in-exercises-27-and-28-a-and-b-are-n-times-n-matrices-mark-each-statement-true-or-false-justify-each

Question

In Exercises 27 and $28, A$ and $B$ are $n 	imes n$ matrices. Mark each statement True or False. Justify each answer.
a. A row replacement operation does not affect the determinant of a matrix.
b. The determinant of $A$ is the product of the pivots in any echelon form $U$ of $A,$ multiplied by $(-1)^{r},$ where $r$ is the number of row interchanges made during row reduction from $A$ to $U .$
c. If the columns of $A$ are linearly dependent, then det $A=0$ .
d. $\operatorname{det}(A+B)=\operatorname{det} A+\operatorname{det} B$

Chris Trentman · Accepted Answer

were given statements about to end by an matrices A and B and whereas to march mark these statements as true or false and justify our answer. So the statement in part A is that a row replacement operation does not affect the determinant of the matrix. This is true to see why refer to fear um, three of the book. This is one of the properties listed in part beef. This statement is the determinant of a is the product of the pivots. In any echelon form you of a multiplied by negative ones. They are where r is the number of row interchanges made during row reduction from a to the echelon form you. This is also true. And to understand why the book actually explains this pretty well in the paragraph after example two of this section to summarize what that paragraph says though, when we&#x27;re putting the matrix a into an echelon form you, we&#x27;re really only performing row replacement operations and making row interchanges. And so the only effects would be their own to changes on the determinant which for each wrong to change, we&#x27;ll have to multiply by negative one. So if there&#x27;s are running to changes that results in negative ones. They are times the determinant. Then in part c, our statement is if the columns of a are linearly dependent in the determinant of a is zero. This is also true. Once again, this was actually explained in the paragraph this time following the&#x27;re, um four. But to summarize, we know that if the calls of Ireland nearly dependent mhm, this would mean that one of them is a scaler multiple of all the other problems. And so if that&#x27;s true, then using the permanent properties we can reduce that column to zero and then we have a zero column and our determinant, which means the determinant would be zero and finally in part d. The statement is the determinant. A plus B is equal to the determinant of a plus the determinant to be so claiming that the determinant is additive. This is false. Actually, Now we do know the determinant is multiplication. However, there was actually an example following example five where they give an example of two matrices such that there&#x27;s some. The determinant of their sum is not equal to the sum of their determinants. So if you want. Think of your own example here, or I could give you one. Consider the matrices the identity matrix and also the matrix. 0110 So we&#x27;ll take a equal to 1001 and be equal to 0110 Actually, it doesn&#x27;t work. What on? Actually. Instead, let&#x27;s take a to be the matrix 1101 and we&#x27;ll take B two b the matrix 10 11 Then we have that the determinants of a plus. The determinant to be is equal to Well, this is going to be one plus one or two. On the other hand, the determinant of a plus B Well, this is the determinant of the matrix 21 12 or four minus one, which is three. So we have here an example where the determinant of a plus the determinant of B is not equal to determined to be a plus B

James Chok · Answer

okay. In this question, we&#x27;re just gonna answer with a series of true and forces. So if three roads interchanges unmade in succession than the new determinants equals the old determinants, this is true. The reason why if we have a make it a and we perform one row matrix one road into change, we&#x27;ll stop right in is right in. This is equal to negative determines off this new matrix C. Yeah. Now we&#x27;re gonna make another interchange. Where was I wrote him. We stopped at with row L. This is going to be equal to the negative off. Negative with determinant. Obsolete. Don&#x27;t new matrix where we saw him. So this is determinant. Oh, see? So yes, this original term, it is legal to do it. The determines after you perform three broke into changes. So be the determinants Off a is a pollock off diagonal entries in a This is false reason why it&#x27;s false. I&#x27;ll just show this major to see you 12 Teoh em for so in this matrix the wrote the columns are linearly dependent on each other because this one time time to give you this one so determinants off. This gives users. And you can actually you can actually, through that yourself interment off a easy. This is here, but though the diagonal entries one times four is for so obviously this is false. If determinant off 80 0 then two roads or two columns are the same or a road or Collins is equal to zero. So once again, this is also false. I&#x27;m just clearing example whether determinant is equal to zero but no two roads north Pollan&#x27;s by equal to zero or even the same. So this is false. So determined off eight in verses, Eagle to negative determines off a This is causing false. So that&#x27;s proof that a bee that I&#x27;ve been to the Matrix then a Indus a invest is also 10 the identity matrix. But the determinants off, eh is also equal to determine its off in this. So this is also false.

Amy Jiang · Answer

Okay, so a is false because this is just the absolute value to determine it. He is also false cause determinants of a is equal to the determinant of 80. We notice from my term. Uh, but see, it&#x27;s joke. Murthy is false.

Sam Sohn · Answer

alright. So Part A is claiming that the determinant is the product of the diagnosed entries in the Matrix. So as a counter example, I am going to give you this matrix to one six three. Now the product of the diagonal entries is going to be two times three, which is equal to six, however, but however, the determinant is actually going to be a D minus B c, which in this case is going to be equal to two times three minus one time six, which is equal to zero. And since sex is not equal to zero, that gives part A is false because one counter example is enough to prove that a statement is false. Part B says that an elementary operation on a does not change the determinant. All right, so let&#x27;s look at, for example, the Matrix. Uh, let&#x27;s say 2110 and I&#x27;m going to say 2120 All we&#x27;re doing is doubling the second row right? The determinant for the original matrix is zero minus one, which is equal to negative one, and the determinant for the second matrix is zero minus two, which is equal to negative two and we can clearly see that the determinant did change. So this is also false. Now, Part C. This one is true and I&#x27;m going to link the proof for why that is true in the answer. But a simple way, Thio. A quick summary is we have the determinant of a times. The determinant B is equal to the determinant of a B. And that&#x27;s what we&#x27;re trying to prove. We&#x27;re going to split this into two cases. Case Case one is the determinant of a is equal to zero. This can also be either the deterrent B is equal to zero. Basically, we&#x27;re looking at the case where one of the matrices is not in vertebral one or more, I should say In that case, the product of a B is also going to be non in vertebral. So since a is non in vertebral, that means a B is none in vertebral, which means the determinant of a B is equal to zero. So that&#x27;s the easy part of the proof. And then we look at the quote unquote harder parts where the determinants up a and the determinant of B are non zero, as in We&#x27;re looking at the case where A and B are both in vertebral, in which case a can be expressed as a product of multiple linear operations on matrices. So let&#x27;s call them E one e to all the way up to E n. And these e to the I. Major cities are just linear operations, for example, multiple multiplying the one of the rose or swapping to other rose or adding a multiple of a row to another row. And we&#x27;re gonna look at all of those. So since we&#x27;re looking at a B, that&#x27;s gonna be either one, either the to all the way up Thio Eat the end Be and keep in mind the sense matrices are associative. We can write this as either the one times the product determined by eat it, too, times the product determined by either three times the product determined by and so on until e to the end. Times be time and let&#x27;s close up all the pro disease. Now, let&#x27;s assume either the end or rather either the I is a rose swab operation. Then the determinant of either I is going to be negative one, and when we swap the rows of the Matrix. The determinant is going to change the sign and that&#x27;s gonna be the same thing that happens with the E f and B if if the e ev en is a roast flop operation. So, in other words, the e of i B is gonna be negative of Yep, which is basically the same as the determinant negative to determine. Sorry. The determinants of either the I since that determined it is negative one times the determinant of beat. So we know it works there. Nice case to be is going to be if the either the I is a road product. So basically, that&#x27;s like saying Let&#x27;s say the forth road gets multiplied by five. And on Lee, that road changes. In that case, the heated I I just passed my real product by ah mm Let&#x27;s say then the determinant of either the I is going to be m. And if you multiply the role of a deterrent over matrix by m, the determinant of that matrix gets multiplied by m. So either I be ends up being am times the determinants of B which is equivalent thio the determinate up. Either the I times the determinant a b so nice it works there as well Part to see ISS. Either I is a linear operation of one row being added to another row. So we&#x27;re going to write that as are a to our A plus RB. In this case, the determinant doesn&#x27;t change it all. So the determinants of e f I is going to be equal toe one. And you also notes that if we&#x27;re just adding a multiple of a row into a different row, then the determinant stays the same. So R e f I B is just going to be the determinant of be itself which is the same as that the determinants of either the i times the determinant a b so nice, no matter what type of role operation we&#x27;re doing for either the I that it does not change the determinant in any unexpected way so we can see that the determinant of a B, which is determined by the all the determinant operations done by e one e one e two all the way to e n, which is the determinant of a times that determine its A B. So that does end up being the determinant a times, the determinant a b again. I recognize that this was not necessarily the best explanation. So I&#x27;m going to link the sites that help me with this proof now for party, it says, If landed plus five is a factor of the characteristic polynomial of a than five is an Eigen value of A and that is false. Because if land a plus five time something times, something is equal to zero and that means land of plus five. Being equal to zero is one of the solutions. And as you probably know, that means Lambda has to equal negative five not positive five. Therefore, party is false. Yeah, we&#x27;re done.

James Chok · Answer

Okay, so we want to see if the perfect expansion over determinant down a column equal to cope at the expansion along a road. This is, of course, true, but the reason why we sometimes expand along a row or column is typically because it has more zero. Other than that, it&#x27;s simply just personal preference. If you choose to expand along a roll call now, determinant of a triangle of matrix is a some off the interest on the main. So we first have to understand what triangle matrixes? A upper triangle matrix is the matrix, which has entries once, you know, which has entries really in the heart. So this is a upper triangle majors and 01 a lower triangle matrix. We&#x27;ll have all entries in the body. Yeah, So the determinant off these, however, is not going to be the son. It&#x27;s going to be the pilot to see why, if we could have to expand the first this this type of expanded on along with first right, it will be a 11 times by the determinants off a two to go a and zero on dhe. Someone was you. Yeah, which is equal to a 11 contractor expanding by the first row again was that a 22 by a 33 and whatever eight. And and then this planet continues. And you you just have a 11 times 8 to 2 times a 33 and 18 1 on a in it, so it&#x27;s the color.

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.

Problem

In Exercises 27 and $28, A$ and $B$ are $n \times…

Problem 27 Easy Difficulty

Answer

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

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

Linear Algebra and Its Applications

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Absolute Value - Example 1

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a. Trueb. Truec. Trued. False.

Linear Algebra and Its Applications

Catherine R.

Kayleah T.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Kayleah T.

Maria G.

Michael J.

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

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

Linear Algebra and Its Applications