in-exercises-23-and-24-mark-each-statement-true-or-false-justify-each-answer-a-every-matrix-equation

Question

In Exercises 23 and $24,$ mark each statement True or False. Justify each answer.
a. Every matrix equation $A \mathbf{x}=\mathbf{b}$ corresponds to a vector equation with the same solution set.
b. Any linear combination of vectors can always be written in the form $A \mathbf{x}$ for a suitable matrix $A$ and vector $\mathbf{x}$ .
c. The solution set of a linear system whose augmented matrix is $\left[\begin{array}{lll}{\mathbf{a}_{1}} & {\mathbf{a}_{2}} & {\mathbf{a}_{3}} & {\mathbf{b}}\end{array}ight]$ is the same as the solution set of $A \mathbf{x}=\mathbf{b},$ if $A=\left[\begin{array}{lll}{\mathbf{a}_{1}} & {\mathbf{a}_{2}} & {\mathbf{a}_{3}}\end{array}ight]$
d. If the equation $A \mathbf{x}=\mathbf{b}$ is inconsistent, then $\mathbf{b}$ is not in the set spanned by the columns of $A .$
e. If the augmented matrix $[A \quad \mathbf{b}]$ has a pivot position in every row, then the equation $A \mathbf{x}=\mathbf{b}$ is inconsistent.
f. If $A$ is an $m 	imes n$ matrix whose columns do not span $\mathbb{R}^{m}$ , then the equation $A \mathbf{x}=\mathbf{b}$ is inconsistent for some $\mathbf{b}$ in $\mathbb{R}^{m}$ .

Bobby Barnes · Accepted 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.

Runpeng Li · 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.

Doruk Isik · Answer

in this problem. We&#x27;re given a few statements and were asked for work. It&#x27;s that where it is true or false. So the first statement, uh, schools, Because the set of backers not only should be linearly dependent, it should also coincide which age? Hey, you could check the definition off base base. Second statement is true, and I&#x27;m gonna report you refer to the, uh, spending said here. Now, the statement is statement, you see, is also true. And this time I would refer to the section named two years off a basis. Uh, Indy, begin the statement. This falls because this standard method of producing a spend settle now a is actually always Cordish iss a re nearly Bennett sent. And it actually always produces a basis for no hey and less Damon. ISS also pulls because we need Thio look at D or use the different elements or give its Collins off matrix A not the magics B, which is the Rebecca Formal, eh? Because some columns of B mike nights be in the column space. Oh, okay. So we should also we should always use the pivot columns off our orginal matrix

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

Runpeng Li · Answer

problem problem. 18. We start with part. Hey, it says, um if bees any edge inform off A than the people call himself, it be former basis for the column. Space of A. So this Damon actually is false. The reason is that the column space the basis optimum off the com space off A is actually not people Columns A B is the corresponding columns in a or his funding columns in a. So, actually, we&#x27;re taking the columns of A instead of the national home. Okay. Okay. So the next R B, we have Roe operations preserved the win your dependence relations about the rose off, eh? When it isn&#x27;t stroke, all&#x27;s well didn&#x27;t steal false. Why? Well, because we we&#x27;ve performed the, uh We performed the road operations we interchange. You didn&#x27;t inter change in the rose. Then everything will be messed up. Interchanging, rose, mess up things right? Because we are interchangeable. Rosen. We cannot tell which Rosie. It&#x27;s exactly the the linear dependence relations we want we&#x27;re looking for Okay. Hard to see so exas. The dimension of the non space of a is the number of columns of a that are not people. Columns Well, this this statement is true. You guess by our wrecked the room we have mentioned up in all space off a is and minus drank a So if we know the dimension of the no space of a number of the columns off, eh? There are not people columns, but rank of A actually keeps our the number of people columns. So those are not people. Columns is actually the columns that the total comes minus the people call him. So this is a you correctly that not people columns. All right. So hearty, it says means the rose space up. Hey, transposed the same as a common space of a Well, this is also true because that&#x27;s how we perform the transpose operation. Just a bite transpose operation. Okay, Because by the transpose operation, the rose becomes the columns, columns becomes the rose. So the the rose face the game will be the same as a con space. When you do the transpose. Okay, our e So it says if a and B are really equivalent, then they&#x27;re roast bases are the same. But it&#x27;s also true. The reason is that you know the what we have to really prevalent people in the matrices that it&#x27;s actually how we do how we find tree, how we find Rose face so we don&#x27;t

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e. FALSE
f. TRUE

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

Linear Algebra and Its Applications

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Alayna H.

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Michael J.

Absolute Value - Example 1

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a. TRUEb. TRUEc. TRUEd. TRUEe. FALSEf. TRUE

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Maria G.

Michael J.

Absolute Value - Example 1

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

Catherine R.

Alayna H.

Maria G.

Michael J.

a. TRUE
b. TRUE
c. TRUE
d. TRUE
e. FALSE
f. TRUE

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

Linear Algebra and Its Applications