. So to do that, But it&#x27;s gonna divide this top row by three, and that&#x27;s going to give us a one negative, for there&#x27;s 2/3 and this girl and then the rest of these air also still Syria because we didn&#x27;t do anything to now to get our equation that don&#x27;t define our system for us, you take this top row right here and write it down an equation for him. Like I mentioned before with these three variables Unequal sign something on the other side of the equal sign. So the first row has a one in that row represents X one. So it&#x27;s one times x one, and we have a negative 4/3 negative. 4/3 on that row is for X two, and then our last row is for X three when we got a 2/3 in that one. So to Parents X three and then we&#x27;re gonna have an equal sign and equal zero. So now you have this with equation. This is the equation that governs the relationships between our three variables. And so now you could define anyway wanted. You could solve it for any of the valuables. But since we set this one x one upto have it Oh, have a constant of one on the outside. It&#x27;s gonna be easiest for us to solve for explain. So all we&#x27;re gonna do is just move the x two and X three terms over to the others upside by adding and subtracting, we will get 4/3 next to minus 2/3 brooks you, I urge x so x one will always equal 4/3 times x two minus 2/3 the next three. And since we got these bottom two rows equaling zero, we know that X two and next three. The other two variables are equal, are equal to whatever they want to be. Whatever you decide that can be taken people to anything, and the system will still hold true. And X one was to have be able to have any value depending on what the values R of X two and X three and that means the next to the next three are what we call free variables, meaning that their value it&#x27;s not constrained. That&#x27;s free to be whatever it wants, and we&#x27;re ready only do is use those to free variables to define the last variable, and that is your final answer.

Carson Merrill · Answer

So we&#x27;re gonna be solving an augmented matrix asking me 1339 47 on 76 So we&#x27;re going to reduce this. So we have our one that&#x27;s good for a pit position, but then we want to get rid of that three down below. So we&#x27;re going to replace road to by subtracting. Three are one, which is row one. So get one here. Zero, three zero for this will be a negative five, and then we&#x27;ll get seven. Um, negative 15. Then we have one more pivot position. We&#x27;re going to, um this is our other private position. Negative five. So we&#x27;re going to reduce. We&#x27;re going to scale this by a negative 1/5 are, too. So then we&#x27;ll be left with 13 four, seven negative. 15. And this will be one. Um, and since we scaled the whole row is actually gonna be a positive three. Then we have our zeros, and we want to, um, reduce it by making sure that since we have a pivot here, we don&#x27;t have zeros there or we have a zero. They&#x27;re So that means that we&#x27;re going to replace the first row with by subtracting four times the second room. So now we&#x27;ll get 1003 10 And then this will be three. This will now be a negative five. So what we&#x27;re left with now is that let&#x27;s say it was easy. We&#x27;ll call this Z Y and X. We see that Z is equal to three. And then, um, we see that uhh X plus three y equals a negative five. So from there, we have general solutions. Um, but we know for a fact that Z has to equal three.

Problem

Exercises 15 and 16 use the notation of Example 1…

Problem 14 Easy Difficulty

Find the general solutions of the systems whose augmented matrices are given in Exercises $7-14$ .
$\left[\begin{array}{cccccc}{1} & {2} & {-5} & {-6} & {0} & {-5} \\ {0} & {1} & {-6} & {-3} & {0} & {2} \\ {0} & {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {0} & {0}\end{array}\right]$

Answer

$\left\{\begin{array}{c}{x_{1}=-7 x_{3}-9} \\ {x_{2}=6 x_{3}+3 x_{4}+2} \\ {x_{3} \text { is free }} \\ {x_{4} \text { is free }} \\ {x_{5}=0}\end{array}\right.$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

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Michael J.

Absolute Value - Example 1

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$\left\{\begin{array}{c}{x_{1}=-7 x_{3}-9} \\ {x_{2}=6 x_{3}+3 x_{4}+2} \\ {x_{3} \text { is free }} \\ {x_{4} \text { is free }} \\ {x_{5}=0}\end{array}\right.$

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications