in-exercises-7-12-describe-all-solutions-of-a-mathbfx0-in-parametric-vector-form-where-a-is-row-equi

Question

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. 
$$
\left[\begin{array}{cccc}{1} & {3} & {-3} & {7} \ {0} & {1} & {-4} & {5}\end{array}ight]
$$

Michael Jacobsen · Accepted Answer

in this example, we have a matrix equation, A X equals zero and we have some work that has already been done on the coefficient matrix. Say it&#x27;s Rory equivalent to the following matrix we see displayed here using this information, provided we&#x27;re going to completely solve this system. Well, that means we also have to write to the Matrix A and Roll reduced echelon form. So let&#x27;s look at our pivots for this pivot column. We have a zero below, and so we&#x27;re all set. But in this pivot column, we have one row operation to make, which is going to be to eliminate that three you see here. So let&#x27;s start off by reading the following The Matrix A, which is Roy Clint to the one displayed, will also be Royko Flint to the following. I&#x27;m going to copy Row 201 negative 45 and if we multiply? Wrote to buy negative three Added to Row one. We obtain one than zero nine and a negative eight. Let&#x27;s say what this tells us so far. Recall we&#x27;re taking a X equals zero vector. We never augmented this matrix here. It&#x27;s the coefficient matrix, but still we can imagine that the zero vector would be there and no one then tells us that x one plus nine times x three minus eight x four equals zero. Row two tells us that x two minus four x three plus five x four Looks like I forgot my variable X here on the x four. And so we have this equation set to zero as well. Now notice that the variables X three and X four are both free variables. I don&#x27;t describe this solution set. First, we&#x27;re going to say that the solution set is a vector X, which is equal to x one x two x three x four and we can do the following next. If we go to the equations that we have displayed here, will be able to solve for X one by subtracting nine x three and adding eight x four. Then do the same for X two for X to weaken solve in this equation for x two. By doing the following at four x three to both sides of that equation and subtract five x four from both sides. Now x three and x four are free and that what that means is that X three is just equaled. X three saw a copy of here. Likewise, X four equals x four. I don&#x27;t copy it here for the sake of organization. Now we want to write this in Parametric Vector form and this turns out to be a technique that is very valuable towards other ideas in linear algebra. So we&#x27;re going to work this out step by step. First we put down the scaler x three and X for these air the unknowns. But they&#x27;re considered parameters now and these parameters multiply. Plus, there particular called vectors. How we find the column vectors is once again a similar theme in linear algebra. We just take coefficients. So start here for the extra variable that coefficients are negative. 941 and zero Because we have it missing down here. Then for the x four variable, go and pull these coefficients, which are eight negative five than a zero and a one. Then finally, this is the Parametric vector form for this equation. We should say one extra thing, however, just to remind herself that x three and X four parameters will say x three x four are in our So as we vary those two parameters over all real numbers, then this is the prayer metric vector form for the complete solution set.

Runpeng Li · Answer

All right. Um so for these problems were given the Matrix one connected to negative nine and five, and the second row is 01 to negative six. Now we need to transform this, transforming its matrix into the reduced rations for So that is, we had had the first rule by twice up to the second roll. So that&#x27;s one neck of two plus two. So that is zero neck neck of nine plus four. That&#x27;s negative. Five and five and five plus negative. Six times, too. So these five minus 12 it is, uh, neck of seven. And the second rule would still be you. 012 negative six. Okay, now we&#x27;re plugging our variables. So that is X one minus five x three. Find a seven next four, which is so and next to plus two x three minus six x four because zero. All right. So observed that there are two for your valuables here because X one is determined by five x three plus seven x four and x two is also determined by X three ANC exports. That&#x27;s a to X three, first six x four. Okay, so we first right down solutions That will be X one substitute by if this expression. So that will be five x three cause seven Next four second x two will be substituted by this expression. So that is to x two x three. Sorry we missed the X three here and plus six x four and we still have X three and export. Now we want to separate this expression. So we first excuse me. So we first have five x three and two x three and x three and zero plus seven. Next four and six x four zero Next four. Now I noticed that it&#x27;s just a separated These separate the inspectors by the sum of two different vectors. So that&#x27;s Ah five x three plus seven. Next four that it&#x27;s a that it&#x27;s the first entry and two x three plus six export at this that it&#x27;s a second entry and extreme plus zero that is 3rd 1 and zero plus exported. That is 4th 1 So the next we take out the x three. So we have the vector 52 one zero Next three for us 76 zoo one and times exports. So this it is exactly the parametric a defective form off the solutions off exit actually cost zero. Now we&#x27;re

Runpeng Li · Answer

Okay, so for this problem work even the matrix 130 elective negative four. And the second row is 260 Negative. Eight. Uh, excuse me. So in this case, noticed that the the first rule is exactly the same as the twice as the 1/2 off the second row. So we can&#x27;t keep the first roll and subtract Subject the first price up The first rule from the second row. So that is 00 00 all zeros on the second row. So this is our reduce station. Now we can take we can take it back to over. A bearable is that it&#x27;s what? X one, us three. Next to minus four. Ex war so x one is determined by negative three x two plus four x four So our solutions will be eggs. We substitute x one by this expression that is connected three x two plus four x four and x two is a pre variable same S x three on next war. Can we need to separate this vector? So would be active. Three x two x two 00 US core X four zero zero x four and we still have next three, so it will be 0010 and time sex three. So we add them together. That will be active. Three x two plus four X War, Mr First Entry and X two plus zero plus zero and sorry, This should be ex, too. I&#x27;m to rewrite this and you were right. It&#x27;s cut off confusing. So his ex too Sex to plus zero plus zero. It&#x27;s the second entry x two and zero plus zero Press x three. That&#x27;s a second That&#x27;s 1/3 entry and zero plus x four zero. So that&#x27;s the port of entry now take out X two So this neck of 3100 times x two plus or 001 times x two x four US 0010 times, x three That&#x27;s our final solution.

Michael Jacobsen · Answer

in this video, we have a matrix equation given to be a X equals zero. So it&#x27;s a homogeneous system of equations. And we&#x27;re also told that a ISRO equivalent to the matrix we see displayed here well, using that information, the idea that we would like to pursue is what is the entire solution set to this system of equations? Well, to do that, we&#x27;re going to have to roll, reduce. Let&#x27;s start back on this matrix A that we see displayed. And I&#x27;m going to make my first role operation to divide Row one by three that obtains one naked of three and two and below. We have negative 13 and negative, too. We noticed. Now that Row two is suspicious. It&#x27;s just row one multiplied by negative one. And when that happens, we can immediately say that this is row equivalent to one negative 32 with a zero down below. Or in other words, if the rose are a non constant multiple of each other, one of them is eliminated. So now that we have a in its role reduced echelon form, we&#x27;re ready to go back and consider the solution set to this system of equations indicated here. We have to be careful, though, because our Matrix say has not been augmented. So when we go to Row One, no one is thick. Waysh in x one minus three x two plus two x three equals zero. Row two has no new information, so we&#x27;ll just record that x two is free and x three is free. These air free variables since the Onley pivot column is column one corresponding to variable x one, making both x two and X three free variables. Now we&#x27;re ready to describe our solution set to this system by a step by step process you can write to. The X is equal to x one x two x three So this is our set up and we&#x27;ll aren&#x27;t going to do the following first go to the equation for X one and solve it for X one. We would be forced to add three x two double sides, so that would go here minus two x three. So now if you read this factor equation, we&#x27;re saying X one equals three x two minus two x three And that&#x27;s no different from this equation displayed here. Now the trick is what to do with the free variables X two and x three. Well, there&#x27;s no restriction on next to since it&#x27;s free, so we can say X two equals x two. But the vector equation we have here we would just write in an X two in this place. Likewise, X three is free means there&#x27;s no restriction so we can write the trivial equation. X three equals x three and that means X three gets placed here in that way. Now, with this set up, we&#x27;re ready to begin describing the Parametric Vector form. Now. One trick in the vector form is the free variables are going to be equal to the scale. Er&#x27;s in this linear combination. Let me start out by writing x two and a plus x three here. Now we just need to determine what the vectors that these scale er&#x27;s X two x three would be multiplying well for the first vector. It&#x27;s associated with the X two variable, and we just pull the coefficients, which are 31 and 00 because there is no value here or if you prefer, you could have called it zero times x two just for the sake of organization. Now for the X three variable, we go to the x three column here and take negative too. Zero and one here X two and x three are parameters or free variables. And although those things are just different ways of saying the X two x three are in our so any real number with no restrictions. So this is the parametric vector form for this system of equations.

Michael Jacobsen · Answer

in this example, we have a homogeneous system of equations written in this matrix form eight times X equals zero, and we&#x27;ve been told that a, which is just the coefficient matrix, it hasn&#x27;t been augmented with the zero Vector. It&#x27;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&#x27;s analyze the pivots to get started in column one we have zeros below, so we&#x27;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&#x27;s start by copying a and I&#x27;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&#x27;re ready for a row operation, but I&#x27;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&#x27;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&#x27;s a lot of copying, but it&#x27;s good practice. So first, let&#x27;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&#x27;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&#x27;re multiplying this by negative three so that&#x27;s positive. 12. Subtract seven and we get a value of five here. Hopefully, we&#x27;re about there, but let&#x27;s look at the pivots. Once again, this is a pivot column with zeros below and calm. Three&#x27;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&#x27;s indicate the free variables as well, So that would be X two, since it&#x27;s a non pivot column x four and x six. So I&#x27;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&#x27;re ready to begin describing the solution. Set first to get Parametric Vector form, Let&#x27;s organize and get their step by step. Let&#x27;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&#x27;ll be a victor in our six containing x one x two x three x four x five and x six. Didn&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;ll be negative. 501 zero, four and one. So this is our Parametric vector form for the solution set

Runpeng Li · Answer

All right, so in this problem, work even a a four by six men Trees, which is 152 neck of 69 zero. Then 001 naked. Seven foreign. Activate There was there a one night 74 negative eight. And the third row is also yours. But the last country, it&#x27;s one. The last rule is all zeros. Okay. All right. So right now when you need to use the culturing divinations, too, to transform this matrix thing. Thio reduce station of form. So right now we can subtract, um, been subtract subject twice off the second row from the first row. So that is 15 zero and so negative six minus negative. Seven times too. So that is positive. 14 14 minus six. That that gives us up eat and nine minus four times too. So that is one and, uh um, minus negative 16. So that would be pumped up 16 and have 001 active seven. And we can also subtract a eight times the third row from the second row, so next aid will be canceled. So daddy&#x27;s one set of 74 zero and 000001 and all zeros in the best fool. And that&#x27;s rude, huh? Excuse me. All right. So now now we can use these corporations to few our equations here. So that is X one plus by next to plus eight explore plus x five 16 x six, which is Darryl and we have X three minus seven x four, US for X 50 and act ex sex. It&#x27;s always there. So we could the barrier of X six back to our first equation. So that gives x one us by ex too plus eight x four plus x five is zero and x three minus seven x four plus four x 50 in the next six. He&#x27;s always d&#x27;oh! Okay, Um, right now I can write down the solutions in the form of a factor, which is X one can be. It&#x27;s expressed in form of X to explore and exploit, So that is effective. Fire X two minus aid X or wireless Thanks fi and X two is a pre variable, and next three can be expressed in form of export and ex fights. That is seven x four minus or an x five and x four it&#x27;s a free bearable. Same S X five and xxx. It&#x27;s always zero. So put zero in the last country and right now we can We&#x27;re trying to separate. These is a vector. So we first right down all the entries that contain X two. So it&#x27;s connected five next to and x two and 000 I do. Then we write down all the all the entries of the contain x four. So that&#x27;s a negative eight x four zero seven Next four and next four and 00 Okay. And, uh, at last, when you to write down all of entries All the entries that contain X five. So he&#x27;s negative X five zero Onda elective for X five and 00 I&#x27;m sorry. Should be x five here, and you&#x27;re okay. All right. Just in case we that&#x27;s double check our answers and our answer gives toe negative five times x two minus negative, minus eight for it explore minus X five. So that&#x27;s exactly the first entry. The next to proceed with a series x two and seven x four minus for X five. That is the third entry and zero plus export was at zero This export on zero plus zero post ex wife. That is the fifth entry and six entry is always zero. All right, now we take out the ex the variables here, so that would be connected. Five one and all zeros time sex too. And active. Eight 01 Authority. There are 7100 time sex for and that&#x27;s one ISS negative. One zero Next, four 01 Do times X fine. Sony, this is our final answer.

Problem

In Exercises $7-12$ , describe all solutions of $…

Problem 7 Medium Difficulty

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.
$$
\left[\begin{array}{cccc}{1} & {3} & {-3} & {7} \\ {0} & {1} & {-4} & {5}\end{array}\right]
$$

Answer

$\mathbf{x}=\left[\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}} \\ {x_{4}}\end{array}\right]=x_{3}\left[\begin{array}{c}{-9} \\ {4} \\ {1} \\ {0}\end{array}\right]+x_{4}\left[\begin{array}{r}{8} \\ {-5} \\ {0} \\ {1}\end{array}\right]$

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

Linear Algebra and Its Applications

Grace H.

Catherine R.

Alayna H.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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$\mathbf{x}=\left[\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}} \\ {x_{4}}\end{array}\right]=x_{3}\left[\begin{array}{c}{-9} \\ {4} \\ {1} \\ {0}\end{array}\right]+x_{4}\left[\begin{array}{r}{8} \\ {-5} \\ {0} \\ {1}\end{array}\right]$

Linear Algebra and Its Applications

Grace H.

Catherine R.

Alayna H.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Catherine R.

Alayna H.

Maria G.

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

Linear Algebra and Its Applications