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Determine whether the following statements are true or false giving brief justifications. Any given matrix is over an arbitrary field.(a) Any system of linear equat...

Question

Determine whether the following statements are true or false giving brief justifications. Any given matrix is over an arbitrary field.(a) Any system of linear equations, given by the matrix equation $A x=0$, is consistent.(b) Any system of linear equations, given by the matrix equation $A x=b$ for a fixed but arbitrary column vector $b$, is consistent.(c) Two systems of linear equations described by matrix equations, $A x=b$ and $B x=c$, are equivalent, if $B s=c$ for any solution $s$ of $A x=b$

Determine whether the following statements are true or false giving brief justifications. Any given matrix is over an arbitrary field. (a) Any system of linear equations, given by the matrix equation $A x=0$, is consistent. (b) Any system of linear equations, given by the matrix equation $A x=b$ for a fixed but arbitrary column vector $b$, is consistent.(c) Two systems of linear equations described by matrix equations, $A x=b$ and $B x=c$, are equivalent, if $B s=c$ for any solution $s$ of $A x=b$. (d) Two equivalent systems of linear equations must have the same number of equations. (e) The augmented matrix of a system of linear equations can never be a square matrix. (f) A system of linear equations consisting of a single linear equation involving $n$ variables is always consistent if $n>1$. (g) If $s_{1}$ and $s_{2}$ are solutions of the system of linear equations $A x=0$, so is $s_{1}+s_{2}$. (h) If $s_{1}$ and $s_{2}$ are solutions of the system of linear equations $A x=b$ for some non-zero column vector $b$, then so are $s_{1}+s_{2}$. (i) For a square matrix $A$, any system of linear equations $A x=b$ has a unique solution if $A$ is invertible.



Answers

In Exercises 21 and $22,$ mark each statement True or False. Justify each answer."
a. In some cases, a matrix may be row reduced to more than one matrix in reduced echelon form, using different sequences of row operations.
b. The row reduction algorithm applies only to augmented matrices for a linear system.
c. A basic variable in a linear system is a variable that corresponds to a pivot column in the coefficient matrix.
d. Finding a parametric description of the solution set of a linear system is the same as solving the system.
e. If one row in an echelon form of an augmented matrix is Π0 0 0 5 0 , then the associated linear system is inconsistent.

In this problem were given four statements and asked to determine if they're true or false for the first statement in part A. We have to ask ourselves if elementary cooperations change the solution Sava system And the answer is that no, they don't. Hey, simply allows toe algebraic we solve a system. So the answer to the statements and parts eight is true. Since elementary rob orations do do not change the solution sound system. Next for part B, we have to ask yourselves what it means to be broke for away when two matrices air bro equivalents. What it means is that there exists the sequence of raw operations that transform one matrix into the other. This is not the same thing as having the same number of rows, because that doesn't guarantee that one matrix can be transformed into enough into another. So the answer to the statement in Part B is false. Next report. See, we have to ask yourself one of inconsistent system means, by definition, wanted systems and consistent. It has no solution. So the statement in part seem, which says that any consistent system has one solution is false, and lastly, four parts D. We have to ask yourself to what it means for two. When your systems to be equivalent equivalent systems are simply by definition systems that have the same solution set. And because of that definition, the statements for Part D is true.

For this problem. We have a Siris of five statements and we need to show whether they're true or false, and we will justify each answer. So let's start with the first one. In order from Matrix B two b. The inverse of a both a times B has to equal I and B times A has to equal I Is that true? Well, yes, yes, it iss if both a times B and B times a result in the identity matrix, then that means that B is the inverse of a That is how we're defining the inverse. Um, you can see that specifically if you look at the text on page page 105 you'll see that it's highlighted there for you. And as long as both ways give you the identity, it is the inverse. Now you do need to show both. So it has to work both ways. But this one is true. Okay, Our second one says, if a and B are both square matrices and by ends and in vertebral, then the inverse of a times B equals the inverse of a times the inverse of B. Now this one is false and to see that you could look a the're, um six in your book, dear, um, six tells you what the order should be for this, which is, if a and B are convertible matrices when you take the inverse of the product, it should be the inverse of a times B is the inverse of b times the inverse of a This is what it should be. And as you can see, these are backwards and the one that were given. So the way it is in the in this problem is false. We need to flip thean versus being a on the right hand side. That would make it true. Okay, Next, I have a two by two matrix on the entries. They're going to be a, B, C and D. Now, if a B minus c D does not equal zero, then a is in vertebral in vertebral. Is that true? No, it's false to see that you can look at the're, um, number four, which tells you the correct order. It should be a D minus B. C. We're looking at these diagonals so that the product of aid times D minus B times C. That's what we're comparing 20 If that isn't zero, then a is convertible. In this case, though, we have the variables in the wrong order. So this one is false. Hey, next part D says if a is an in vertebral and by N matrix, then the equation a X equals B is consistent for each be in my, uh, in my domain and that one is true. In order to see that you can look at the're, um, number five from the book and the're, um number five tells us that this equation has a unique solution for each be within, uh, within my real numbers there. So this one is indeed true. Now, our last one, each elementary, each elementary matrix is in vertebral. And yes, this one is also true. In order to see that, you can turn to page 109 in your text and you will see fairly close to the top. It says that each elementary matrix e is in vertebral. And not only that, but the inverse of e is the elementary matrix of the same type that transforms e back into I. So yes, every elementary matrix is indeed convertible. So these are our five statements and which ones are true and which ones are false.

Problem. 11. We want for each a statement to say that if it's true or false and if there is a statement that the bends on another statement, then we will implicate that this answer is related. Toe another statement. Let's start by statement a a statement a say, is that if the equation X equals zero as only the trivial solution, which means that the metrics A is inverted, then Israel equivalent. So the end by an identity metrics mhm his role equivalent to the end by end identity metrics is this is true. Yes, this is true because as long as he is in veritable and by applying the inevitable metrics theater, um, we have a Israel equivalent to the n by n identity metrics For a statement to be, we have the statement that says, If the columns of a spend all and then the columns are linearly independent because mhm spin the columns of a span are the bottom end. This means that a columns are linearly independent and the statement is true for boxee we have. If a is an n by n matrix, then the equation X equals B as at least one solution at least one solution for HIV in order to the war of end. This statement is true only if the metrics A is inverted. This means that the statement see depends on the statement. A. This is true because statement A is true for birdie. If the equation X equals zero has a nontrivial solution. X equals zero as nontrivial solution than a has fewer than in pivot positions. This statement is true because as long as it has a nontrivial solution, which means it's not mhm is not in vertebral and by the convertible metrics, Terram number of bids will be less than and for birdie. If a trance was is not in vertebral, then he is not in third. If we switch it, the variable Toby be as b equals a transit balls and this is the same metrics as it transports. Then be is not in vertebral and by the in virtual metric serum, as long as B is not convertible, be transposed is also not convertible and we know that be transposed equals a transfer Valls all transports which is a and we've said that be transposed is not in vertebral then e is not in veritable, which makes this that went through. This means all the statements or true and for birds. See, it's true because it is true on, and this is the final answer off our problem.

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'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'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's how we end up getting true for B. See, I believe this one is also true, but let'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'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'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's go ahead and write all those out. So eso let'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'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'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's. So this is some. So that's good. I guess we should have also did the zero vector. But that one's kind of true just by definition. So this is zero element of Norway. Alright, so that'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'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't know if you really need to be that exact, but, I mean, it doesn'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's not necessarily just for the solution of one. It's supposed to be the span of all of the columns I'm sorry of. Ah, yeah, the span of all of the columns. Um, so, yeah, that's why this one isn't necessarily true. Because it'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'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's come down here and do that, right? So let'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't be the case. Um, so we have, like, 123 uh, then 567 13, 14, 15. And then, I don't know, negative 112 So hopefully if I reproduce this year Ah, this gives me 111 000 And if it doesn'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'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'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're true. And then the last two are going to be false.


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