in-exercises-23-and-24-mark-each-statement-true-or-false-justify-each-answer-a-if-mathbfx-is-a-nontr

Question

In Exercises 23 and $24,$ mark each statement True or False. Justify each answer.  
a. If $\mathbf{x}$ is a nontrivial solution of $A \mathbf{x}=\mathbf{0},$ then every entry in $\mathbf{x}$ is nonzero.
b. The equation $\mathbf{x}=x_{2} \mathbf{u}+x_{3} \mathbf{v}$ , with $x_{2}$ and $x_{3}$ free (and neither $\mathbf{u}$ nor $\mathbf{v}$ a multiple of the other), describes a plane through the origin.
c. The equation $A \mathbf{x}=\mathbf{b}$ is homogeneous if the zero vector
is a solution. 
d. The effect of adding $\mathbf{p}$ to a vector is to move the vector in a direction parallel to $\mathbf{p}$ .

Runpeng Li · Accepted Answer

Okay, So this problem, we&#x27;re going to check whether the following statements are true or false. So the first statement says, um, if axis a nontrivial solution up off X equals zero than every entry in access none. Zero. So this is clearly false because we can consider a solution. Consider such a solution that it&#x27;s x three one and zero x three. So in this case, ex three&#x27;s a pre bearable. But, um, the specter asked, contains zero, which is the second entry here. So what? That means our our first aim in this wrong. So that&#x27;s we will mark falls here. All right, the second statement, Now we have ext because x two times you us x three times fee. So we need to check that, um, whether this is the plane that that is passing through the origin. So now remember how we describe a plane as how we describe a lying in ah are to space. So that is acts equals you times you plus tee times feet where you and V are vectors and tease is ah, real number. So similarly, we can express playing in our three space. That is you times as a plus B plus the ice tea class W your W ys accounts tend the casting term, and we have two variables here that determines the plane playing in the, uh that determines the plane are three space. But right now, the key thing here is the casting term. W now recall that the constant, constant terms the cats in terming are two space that represents the The point is that the line is passing through, and so in the arteries are three spacing the plane. So if we have the constant that is zero, that means that means that this plane is passing through the already origin. So that means in this case, we have x two times you and x x three times V. But we don&#x27;t have any. We don&#x27;t have any constant terms. So that means the council terms is the reflector. So that that implies this plane is passing through the origin. So we&#x27;re done. All right, this third a statement. So excess the equation. There he acts equal speeds homogeneous. See if the zero vector is a solution. No. Now recall the detonation of homogeneous equation. That means, uh, we have a homogeneous equation. So That means the baby has to be zero because we don&#x27;t allow any consented. None zero constant. On the right hand side, we have a homogeneous equation, so as chatty If this is true, so zero is a solution. So that means we plugging zero. So a times zero vector will be zero. And that is exactly be, because no matter better what the matter was solution. We put in our equation in a basket this council be, but, uh, here we put in zero, so it must be zero or so. So that implies he is a zero constant. And so and so we have a tax equals zero. So this is a homogeneous system, and we&#x27;re done. All right, The fourth statement. So it says, um effect off Adam P two a vector is to move the with the factory in a direction parallel to Pete. No. Oh, before we start 1/4 1 out just right. Right down. True or force here. So 2nd 1 is true? Yeah, the first ones Paul&#x27;s 2nd 131 are all true. So the 4th 1 So right now here we&#x27;re given I say we&#x27;re given a vector. Say, b how Drove it in pictures. Um, I&#x27;m trying to be this way. So we&#x27;re given the vector the and also we are adding p to this vector. Me. So let&#x27;s say P is, uh is this this is Pete. So to perform the vector edition withdrawal such a thing here. And finally we will get this and this is the plus peak. And we can we can see from a picture that we&#x27;re moving be from this point directly to this point. And it is along the direction off p. We&#x27;re moving straightly, right? Poor. My word in this direction the direction of P. And so that means it is It is exactly just like a translation here. So it&#x27;s true.

Bobby Barnes · Answer

right. So they want us to determine if these statements are going to be true or falls. Um, so this first one, it says essentially, that we have the zero vector and H. So this is a subspace of our in and this is going to be false because it doesn&#x27;t say anything about the other two properties that we need to have, which is it Closed under sums and closed under scaler multiplication. So false. Because we do not No. If actually, let me beat the Patriots. Bigger. Do not know if closed under some and scaler multiplication. Okay, so we have a there now for B. They say we have this set of factors, and the set of all linear combinations is a subspace. And so this one is going to be true, because if we just think about, um, what this is so it would be something like see one V one plus c two, b, two plus dot, dot dot all the way up to C p v p. So we can get the, um zero vector by saying all see, I equal to zero for every eye. So that&#x27;s fine. Um, we can get any scaler out of this by saying, see, hi equals zero except for some value. Um, but I is equal to and where it is, like the vector that we want. So that gives us the scale of property and three, some wolf. I mean, some just kind of comes from the definition. So some is, um, true by definition of linear combination linear combo. So that&#x27;s how we end up getting true for B. See, I believe this one is also true, but let&#x27;s go ahead and kind of draw it out. So if we have over here some Mbai in Matrix, remember what the knoll space really is? It&#x27;s saying, Well, what collections of vectors here output the zero vector. And, um, in order for us to multiply these we needed in by one matrix, which means we would have in entries here and then this is an element of our in. So then the null space has to be a subspace of our because it will have the zero vector on. We&#x27;ll also have where if we add two things by the linearity of matrix, that should be fine. And then if we just multiply this, that should also be fine. Um, well, actually, let&#x27;s go ahead and write all those out. So eso let&#x27;s say just so we can kind of be a little bit mawr exact right? So Ah, we have some matrix A with the vectors. Let&#x27;s just call it X Y elements of the knoll of a All right. So if we want to do the some property So let&#x27;s look at a Times X plus y Well, then this is going to be a X plus a y and by definition of both of these being in the null space, this is the zero vector plus zero vector which is zero vector. And so then that implies X Plus y is also an element of nor a Okay, so we have that one and then first scale er&#x27;s. So this is some. So that&#x27;s good. I guess we should have also did the zero vector. But that one&#x27;s kind of true just by definition. So this is zero element of Norway. Alright, so that&#x27;s good so far and then lastly we need these scaler. So we have a times c of X, remember axes so in the north space. But we can go ahead. And do you see? Actually me, Right This over here. So this is going to be C a X like that so we can apply community since he is just some constant. And then we&#x27;ll that c times zero vector, which is equal to zero vector. So then that implies C X is also in the knoll. So then the constant multiple or the scaler also holds. So, um, I don&#x27;t know if you really need to be that exact, but, I mean, it doesn&#x27;t really hurt now for D the column space of Matrix A is the set of solutions. A X is equal to be all right. And so this is actually going to be, uh, false. And the reason why this is false is because it&#x27;s not necessarily just for the solution of one. It&#x27;s supposed to be the span of all of the columns I&#x27;m sorry of. Ah, yeah, the span of all of the columns. Um, so, yeah, that&#x27;s why this one isn&#x27;t necessarily true. Because it&#x27;s not talking about a solution set and then for E. So this is saying that B is the echelon form of a matrix A in the pivot columns of B form a basis for a And so this one, um, is also false Because it&#x27;s not the pivots of B, but the pivots of call a Arab the pivots of a instead. So this should actually say a here on beacon, actually kind of show that off on the side really fast. Eso Let&#x27;s come down here and do that, right? So let&#x27;s say we have some matrix here. And so this is kind of like you can use, like, a counter example as to why this wouldn&#x27;t be the case. Um, so we have, like, 123 uh, then 567 13, 14, 15. And then, I don&#x27;t know, negative 112 So hopefully if I reproduce this year Ah, this gives me 111 000 And if it doesn&#x27;t just kind of tweak this so it does. But I believe this will give us that. Okay, well, if we look at it, this is a and then this is the echelon form. Be over here. Well, if we were to look at this, though, over here, this is saying if we take each of these are fourth position of our vector is always going to be zero. But over here, notice There are numbers here. So we wouldn&#x27;t even be able to get out the, like, original columns here. So because of that, um, we would need to use the columns for a as opposed to the columns of being So Yeah. So that&#x27;s kind of like the rationale behind. Why that? Why that ISS? Yeah. So s it looks like the first statements. False second statement. True, They&#x27;re true. And then the last two are going to be false.

Ashley Boni · Answer

eso part, eh? You don&#x27;t be minus b dot You equal zero. Well, we know that the dot product is commuted s. So that means that you dot b is equal to read about you. And these two things are part B for any scaler. See, uh, the norm of CVI is equal to see times the normal b Well, so we want to see if, uh, we can pull a scaler out of the norm. And if this holds true, So let&#x27;s take c t equal Negative one on our vector Be to just be, say, 11 Let&#x27;s compute uh, these two values So the norm of CV is equal to the norm of negative one negative one which is equal to the square root. Um, negative one squared postnegative one squared, which is a squared. But now, if we take, uh, c times the normal V, we get negative one times the norm of 11 which is negative one times the square root of one square post one squared, which is negative, squared or two. So these things aren&#x27;t the same once he&#x27;s negative. So this one iss who else? This person is true. Uh, next for part C. If X is or thought inal toe every vector in a subspace W then X is in, um, the Ortho Wagnalls base to W. So that&#x27;s just the definition. Um, the space here, This just means that all the vectors that or thought minal to the space w closes just true. By definition, party says, if the normal view squared plus the norm of these squares is equal to the norm of you plus b squared then you movie or are orthogonal. So if we know that this is true, let&#x27;s write out what this means. So the normal view squared is you, don&#x27;t you? Normal v squared is we Dock B now on the right We have you close to me dot You cost me So we have you, don&#x27;t you? Plus B, that me is equal to, um this very hand side we can essentially like foil it out So we have you got you Plus to you don&#x27;t be plus me dot B um, so you don&#x27;t use could cancel be Darby&#x27;s can cancel. So we get zero equals two times you don&#x27;t me. So this means that you don&#x27;t be has to equal zero, which means that you and v our orthogonal. So this is true. And lastly, part e ah, for an m by n matrix A vectors in the null space of a our earth Ogle two factors in the roast base of it. Well, if we look in the section on here, um, three. It states that everything orthogonal to row space of a is equal to mall space of a. So this is just another way of saying exactly what the statement says. Um, so this is true because this is just stating they&#x27;re on three, which is on page 3 55

Doruk Isik · Answer

in this troubled world. Give a few statements. I was asked to decide whether to give a statement is sure falls. So first thing is true. Envy can ease. You&#x27;re in two to show that it is to know statement importante. Well, Sylvia metrics uh, size and my answer. It has enrolls and then calls. So each column in colors, uh, and Rose. Right? So this means that each one you tell them is a subspace in our m so that the given statement is true. Some party you could use a serum. The therapy is it tells us that whole space is the set. Be well, the system execute be or some eggs king are the end. So it means that Dignan statement is oh, very party. Well, Karen, a local man is a suitable likes such light FX is zero and you&#x27;re in a local map actually, is the definition of panels So it means that given statement is true The levian just permission. Um they say value from where to space be. So they saw you For what? This space B and then turn from that into what to Space Deli And the range of D is this settle. Oh, um, Ranger T as the all vectors being doubly so I don&#x27;t be given statement is sure. Now let&#x27;s assume in partner that we have a homogeneous differential in questionable and degree. And we can like this one is a n the air for V s plus a n minus one be in minus one over the X and minus porn. Doctor, Doctor, they warn the f O. R D X plus is zero. So we can say this is equal to a n D. And over the X men, plus a n minus one the n minus form. He had someone&#x27;s one lost dr dot a worm the ODS plus one and this whole operates on. So let&#x27;s say this part. These are radio D. So probably near transformation. We didn&#x27;t know that the new transports transformation disclosed Thio Vector addition. So it&#x27;s a linear transformation expose. So as this be, since it&#x27;s on a freighter acting on F. So it means that we have system working like the F zero and this is the bait. This is basically the definition off. The colonel said. This means that suitable solutions off Virginia&#x27;s linear the French oppression is the colonel off a linear transformation. So it means that the government statement is two

Chris Trentman · Answer

Well, it&#x27;s like a different album. Were given a statement and were asked to mark each statement True or false and justify our answer. V is a vector space. Mm. In part a, the statement is Yeah, our square. It is a two dimensional subspace of are acute dominance of Yes, So they&#x27;re in there. They&#x27;re like, you know, podcast. They got a very dialectic. This is kind of a tricky question, but we know there are two is not a subset of our three because the vectors in R two of two entries and vectors in R three of three entries, however, are too. Is I so more thick to a subset of R three? Therefore, since our two is not even a subset, the statement is false. Then in part B Mm. The statement is the number of variables in the equation. X equals zero equals the dimension of the null space of a Mm. Oh, well, we know that the dimension of the null space of a is actually equal to the number of three variables. Therefore, if the number of free burials variables is not the same as the number of variables which is possible, the statement is false. So the statement is false. So get dick kebab. Pray in part C. Something like statement is a vector. Space is infinite dimensional. If it is spanned by an infinite set, read it. Suck dick Just legal out Grace. Well, in fact, I actually want it&#x27;s true that any and finite character doesn&#x27;t make any sense. Vector space is spanned, and it&#x27;s not like I don&#x27;t I don&#x27;t bad job by an infinite set as well. Green Any non empty finance sector space, I should say. Therefore, to solve your requirement is that an infinite vector space cannot be spanned said by a finite set. And in this way, an infinite vector. Space is different from a finite vector space because finite victor spaces must be spanned by a finite set. Therefore, the statement is once again false. What happens to it finally in part? Mm. Actually, that was part C in part d. The statement is if the dimension of V equals n and s spans V then s is a basis of the Yeah, but you say types that we&#x27;re talking about fucking face. Yeah, this is actually mhm. Not necessarily true. Well, if the dimension of V equals n and s is equal in s spans, V then it follows. That s is a basis for V if and only if the number of vectors and s is also in. Therefore the statement could be false. We could choose a set s which spans V, which has more than or less than n vectors. In fact, it has to have more than n vectors. Finally, in part E were given the statement. The only three dimensional subspace of r three is ours. Three itself. Mhm. All the well, we know that the dimension of our three is three. Yeah, therefore span of our three must contain three. Mm, but nearly independent factors. And therefore he didn&#x27;t finish mhm r three. It&#x27;s happening right now, right? Okay, that&#x27;s true. All right is the only three dimensional subspace of our three. Therefore, the statement is true. Would have

Arden Mccarthy · Answer

for this question were asked to solve the given set of the given linear system using any methods were going to use the Gas Ian elimination method. So that means that we are going to first take our system of equations and put into matrix form and then put that matrix into reduce echelon form and then solve for the variables. So let&#x27;s first make our matrix from the equations that are given to us in the one your system. So the first equation reads zero x one zero x to because there are no excellent or X two terms but sex three close x four was expired, equals zero. Then we have negative X one. Find his ex too plus two x three minus three x four plus x five is equal to zero, then our third equation. Reed&#x27;s Ex Gordon, plus X two minus two x three minus x five is equal to zero and lastly or lost question is to x one plus two x two minus x three. But sex five is equal to Sierra. All right, so from there we are going to roll reduction to get this matrix into reduce special informed. So here What we see is ah Siri&#x27;s of road reductions and we&#x27;re gonna figure out what happened between each of these matrices. We&#x27;re gonna put that on top of the arrows. So if we look at the first Matrix versus Second Matrix, we know that the only thing that&#x27;s different about them is that these two rows are switched. So that means that our role reduction was setting our one equal. Two are four and our current or for evil or our old are for equal to you are one and not give us this matrix here. So now we want to look at the difference between our two and our three. What jumps out is the second row. So those negative ones were eliminated so that this too could be the Onley non zero injury in the rose. That that was the first step in doing so. So in order to do that, we had to take our two and said it equal to 1/2. Our one bus are too. And I gave us our third Matrix. So now we want to see the difference between our Third Matrix and our fourth matrix. We look and we see that again, these ones have become zeros. So in order to do that, what has happened is a change toe. The third rope so are three was set to equal 1/2 or one. We added out to our three. So I just copied, uh, our fourth matrix here to the side. And now we can see that the only difference between these two matrix matrices is the fourth row. So we see that this one was eliminated to be a zero. So there was a change to our four, and it was sent to equal or four was 2/3 are three. All right, it down here so it doesn&#x27;t run into the Matrix. Okay. All right. So now what we want T. C is the difference between our fifth made drugs here and our six majors major here. So we can see from our fifth Matrix that we have our kibitz. But to get them into produced echelon form, we want to make sure that they&#x27;re all equal toe one. We can also see that the second row was also scaled to be able to run tau one. So, in order to get the first entry in the first sort of equal to one. We have said Arm One equal to 1/2 are, too, for the second row we had said are two equal to 2/3 partner. Sorry. So for the 1st 1 for the first room, we said an equal to 1/2 or one for the second row. We had to send people to 2/3 times are too third row. We had said unequal to negative 2/3 turns the third room, and since this was already a one, we didn&#x27;t have to do anything to the fourth row. So that&#x27;s how we got our matrix here, which is now in produced echelon form. So now we want to solve for variables. We have x one x two x three, explore an x five So we want to figure out what our free variables are. First we see that X one is a pivotal column, so we know that X one is not a free variable. X two, however, is not a pivotal problem. So x two is every variable. X three is a pivotal column as his ex four, so neither one of those different variables. X five, however, isn&#x27;t so. It&#x27;s too and I stopped for our free variables. So we want to solve for X one x three the next score in terms of X shoe, the next five. So we&#x27;ll start with solving for export so we can look Thio, the fourth row and see that we just have explore is equal to zero Hippolyta. So we moved to x three and we can use the third row to see that we have x three plus X pi is equal to zero, So x three is equal to negative X five Now for X one, we use the first row we see that x one about sex too minus 1/2 x three plus 1/2 x five is equal to zero And since we wanted in terms of our three variables which is X two, an x five, we want to but in the substitution for extra that we found here, which is in terms of x five. So we get that X one is equal to negative x two since negative 1/2 x three is equal to 1/2 x five. We know that this is minus 1/2 x five and then moving this to the other side of the equation. We have another minus 1/2 x five, so bring not to a cleaner page. Well, rewrite that really quickly. We have X four is equal to zero. Exterior is equal to negative x five and then we have this. X one is equal to negative x two minus 1/2 x five Linus another 1/2 x five We combine these like terms to find that X one is equal to negative X two minus expire and then we can write this altogether x one x three and explore which one right? He&#x27;s in terms of x two. Tom&#x27;s a vector plus x five times a vector so we can see that X one has a coefficient of negative one for x two and negative one for X five. We could see that X three has no value for X two. So there&#x27;s a zero there and negative one is the coefficient for X five. Since export is equal to zero, we know that there is no X to term and there is no x five

Problem

Prove the second part of Theorem $6 :$ Let $\math…

Problem 24 Hard Difficulty

Answer

a. False.
b. True
c. True
d. True
e. False

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

Linear Algebra and Its Applications

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Kayleah T.

Kristen K.

Absolute Value - Example 1

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a. False.b. Truec. Trued. Truee. False

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Kayleah T.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Alayna H.

Kayleah T.

Kristen K.

a. False.
b. True
c. True
d. True
e. False

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

Linear Algebra and Its Applications