in-exercises-39-and-40-a-is-an-n-times-n-matrix-mark-each-statement-true-or-false-justify-each-ans-2

Question

In Exercises 39 and $40, A$ is an $n 	imes n$ matrix. Mark each statement True or False. Justify each answer.
a. The cofactor expansion of det $A$ down a column is equal to the cofactor expansion along a row.
b. The determinant of a triangular matrix is the sum of the entries on the main diagonal.

James Chok · Accepted 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.

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.

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.

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.

Chris Trentman · Answer

were given statements and were asked the label them Sure false and to explain in part, a statement. Is that an n by N matrix that is orthogonal e Dagnall Izabal must be symmetric. This statement is true. This follows from here Um too from this chapter which says that and then by n Matrix A is or Thorgan aly diagonal Izabal if and only if a is a symmetric matrix part B The statement is if a transpose equals a in defectors U and V satisfy a U equals three you and a V equals four d then you dotted with V equals zero. This statement is true noticed that since a transpose equals a we have that is symmetric therefore by theorem one from this chapter it follows that Eigen vectors from distinct wagon spaces are orthogonal to each other. And we have that since a U is equal to three. You and eighties equal for V. We have that you and the are Eigen vectors of a associated with the Eigen vectors for Eigen values three and four, respectively. These are distinct. I&#x27;d in values so by fear and one it follows that you one is orthogonal to or use or Foggin all to be. And so it follows that the doctor product of U and V is going to be zero in part. C Statement is an n by n. Symmetric matrix has end distinct really Eigen values. This is false Indeed. An example in this section included a symmetric matrix a which did not have and distinct rial agon values. This was example three. From this chapter, a true statement would be that the end by in symmetric matrix has and realizing values accounting for Multiplicity. Statement. D says that for a non zero vector V in RN the matrix be transposed. It&#x27;s called a projection matrix. Need to make sure this is so. There&#x27;s actually attend. Um, this should be matrix vv transfers Kidding. And this is false. See why I recall the definition of a projection matrix we have. The projection matrix is a vector. Well, the matrix such that the projection matrix times a vector and are in, is the orthogonal projection of that victor onto the subspace spanned by you. And the reason this isn&#x27;t true is because we need you. Could be Ortho normal Eigen vector. Simply not the that This would be true if the Waas and a member, uh, a set of or so normal I in vectors of the Matrix A Not just any matrix, but a symmetric matrix, but this is not true.

Subham Jyoti Mishra · Answer

in this question. We have a Marcus. This came close and metrics. And we have a set off statements which I read up to Workforce. You will check them one by one. What statement is the rule? Space off is the same as the common space. Off transpose. No, those space off is a total carbon. Yes, off a transport. This is true. Hmm. Programs was off road space off. Is the total number off non zero this panel? They&#x27;re not non zero, though. Viktor&#x27;s. Uh huh. Metrics. But when is transports toe transport that always become columns? So therefore, this is equal and well by work. Government. So the So there is this true second statement? Is he? He&#x27;s in a form of here and his 39 year groups. Because do those people on the beach who a now here, living under be used, uh, take alone normal here. Not really different up here. And and yeah, and that number off non zero of those visible the tree, so it may not be necessary that the Post she rose? Uh huh. Norm zero. Those two roles, maybe, uh, independent. So therefore, we cannot create necessarily say back. Uh, well, studios mama B. C&#x27;s off the circus sports? No. From the whole shipment we have when the dollar statement. We have the diamonds channel. The rose space up here is equal to payments. Um, off column space off here, given he iss not a square metrics. And it is true from the rank here, Um, and also the diamonds and off role space here is equal number. All none. Zero rules in the assist. Ekotto&#x27;s number four columns. So therefore, number all favorite problems. So that days ago, diamonds and off problem space, then in the pork statement we have is some of dimensions of the rules. No space off going toe, remember, Oppose. Mm. Which is diamonds and, uh, no. Is this dimension of Cruz, please? Here has your work diamonds and most is here less diamonds and all column space care. Since dimension of those places, dimensional space here has thank off by rank and which is number off columns is a nautical EMS response with this statement. Mhm miss. On a computer role, prisons can change the train called metrics. You are the true on computer He did can be his grounding. Yeah, So the word is true There in the the rank of metrics can change enroll prisons and computer

Problem

Let $\mathbf{u}=\left[\begin{array}{l}{3} \\ {0}\…

Problem 40 Hard Difficulty

In Exercises 39 and $40, A$ is an $n \times n$ matrix. Mark each statement True or False. Justify each answer.
a. The cofactor expansion of det $A$ down a column is equal to the cofactor expansion along a row.
b. The determinant of a triangular matrix is the sum of the entries on the main diagonal.

Answer

(a.) TRUE
(b.) FALSE

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

Linear Algebra and Its Applications

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Anna Marie V.

Alayna H.

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Absolute Value - Example 1

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(a.) TRUE(b.) FALSE

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Alayna H.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Anna Marie V.

Alayna H.

Maria G.

(a.) TRUE
(b.) FALSE

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

Linear Algebra and Its Applications