in-exercises-1-4-determine-if-the-system-has-a-nontrivial-solution-try-to-use-as-few-row-operation-4

Question

In Exercises $1-4,$ determine if the system has a nontrivial solution. Try to use as few row operations as possible. 
$\begin{aligned}-5 x_{1}+7 x_{2}+9 x_{3} &=0 \ x_{1}-2 x_{2}+6 x_{3} &=0 \end{aligned}$

Runpeng Li · Accepted Answer

All right, so in this problem were given to eke wishes. So that&#x27;s it. They are negative. Five x one seven x two plus nine x three He was sterile and the second equation is x one minus two, x two class six X x three and it&#x27;s one that is zero again. We need to apply the gushing eliminations here. So we&#x27;re set down the, uh, corporations for each question that is Echo 579 and one elected to six. We first interchange interchanged the first rule in the second row. So it will be one elective to six connected 579 And we at the throwing a rock at the second row by five times the first roll. So it will be Excuse me. So that will be one active to six on the first roll and, uh, naked fire plus five times one that is zero and seven plus neck of two times five. That is 10. Minus that it&#x27;s seven minus 10. So it&#x27;s a negative. Three or end. Ah, nine plus five times six. That is 30 30 plus night. So that&#x27;s Ah, 39 and we can simplify the second roll so that is one nectar to six and zero one and 13 connective 30. All right, so right now we have x one connective two, x two plus six Next three, which is zero and x two minus 13 x three We just do so you can see that X two is dependent on X three and X one is dependent on X to an x three. So that means X one is also dependent on X three. That means X three. It&#x27;s a free bearable. And since we have a free bearable here, so that means there are there. There exists a nontrivial solution to this equation, so we&#x27;re done.

Michael Jacobsen · Answer

in this video, we&#x27;re dealing with a homogeneous system of equations. We call it homogeneous because the right hand site contains Onley zeros, and we know for such a system X, which is equal to x one x two x three So equal to all zeros, which is equal to the zero vector is automatically a solution. This is return for two as the trivial solution. So now the question becomes, Are there non trivial solutions? Well, to determine that lets right first the coefficient matrix for this system. I&#x27;ll call it a matrix. A. Just so it has a name and the coefficients start with negative three negative six for X one, five and seven for X two Negative seven and 14 x three. We&#x27;re dealing with the coefficient matrix, since if we&#x27;re looking for non trivial solutions, then that boils down to whether or not there&#x27;s going to be a free variable. But let&#x27;s analyze this set up here for a moment. There&#x27;s two rows and three columns. So what&#x27;s the best case for pivots? While best cases we have a one pivot in calm one or row one and another pivot in the second row and so this tells us that there is Let me convert to a pen. There is, at most to pivots, but if there&#x27;s at most two pivots, that means there is at least one free variable. So now we know that there are free variables. There&#x27;s at least one for this particular system, but as soon as we have a free variable for a consistent system, it means automatically that there are infinitely many solutions. So the only solution that&#x27;s the trivial solution is this one, and that tells us that there are nontrivial solutions and we immediately have a nontrivial solution due to the free variable.

Runpeng Li · Answer

All right. So for these problem working, then the set of equations that x one minus three x two plus seven x three As the first equation and connected to x one plus x two minus four x three That&#x27;s the second equation and the last one this x one us two x two plus nine x three So we need to find whether they really were there, whether there exists the trivial solution. So we need thio need to use the culturing elimination here We set down the corporate himself, uh, set off equations. It&#x27;s one neck of 37 and connected to one next four one to you nine. Okay, All right. By using gushing elimination, we first used a the third row mine Mr First Row. So that will be one next 37 and we change. Interchanging the rope from interchange the third row two and the second roll. So that will be 129 on the second roll and next to one minute it or And we use the second roll minus the first roll. So it will be one negative 37 on one minus 102 minus neck of three is five nine minutes, minus seven. That is, too. And also you can use the third rule at by the first rule, two times the first rule. So he&#x27;s acting too. Plus two times one that is zero and one plus two times negative. Three. That is negative. Five. And next. Four plus two times seven. That is 14. So that&#x27;s 14 minus four. That is 10. And simplify it. It is one elected 37 zero one to fifth on zero one. Negative, too. All right. On, there are something more we need to do here. So here we have on negative three, seven 01 to fit and 01 connected to now. Next, we need to use the thorough minus toe, second row. So that is one minus 37 and 01 two and one minus one is zero negative to minus to to fifth. That is, um, tell Booth connected to pit minus 2 50 So that is, uh, negative. 12. 50. So that&#x27;s next 12. Now, simply by it. So it&#x27;s one naked 3701 to fifth 00 one. Just multiply that. They&#x27;re rode by Bible were 12 by negative by overto so you will get a positive one. All right, So now, now we can, um, used our apply over bearable right now. It&#x27;s ah, x one minus three, x two plus seven. Next three, which is zero and x two class to fifth x three, which is zero and x three is there. So excuse me. So now we know x 30 and plug you to the second equation. We will get X to it zero to and since x two and x three are zeros, so x one must be zero. So the only solution here, x one next to x three are all zero. So in that case, this is a This is trivial solution. So there&#x27;s no none trees turning nontrivial solution for these equations. And so we&#x27;re done.

Michael Jacobsen · Answer

in this example, we have a homogeneous system of equations. Recall this homogeneous because the right hand side is identically, always equal to zero, and we know that we have a trivial solution. X equals x one x two x three Set equal to zero or just simply the zero vector will automatically solve this system. The question, then is. What about a nontrivial solution? That is a solution X that does not contain all zeros well to determine if this has a nontrivial solution, Let&#x27;s consider the coefficient matrix A The Coefficient matrix A has coefficients to native to four for X one variable, then negative five negative 72 for the X two variable and 817 for the extra variable. So we&#x27;re considering the coefficient matrix because if we ro reduce this to echelon form, we can determine if there is a free variable or not, which will help us answer the question about whether or not this system has a nontrivial solution. So let&#x27;s begin roll reduction. We&#x27;ll use this to as a pivot and eliminate the entries negative two and four below. So first I&#x27;ll copy Row one, which is too negative. Five and eight. Now, if we add Row one to Row two will attain zero negative 12 and nine. Then if we multiply row one by negative to add it to row three will obtain zero positive 12 Negative nine. Now notice that Row one and Row two are the same except that they&#x27;re off by a negative sign. This means that we can eliminate that row and right that there is one more matrix to negative 58 zero negative 12 9 and 000 that we needed to produce to get to this echelon form. Now that we&#x27;re in row echelon form, we find that there is x one and x two for basic variables. However calm three, which corresponds two x three, has no pivot whatsoever. This means the X three variable is free. But this means that the homogeneous system eight times X equals zero vector in matrix form has nontrivial solutions. If we found no free variables, we would have said a X equals zero has on Lee the trivial solution

Vishal Parmar · Answer

This is our question Number 42. Here we have it on the augmented metrics for the given system of immigration. So here we want to make zero over here and here right now for the second rope or rope creationist are toe minus by our one, which means we need to multiply row one by name, very five, and add into second group right now, our second rotation, That is our three plus for our one right. Which means we need to multiply. Roll one by four and add in tow. Drove of your false choice as it is your secondary seal Negative three three and your negative nine. You&#x27;re Hardaway cereal Tree Mega Do Tree. And your nine, right? Not everyone toe make here by 11 Right for that. Our probation is too multiply like getting one tarred right. So he or foster and told Rory entities your negative to do what? Dina, Get it. Very good. Use a cold one. Treat you edited by Never do three, which is equal to minus one and negative 90 rated by Negative. Tribute is equal to three. Now we want to make cereal or here for that are wrong impression is on three minus three are too right of your Boston second row are as it gives your tone bullies CEO, but so they can see your turn. Roy&#x27;s serum So we can say that the system is consistent right now. We want to make cereal or here so your corroboration is our one. My industry are too night. So your first of all that is 103 on the right side, we have negative seven were second and told Buoys LTD&#x27;s Mike now right, the system of equations. So your X plus threes that is equal to negative someone right? So here, X Plus three that his acquittal negative seven. Now subject X in this aggression. So here exited with a negative selling, then get it through said Now problem the second row we can right here. Why Minor said is a cold three. No subject. Why in desecrations for your visor called three. Plus that. And here&#x27;s that they called that, which means that is three of any of them, right? No, right. Pigeon resolution. So your ex is maybe 87. Thank you. Do through that. You&#x27;re right. These Tripler said on zzz called. Who said this is our general solution. Thank you

Vishal Parmar · Answer

So this is a question number 4 to 3. Here we have the attendee augmented Medics for the given system Off equation. Right now we want to make zero over here and here. Now, for the second rope or rope patients would be hard to minus two Are one right on for the tundra. All our little patient should be our three. Your tree. So here we have negative three are one. Right? So we can make zero are here and here. Night sore for stroll These allergies right now. Second, Roy&#x27;s 07 Your negative for in your fire right now. Here. Thought joys 07 negative for in your say one, right? No, we want to make zero over here, right? So or Prohibition should we are three minus one or we can write only our toe, Right? No need to multiply one over here. So here are four stories. Aditi&#x27;s second royal. Also as cities, you&#x27;re seven minus selling his Tito negative four plus four year olds with you and say one minus y majestic due to write? No. From the last row, we can I digression like this that these you&#x27;re zero explosive. RWE Plus here Ciro&#x27;s ed, right? You&#x27;re zero x Bless you know why less pseudo said is equal to on the right side. You have to write, which is not possible, but anywhere your X y and then because he or zero explosive wipe a zero that cannot be called to to write. So Weaken said that the system is inconsistent right on. The system has no solution. Right? Thank you.

Problem

In Exercises 5 and $6,$ follow the method of Exam…

Problem 4 Easy Difficulty

In Exercises $1-4,$ determine if the system has a nontrivial solution. Try to use as few row operations as possible.
$\begin{aligned}-5 x_{1}+7 x_{2}+9 x_{3} &=0 \\ x_{1}-2 x_{2}+6 x_{3} &=0 \end{aligned}$

Answer

The variable $x_{3}$ is free; the system has nontrivial solutions.

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

Linear Algebra and Its Applications

Alayna H.

Kayleah T.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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The variable $x_{3}$ is free; the system has nontrivial solutions.

Linear Algebra and Its Applications

Alayna H.

Kayleah T.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Alayna H.

Kayleah T.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications