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(10 pts) Determinc all the valucs of a for which thc system is consistent_X[ T2 + 3x3 = -1 581 412 +1383 = -3 281 82 + a13 a | 2A. a +-2B. a = -2.C. a # +2. D a = _...

Question

(10 pts) Determinc all the valucs of a for which thc system is consistent_X[ T2 + 3x3 = -1 581 412 +1383 = -3 281 82 + a13 a | 2A. a +-2B. a = -2.C. a # +2. D a = _2[2 (10 pts) Let A =~4 =5 ~1 ~10 12which of the following is a basis of the columnspace of A?10|1012-5 ~10120 =

(10 pts) Determinc all the valucs of a for which thc system is consistent_ X[ T2 + 3x3 = -1 581 412 +1383 = -3 281 82 + a13 a | 2 A. a +-2 B. a = -2. C. a # +2. D a = _2 [2 (10 pts) Let A = ~4 =5 ~1 ~10 12 which of the following is a basis of the column space of A? 10| 10 12 -5 ~10 12 0 =



Answers

Solve each system, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\{\begin{array}{l} 0.02 a=0.02-0.03 b-0.01 c \\ 4 a+6 b+2 c-5=0 \\ a+c=3+2 b \end{array}\right. $$

We need to find the change of basis metrics from B to C. So first we need to equates a fair specter of basis. Be the cheese to negative by zero on. Find the victor that has three components C one c two C three that converts the or the basis see to be so Here we have C one supplied by fair Spektr on one plus c two 20 one plus c three 013 So here your cheeks. Why, I said. And then X is a call to see one by, except the first victor see to my way the first x of the second victor and similarly for the third victor. So we have a system of equations. Three equations see one thus to see to close to two negative C one plus C three, because to negative fine. I see one plus C two plus three seats. Three is close to zero. Here, you can just use one of thes a different color. So you have C three close to C one plus C minus five and then replace C three. Come over here, or we can also use because in elimination and documented metrics to solve it. It's ah, just a Z self aversion to solve. So we'll find. See one close to four C to close to 91 c three because to unit one on the first Pictoris for a 1 81 So this was the first factor you need to, right? The other one. Jeez, the second Muktar reaches 305 and same s. We did see one by one and round one plus C two Choose your one C three video 13 So should not leave you right. This isn't questions for the second victor se see Juan Plus to see to close to three, then the for why he writes native see one plus c three because +20 on lastly, see one plus c two plus three c three these because to fight here we can use C three because to see one So it's very simple. You just replace, uh here and then we have system of questions off two equations so it's easy to solve on you'll find C one c two c three to be one one I want So the second victor would be what is wrong now we need to write the third victor on Equated to the other basis. See? So we have h needed to maybe nine. It goes to I see one bye bye. One the good one plus C two by 20 born plus c three 013 And we write this stuff equations Your I see one plus to see to because to age And if you want. Plus, C three is, of course, a negative, too. See one plus C two plus freeze C three It's close Winning nine and yeah, smart before C three Coster Negative too. Plus you wan on, we can replace it here and we will have this itself. A question to a question So four c one c two C three to recall so they have to by on and maybe four. So the fares victor will be native to five and negative four. So these vectors we have weakened three vectors these to about here each are one Colin off the metrics were looking for. So the final, um, answer which is the change of basis metrics from B to c. R. A. T you b two c is a call to on three columns that we had for native want born one on one and then to by negative for and this is the right now.

This question is a little less computational. Lee Heavy. Ah, See the information that we want from the problem? They just want us to find the dimension of the north space of a given matrix. Okay, well, let's see if our matrix is a negative. 933 negative five. This is a one by two matrix. So it should be out putting five dimensional vector. However, we only have one entry to express leg. The possible solution. That means that five subtract one for therefore, um degrees, which are completely ignored by our matrix. So the null space will definitely be for and the reason why that helped. Because we knew the range waas one. And we know that if we had that I mentioned of the rain and the dimension of the hospice, we should get the total dimension of the cold. I mean,

Yeah. Given the system of equations, we want to solve what we're gonna write this as a matrix equation. First matrix A will be our coefficient matrix. And the first equation coefficients are 1.2 negative 0.3 negative 0.7 In the second equation, their negative .4, 1.30 point four. And then the third equation 1.70 point six and 1.1. We're gonna multiply that by our variables X, y and Z. And that will be equal to our constants. Which we're going to call matrix B negative 0.5 .9 in 1.3. And so when we solve for X, Y Z. That's equal to a inverse B. And so let's use a calculator. Help us here we're gonna call matrix A three x 3 matrix that is our coefficients. And make sure you typed them in carefully storage Matrix B Is a three x 1 matrix that are the constants negative .5 .9 and 1.3. Once they were both story I have to do is eight inverse B and we will round to two decimal places 1.13 point 51.71. Yeah. Which means our solution is approximately 2.13 for X. Y is about .51. NZ is about .71.

All right, So we're solving for this system. I was systems not pretty, right, cause you have decimals. And if I were to use substitution than I would solve for B and I would sell for a here, but dividing by 8.5 and dividing by 1.3 will just give me larger decimals. But I don't want to work with, um the other thing you could do is multiply everything by 10 to get rid of the decimals opinion. Really? I'm working with large numbers and they're just going to get larger, so I'm just gonna keep it as it is. And here we go. So what I did first Waas is I took on the second equation and I multiplied it by 11. Um, what else? I'm going to dio when I multiply this equation by 11 is I'm also going to switch the two terms because I would like to be first and then see, just because bad is the ordered, the alphabet goes so I like to keep it in that order. So I'm gonna do 11 times negative 8.5 b first and I get negative. 93.5 b, then I'm going to take 11 times a positive 1.2 c and I get about 2.2 c and then I'm gonna take 11 times Negative. 24.4 mega negative to 68. Four. Okay, the next thing I'm going to do now is take Equation three and multiply it by four. So you're probably curious as to how did I get these numbers? Well, notice both of these equations have seen. So what I did was on the side. I listed out 1.2. Then I listed out the next one by multiplying it, or by adding adding another 1.2, adding another 1.2. And I kept going until I found a common denominator or common multiplying. Um, I found a common number for C so that they matched up so that when I add or subtract, I can eliminate C and what I found out, that is, if I took 1.2 times 11 I got 13.2, and when I took four times 3.3, I also got 13.2. And so I saw that similarity. And so that's where I got these numbers. So Let's keep going again. I'm gonna take the third equation and multiplied by four. And I'm gonna do the same thing I did with the second equation and slipped flop because A it comes first in the after it. I want it to be the first term. So I'm gonna do four times positive 1.3 a and I get 5.2 a notice. I also didn't line it up with B. I kind of have it off to the left because they're two different variables, so they're not going to mix together. Then I did four times 3.3 and I got 13 point a positive 13.2 c. Now notice my Caesar the same. So now I am able to eliminate that. And I did four times 29 and I got 1 16 Okay, because both of my seas of the same content has have the same coefficient, which is 13.2 that number out in front. I'm going to subtract it. And they're both positive there too, from this attractive. Remember what I like to dio is distribute that negative threw out. This becomes a positive. I'm sorry. Run out of him. This becomes a negative. This becomes a negative. This becomes a negative. I just don't want to forget my subtraction sign with any of my terms. So that's why I do it in the beginning. It's a now we have a negative 5.2 A. No. This, um, if I was smart, will hope on my seas are eliminated A man. These are both negatives. So it's like adding them. But my answer remains negative. If you were so if I was smart, I probably would have flip flopped these so that when I subtracted all my numbers would have been positive, but I didn't. Okay, now, next I'm going to take, um my this equation that I just did, and I'm going to multiply it by five. Okay, so five turns negative. 5.2 IHS naked is 26. Hey, five tens. Negative. 93.5. It's negative. 467.5 and five tens. This large negative. 384.4 comes negative. Okay, No home. That's nice. Is it? Becomes an even number. 1922 naked. If 1922 okay, and then I'm going to take my first equation up here and multiply it by 52. Holy cow, Right, Because I did the same thing. Noticed this equation in this equation. I looked at a f r a would be a little bit easier to deal with, and I did the same thing that I did. That first time here is I took 0.5, which is easy. It's like 1/2 enlisted a mall 0.511 point five, right, as long as I could find something even. And then I took my 5.2 and wrote that out and found out by five I get 26 which matched up to 26 over here. So that's how I found the numbers to be multiplied by five and 52. Okay, 52 times 1/2 of a 26 8 52 times 520.3 is a positive 15.6. Sometimes it's writes well. Sorry to you and 52 times 2.2 1 14.4 See what I was saying earlier? These numbers get really large. Um, and if I had multiplied by 10 than I would have just had 1144 at this point. And 19,220 for those numbers. And I would be subtracting 260 a instead. So it's give or take. Okay, so my aides air the same except different symbols. So I'm gonna add the two so that I eliminate the A's when I add the bees because they're different symbol or different sign ones naked. One positive. It's like subtracting them. And same with my answer here. Okay. And then, believe it or not, when I take this number and divided by both sides, I actually get in even member. So be becomes a simple four. Yeah. You okay, So now, now that I know be, I'm going to go back up to my third equation. My original third equation. Um, remember, I'm gonna put the 1.3 a first. Oh, nope. I can't do that one. I'm going to go up to Mar'ie. Um, my first equations are 0.5 a. Because I think that 1/2 a will be easy to deal with. 0.3 times be which I know is four equals 2.2. I'm gonna multiply the 3.3 and the four 1.2 subtract 1.2 from both sides. I get just simply one over there. Divide by half. I get to Yea. Now I'm going to take what I was telling you. Equation three. But I'm gonna put the 1.3 a first train, and I know a, um I took this equation just because, um I liked how, um positive everything waas right. If I would have taken the second equation. I have a naked of 8.5 b. I just didn't want to deal with that. Okay, So 1.3 times to or if you just double that, subtract it on both sides, and I simply that's 2.9. That's just a 29. So 29 subtract 2.6, 26.4, and then divide by 3.3 humble sides and see also becomes on even eat again. I'm gonna write it out. Is the order triple a be? See? There is my answer to my system


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