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(6) Consjder the given matrix; A =Choost 5o that YO choice ol Makcs tie MAuixAinver- ible . Explain YOIT AnsWer For the choice of Ahove does the Sstem Ar 6 Mve Miqu...

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(6) Consjder the given matrix; A =Choost 5o that YO choice ol Makcs tie MAuixAinver- ible . Explain YOIT AnsWer For the choice of Ahove does the Sstem Ar 6 Mve Mique soltion lor Aeb coln matrix 6 Explain and justily YOHT ASWOT. For the SAme choice of ADove what can YO# SAv about the soltion t0 tle homogeneons system Ar = 0} Eplain And justily: For the SAme choice of above TOW equivalent to & Explain and justily. For the SAme choice ol above caM Abe witten As prodluct of elementauy matrices&

(6) Consjder the given matrix; A = Choost 5o that YO choice ol Makcs tie MAuixAinver- ible . Explain YOIT AnsWer For the choice of Ahove does the Sstem Ar 6 Mve Mique soltion lor Aeb coln matrix 6 Explain and justily YOHT ASWOT. For the SAme choice of ADove what can YO# SAv about the soltion t0 tle homogeneons system Ar = 0} Eplain And justily: For the SAme choice of above TOW equivalent to & Explain and justily. For the SAme choice ol above caM Abe witten As prodluct of elementauy matrices"? Explain and justily. H Yow Answer is YOS, findl the elementary matrices Al write AAs prodluct ol them_ Chioos s0 that YOur choiceof a makesthe matrix A singulat. Explain YOIU AnSWOE . For the choice ol above; does the Sstem Ar 6 Mve uniqque soltion lor AL 2bV coln matrix 6 Explain and justily YOHT ASWOT. For the SAme choice of ADove what can YO# SAy about the solution t0 iu homogencous system Az = 0' Explain And justily. For the SAme choice of above TOW equivalent to & Explain and justily. For the SAme choice of above caM Abe witten As prodluct of elementauy matrices"? Explain and justily. H Yow Answer is YOS, findl the elementary matrices Al write AAs prodluct ol them_



Answers

Let $$A=\left[\begin{array}{ccc} -2 & 6 & 1 \\ -1 & 0 & -3 \end{array}\right], B=\left[\begin{array}{ccc} 2 & 1 & -1 \\ 0 & 4 & -4 \end{array}\right]$$,$$C=\left[\begin{array}{ll} 1+i & 2+i \\ 3+i & 4+i \\ 5+i & 6+i \end{array}\right], D=\left[\begin{array}{lll} 4 & 0 & 1 \\ 1 & 2 & 5 \\ 3 & 1 & 2 \end{array}\right]$$ $$E=\left[\begin{array}{rrr} 2 & -5 & -2 \\ 1 & 1 & 3 \\ 4 & -2 & -3 \end{array}\right], F=\left[\begin{array}{ccc} 6 & 2-3 i & i \\ 1+i & -2 i & 0 \\ -1 & 5+2 i & 3 \end{array}\right]$$ In these problems, $i$ denotes $\sqrt{-1}$ (a) $5 A$ (b) $-3 B$ (c) $i C$ (d) $2 A-B$ (e) $A+3 C^{T}$ (f) $3 D-2 E$ (g) $D+E+F$ (h) the matrix $G$ such that $2 A+3 B-2 G=5(A+B)$ (i) the matrix $H$ such that $D+2 F+H=4 E$ (j) the matrix $K$ such that $K^{T}+3 A-2 B=0_{2 \times 3}$

In this question. We have American Tory's. I just grew by three mattress and who is the most? Call them off the that has You want you to end you three Parnham's You've been given their rank off here is you got one. It implies when rank is one you want is the pieces the column space here It implies you to learn you three dependent you want so they can be a scandal. Our X And why that's that your political X, you one. You're three years ago. Oh, boy, you want? So in this case they will become human ex You one. Why, you one This can be returned. Us. You want you to your tree. I get tired one. Thanks. Why? And is this you, Victor and his The transport. So we have a vector V, which is one X? Why such that it in middle Tennesse? Yeah. So you have a camera is very nice. You wonder certain place is a you won. The transpose of them is The key is when you want is not a zero. You one is a gold zero And you go where become unique and you feel become, Sir, you to there appears be there different 01 is there And Abel in your school we transfers Then do you want you to go 34 0 and you re will only be present in a very require your tree? Review the prosperous Be very procedural, you know one. So in any of the cases he can be expressed us you. I think they're very good transport so that I existed Vector space be

We have been given that the determinant of three by three matrix A is defined as follows. And then we have been clearly, we can see that in the given question. The determinant has been expanded and open and now what we have announced this, we need to verify the if the method of the diagonal rule identify the diagnosed even and multiplying them. Okay, so we need to verify the method using the method of diagnosis character. Not with general if it matches the general general solution of a determinant. So first we can just clearly write it down and I just write it. So what we have been given is A is even one even to a 13 and then you do one, a two, 8 to 3 and then a 31 832 and a 33 Now on the right hand side of this matrix all the determinant. We can just write the first renders, you can come bigger part and second and the third column, so you want to 8 to 2 and then it really and then even three. Thank you too. And a 32 oh Ernest. We need to actually reread the 1st and 2nd columns on the right hand side of the determinant. So now uh from the figure we can clearly see that uh to get the first diagnosed even it is, it would remember deploying these three elements and then I will be changing the color so that it would be easy then for the second diagnosed. D. Two. So the second diagnosed can be obtained by multiplying these three elements and then the third diagnosed can be octane. Bye! Multiplying these three elements which I am crossing by the blue arrow and then to obtain the fourth diagnosed and just change them ahead. Then to obtain the fourth diagnosed, we need to my to play these three elements. And then for the first diagnosed we need to multiply these three elements. And then finally for obtaining the sixth diagonal G. Six. We need to multiply these three elements. Okay I think I didn't change the color but I just mentioned it. This is 45 this is for the six. We just need to multiply these three elements and then uh we can just find even day to day three and subsequently to the values of the six. We just simply need to add the even PSD 2 23 and then subtract Divan D. Four D. Five and D. Six at the addition of the four different A. Six. And then we can clearly obtain the values as similar has given in the given question which is even even 1822833 plus even 2823831 plus even 3821832 subtracted by a 31822 even three plus a 3 to 8 to three, even one plus 8338 to 1 even to We will get the exact same answer we originally. So this is how we solve the question. Thank you, friends. I hope you like the video.

Mm hmm. Were given a matrix B. We're told that B. Is the matrix. Yeah. Yeah, obtained by normalizing the row of a. Didn't know that really Where a. is from problems 7 33. So you recall that AIDS the matrix with entries 1, 1 -1, 1, 3, 4 And 7 -5 2. So my my child in part A. We're asked to find this matrix B for define being 1st. Let's find the norms of the vectors, the rows of A. So the norm of 11 negative one squared. This is one plus one plus one, which is three, Likewise the norm of the 2nd row, 1, 3, 4 squared is one plus nine plus 16, Which is 26. And finally the norm of the 3rd row, 7 -5, 2 square 49 plus 25 plus four Which is ah 78 christmas. And therefore the matrix B has the entries one over the square roots of three, One over the square root of three and -1 over the square root of three. The second row is uh one over the square root of 26, One over the square root of 26. Sorry, that's not one, that's three over the Square 26. And then four over the square root of 26. And finally in the 3rd row, this is seven over the square root of 78, -5 over the square roots of 78. And to over the square root of 78. So this is our Matrix B. See oh yeah, they got this bitch. Then in part the sounds like she tried the charger santa galaxy, we were asked to be as an orthogonal matrix. Well, recall from the previous exercises the rows of a war orthogonal. So it follows that the rows of B are still orthogonal. Mm Hi, since you just multiply them by scholars and are now units vectors. Therefore it follows that B is now an orthogonal matrix. Mhm. Finally, in part C were asked if the columns of B are orthogonal. In fact, we didn't check each of the dot products. We know that the rows of B form an or so normal set. The headline. Alright, here we go. Of vectors. Because B is in fact orthogonal and therefore by a theorem from this chapter, in particular dirham 7 11. It follows that the columns and she of B also form an Ortho normal set. I'm not tired. I can help you move girlfriends like no, seriously, I can help you The end. Therefore the columns of B are in fact orthogonal. Well, that's a bit. I just think you could do and I don't that would kind of

In this video. We're gonna go through the answer to that question 39 from chapter 9.5. Um, so yeah, but given the Matrix A which is written here, we asked Oh! Ah, derive. Uh, yeah, There is an ancestor questions A, B, C and D. I think this is quite hard questions, so I'll try and take us through it slowly. Okay? So question apart a were asked to find that I can values I come back to us basically on dso first. Find ion values. Gotta find the determinant off the matrix A minus I So basically mind seeing are from the leading diagonal. Okay, So what's that gonna be? We can evaluate it around the top itself forever. That's two minus r times by the matrix of at the seven of the Matrix to minus one minus two minus one minus. One times the matrix. December of the Matrix. No, there should be a minus. Other close. One times the matrix. 12 months, huh? Minus two minus two. Okay. Uh, I think I'll avoid boring you with the algebra for this one. Um, you guys can work that through and show that is equal Thio. Well, you can expand it all out and show that it's minus, uh, cubed. Plus three squared minus three. Ah, close one which could be fact arise to be one minus R cubed Get. So that's equal to zero than that shows the the value of ah, which are Ah, I convey values. Eyes equal to one on It's a triple. Really? Because about three here. So it's an egg value with multiplicity three. Okay, fine. Victor, that's fine. I'm vectors. We find a minus one times. I because I value this one times by you equals zero. So what is a minus? I was going to the back there. The matrix one 11 one on one minus two, minus two minus two. It comes by u equals zero. Okay, so if we let the components off, you be X y zed, then each of these rows in the matrix equation tells us the experts Why? Course said is equal to zero. Okay, so then what could solutions look like to this equation? Well, we could have that. The first term at the Ex Capone is zero. Then that would mean that Ah, the wines that components with a sign of each other. We could have that The y component is zero. In which case it'd be minus one one. Or we could have a second potent be zero. Gonna be zero minus one, but one. Okay, but now look at this guy. This equation this factor could be written in terms off this factor on this vector. So you can see that if you, uh let's see if you do this vector minus this factor that will give you this vector. So therefore, this factor is not linearly independent to these guys. So, um, where these guys are linearly independent to each other. So therefore, any item vector must pay off the form. That's, um, constant s times by the first of the linearly independent vectors. First, some constant v times a second of our linearly independent backers. Okay, so if so, have chosen this kind of arbitrarily because we could equally right. We could have equally written this guy as a linear, some off this thing guy and this guy, in which case we would change the inspector's on DDE. That would also be correct. So we've kind of written this kind of arbitrarily the point is that we kind of have only two degrees of freedom on our choice. If I came back that way, can't write the Eiken vector as a some off Constance Well supplied by all three of these. I mean, we could, but it would kind of be, ah, were necessary because we already showed that, uh, to sufficient. Just two of the of the expression, too. Give you all of the possible high in vectors to the Matrix. A. Okay, so hopefully that's Colbert. Why? We just have two vectors here. Okay, So, uh, B, this is just following standard rules for linear started different differential equations from those two aiken vectors. We can write too linearly independent solutions on They're gonna be e to the First Aiken, How you bought the only item value times T c to the t times by the first heart of the Eigen vectors on then. Secondly, Secondly, nearly defendant solution that's gonna be eats the tee times by the second darken vector. See? Okay, let's use the form that they've given us, which is x three, because t into the tea. You three plus a to the t you for. Okay, So what's the derivative of this Well, we can take out you take The city is a common factor. Used the product rule Gonna be one plus t you three plus you for times e to the t. We know that this is gonna be equal thio because what we've assumed that it's that this form is a solution to, uh, the matrix differential equation so we could have tea. That's a you three for us. Hey, you four times by E t t. Okay, so this guy and this guy are equal to each other so we can cancel each the teas and then compare coefficients off. Tease will tell us that a U three is equal to you three itself on and you have a cross compare coefficients here. This tells us that a minus by you three is equal to zero on dde a minus. I you four secret to you three. Okay, so this tells us that you three is an icon vector. That means that it needs to be of the form. Found a pot A which is s times minus one 10 close a times minus ones. They're well you asked me about for so Okay, let's try and figure out what form we're gonna take. Okay? So to figure out what you four is gonna be a we're gonna need Thio. Find out what this matrix is. A minus. Eyes the matrix one 11111 minus two minus two minus two. What time did not buy you for? And then this is gonna be equal to you three. Which, as of yet, we're unsure as to what I infected to choose. So what's gonna help us? Well, if you four has the components X y and said Just try to see you. What's that? You for its components, X. Why? And said then the top road off this equation, it's gonna tell us the expose, what was said is equal to whatever this first component is a man. The second I was gonna tell us the extras Wipers said because he was the second component third component in the third row is gonna tell us that minus two times by extra swipe, all set is equal to Okay, So then now, in order to make this house a any solutions, all we need Well, we need this component on this components were the same because otherwise would have experts watches that is equal to two things which is just mathematically consistent. So that means that, um yeah, I need those guys to be the same. So let's say that we let them be equal to one on one. Let's see whether weaken do that. Uh, yeah, definitely. Can. So that some if s is equal to you. Uh, let's see, that's equal to one. And B is equal to minus two. Then that will work. And then how does that work? That tells us that this 3rd 1 is gonna be minus two, which shows that that's gonna be minus two 11 works because now, diesel, these equations here or class into one equation which settles wipeout set is equal to walk. Forget it. So I've chosen you three to be 11 minus two. So therefore X, with y plus said where we need to choose. Yeah, a basically any backdoor that it satisfies this so we can just easily choose a vector 100 because that satisfies the equation. X plus y equals said Okay, so just finishing that off on these x three is equal to t It's the tea. 11 minus two. I saw x three. Plus it's the tea 100 we saw explore. Okay, then pot de a minus. I squared times you for equals. I was gonna be a minus. I times a minus I you for which is a that which is you three. And then we'd know from hot See, that is equal to zero. Okay, nothing. She's answer a question with the night.


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