a-steam-plant-burns-two-types-of-coal-anthracite-a-and-bituminous-b-for-each-ton-of-a-burned-the-pla

Question

A steam plant burns two types of coal: anthracite (A) and bituminous (B). For each ton of A burned, the plant produces 27.6 million Btu of heat, 3100 grams (g) of sulfur dioxide, and 250 g of particulate matter (solid-particle pollutants). For each ton of B burned, the plant produces 30.2 million Btu, 6400 g of sulfur dioxide, and 360 g of particulate matter.
a. How much heat does the steam plant produce when it burns $x_{1}$ tons of $\mathrm{A}$ and $x_{2}$ tons of $\mathrm{B}$ ?
b. Suppose the output of the steam plant is described by a vector that lists the amounts of heat, sulfur dioxide, and particulate matter. Express this output as a linear combination of two vectors, assuming that the plant burns $x_{1}$ tons of $\mathrm{A}$ and $x_{2}$ tons of $\mathrm{B}$ .
c. $[\mathbf{M}]$ Over a certain time period, the steam plant produced 162 million Btu of heat, $23,610$ g of sulfur dioxide, and 1623 $\mathrm{g}$ of particulate matter. Determine how many tons of each type of coal the steam plant must have burned. Include a vector equation as part of your solution.

Priyanka Sadarangani · Accepted Answer

in discussion that, given is there is a steam plant they advanced to type of poll is interested and B is luminous for each ton off a burn, the plant produced 7 27.6 million BTU off heat, okay, and 3100 g off sulfur dioxide and 2 50 g off particular matter way can say that he is the particular matter and the same in Batum Ina&#x27;s. It will give on the values, uh, 30.2 million video off heat and 6400 g off sulfur dioxide and 3 60 g off particular matter P or politan. Okay, so the question is, how much heat does the steam plant produce when it wants Excellent Turn off e plus export Enough. Be so simple. This will be excellent on off a an expert on or be so excellent and is Onley heat. The question given is only here, So this will be 27.6 Class X two on the heat from the bottom Ina&#x27;s Is this 30 point to and that will be the final product million btu off heat. So this will be the answer off. Caution A. Okay, now the ocean be the same as question A. In the question be it is asked about all the heat, the sulfur dioxide and the bulletin matter. Okay. And that. So I will read the question. Be first, uh, in the output. Supposed the heat cell photo oxidant particular. Enter particular matter. All three are present. Then we have to write this output as a linear combination of factor. Okay, so this will be simple X one A plus. XTO be and excellent. It will be 27.6, heat 3100 sulfur dioxide and 2 50 particular matter plus x two and B is 13.2. Video off million video off heat 6400 s auto on to 60 off particular matter. And that will be questo. Total amount produced in the steam plant. Okay, so this will be over. Answer of question be. And the cautions see, is the next part off portion be in the caution? See, the given is the given Is the total production off this equation? So this equation is 27.6 30. 100 2. 50 plus x two, 30.2, 6400 to 60 and total amount that is produced is given as 1 62 million VT! Off heat 23.610 is 23 1000. 6 months. Zero is sulfur and 16 to 3 is particular matters. So we have to find X one and x two hair. So for this, we will put in these in the metrics. Okay? And this will be like this. 27.6, 3100 to 15. Okay. And 30.2, 6400 to 60. And the result will be 1 62 to 3610 on 16 to 3. So when we reduce this metrics until we find the values of X one next to we will get in the final one. Zito 3.9 and 11.8 on 000 So from Disick wish thesis metrics, we can say the value off X one is hair and extra is hair. So excellent equals two 3.9 and x two equals two, 1.8. It means excellent equals two equals 23.9 ton off turn off entry side A and X two equals two, 1.8 tons off. Bit ominous be. This will be our final answer. Thank you

Salamat Ali · Answer

what ails the problem. The efficiency for each engine is given by efficiencies equal to 0.6 five times. They cannot engine efficiency, which is represented by easy, so the efficiency for a first engine is equal to 0.65 times the efficiency for a first or not that is equal to 0.65 In terms off, the temperature is given by one minus. T l one low temperature one and divide by th item richer, one that is equal to zero point 185 playing the value for a temperature. Similarly, for efficiency into we will do the same procedure. And also we get Geary&#x27;s 0.137 Um, the first engine import sheet from the call will be work is a call to efficiency. One. Uh, Cube You. Which one? Which one? That is equal. Thio efficiency off, injured one times the king coming from a call word hate coming from corporate, and then a w one he&#x27;s read Today&#x27;s using Go Well, first law of thermodynamics. That is a kun uh, which one, uh, hope minus w one that gives us, um one minus efficiency. One times Q. Off the cool. So for a call, I would like to see you then. W to he&#x27;s written. Is e too Kyul bitch too? Which do? That is a cool thio. He too one minus lied to times Cules called. They, uh Excuse me. Ending everyone in w do I will do here. Excuse me, but then eating both worked. Everyone under two w two. We hope that expression which is he one plus e to minus e. One times he to ah, whole much applies We together the heat from the court. So this is Q Z, then the rate off the burning off the cool Uh, this is C c. Great of the burning off. Cool. It can be found by dividing my tea. Then doubly one is also one plus. The Bluetooth is derided by, um uh t t are divided by the efficiencies. The one plus two minus the one turns to two. Then the rape Awful Colburn. We can find it by fighting the values which are given in the problem. Which then the range off Colburn we get here is three point in 20 times. 10 to the power. Ah, nine Jule Jule Per second we can contort into K g by, uh, deriding it by 2.8. So you wanted you. We have 2.8 times. Don&#x27;t power seven jewels by dividing it dancer by the This number gives us 114.4 kg for seconds. So the rate of coal burn real Edie&#x27;s 114.4 kg per second in part B. We are us to find ah, the rate off the water which we can find it by the heat from exhaust. So right to eat eyes equal to the heat from cold minus work. Then wake tonight Heat from exhaust, uh will be used. So this is he D&#x27;oh, Heat the water off, which is a mass off water. Time to specific eat over water times delta t A temperature increase of water. Then I console for, um Moss here and right here The equation in terms off the heat from a coal a minus work. Yeah, and then deriding it by t Here I get the rate off the from us. And here I have Ah, si, especially Kate Off water times Delta t don&#x27;t you then putting the values for All right, inside. Then we get here. 9.78 uh, two dives. 10 to the power. Four kg per second. Then, when working this amount of water into Galan&#x27;s, we get here of 9.3 times 10 to the power seven gallons G A four gallons. But so water is used to cool the power plant. So, uh, we need this amount of water that are thio cool the power plant. End off the problem. Thank you for watching.

Averell Hause · Answer

So we need to find the rate at which call must be burned and then how much the rates at which mass rate of, Ah water in order to cool. So we need we rather we have the efficiency that&#x27;s going to be sixty five percent of the efficiency of the Carnot engine. We have the temperature of the higher region of the first system. This is going to be seven hundred fifty degree Celsius or we can say one thousand twenty three Calvin. We simply need to add to seventy three to the degrees in Celsius. In order to convert to Calvin T of al, one would be four hundred and forty degree Celsius or again, seven hundred thirteen Calvin th sub two is going to be equal to four hundred fifteen degrees Celsius. This is equal to six hundred eighty eight Calvin, and then we&#x27;LL have tea of else of Elsa, T. Said L, too is going to be equal to two seventy degree Celsius or five hundred forty three Calvin. So here we now we also know we have two point eight times ten to the seventh Jules per kilogram of coal. So for party, we need the burn rate of Cole s O. We can say that a is getting so party we have. The efficiency of the first system is going to be equal to the sixty five percent of the efficiency of C too foreign brother, See one. And this could be six point six five. This will be ah one. I will say one minus. And then this will be tea. House of one divided by t Ate someone like that and we can solve. So east of one is going to be equal to point six five, our one minus. And then this will be seven thirteen if I had by ten. Twenty three and this is going to give us point one nine seven east of two is going to be a point six five e carnot too. So this will be point six five and this will be simply one minus t sub l two divided by T&#x27;s of H too. So we have that east of two is going to equal point six five and then this will be times one minus five forty three divided by six hundred and eighty eight. And this efficiency will be point one three seven. At this point, me confined the ah work done by the first system. This would be equal to east of one que ce of h one. And it&#x27;s going to be equal to isa born queue, of course. Yeah, we know that Cue us of one equals cues of H one minus the works of one. So this will be equal to one minus he someone and then times queue. Of course, now we can relate and say that works of two equals east of two cues of H too. Two this Ian equalise of two Q sub l one. And this will equal Liza too. One minus piece of one q. Of course, So we can then add these and say that work plus one works of one plus works of two is an equal east of one Q of coal and then plus isa to one minus east of one Q. Of course, and we&#x27;re getting that this is going to be quits, accuse of cool, and it&#x27;LL be east of one plus east of two, minus east of one times Easter, too. So cue of coal. Rather, it&#x27;s getting new workbook. Queues, of course, is going to be equal to work. Someone plus works up to divided by east of one plus east of two months, east of one easily too. So we can simply say que of coal over tea, which is tryingto we&#x27;re trying to find how much it should be burned trying to find what&#x27;s the Baron rate of coal. So this will be equal to work&#x27;s work. Someone plus works up to per time and then again east of one plus east to minus piece of one easily, too. And we can actually substitute. So ah Q Clu T will equal nine hundred fifty times ten to six or nine hundred fifty million watts. Ah, and then we&#x27;LL have point one nine seven plus point one three seven minus point one nine seven times point one three seven and we need to convert. So the first of all that&#x27;s gonna equal three point zero nine for times ten to the ninth Jules per second and we should convert so cute. Cole of tea will equal three point zero nine for times ten to the ninth Jules per second, and then we can say times one kilogram for every two point eight times ten to the seventh. Jules. So this will equal one hundred and ten point five kilograms of core per second, So it should be the burn rate. Of course, Now it&#x27;s saying for part B, we need to cool it down. So how much water is needed? So part B, we can create a relationship and say that the queue exhaust is going to be equal to the queue of coal minus the work. So this is simply another way of saying cues of l. And so we have Cousteau, coal mine S W. And this can equal the mass of the water times the specific heat of the water times the change in temperature of the water. So we have that the mass of H two o is going to be equal to the Q exhaust divided by cease of H two o times, Delta T of H tio. And this will equal queue of coal minus w divided by cease of H two o Delta teeth A bit too well, at this point, we can divide by time. So if we divide by time, we could say that massive H two o divided by T is going to equal the queue of coal. A lot of I t minus work divided by t all divided by the specific e capacity of water times Delta t of H two l&#x27;s. And at this point, we can actually substitute so we can say three point zero nine four times ten to the ninth and then we&#x27;LL have nine point five times ten to the eighth. This will have to be minus apologize all divided by four thousand one hundred eighty six Jules per kilogram per kelvin And then the change in delta T will simply be four and a half degrees Celsius. So we&#x27;re getting that the mass of H two o needed her second would be, ah one point one three killer Sorry, one point one three times ten to the fifth kilograms per second, so this would be are needed. Now we need to convert to gallons per hour so we can simply just say one point one three times ten to the fifth. This will be kilograms per second and then we can say times thirty six hundred seconds per hour. We know that the density of water one meter squared, divided by one thousand kilograms and we know that there&#x27;s a thousand leaders in every cubic meter and we can say that this is also times one Galland equaling three point seven eight five leaders. So at this point, we can make and simply solve, and we find that the mass of H two o per unit time is going to be equal to approximately one point one times ten to the eighth gallons of water per hour. So this would be our final answer for Part B. That is the one in the mass of the mass flow rate, essentially of the water. That&#x27;s the end of the solution. Thank you for watching.

Ryan Hood · Answer

in part A. We&#x27;re going to use the fact that efficiency is able to work over Cuba. Sewing for Q H. We get work over efficiency and then now transforming this into power. We get th is equal to PW over deficiency, and we know what P. W is. And we know what the efficiency is so we can calculate he ate. And so P. W. Is 1,000 megawatts and efficiency is 40%. And so this gives 2.5 times 10 to the third. Make it once and this is the answer to party. In part be, we know that the he input and one day is going to be this time is the time of one day. Now there&#x27;s 8.64 times 10 to the four seconds a day because they&#x27;re gonna multiply this time by this power to get the toll heat going in. But this is making a lot, so I&#x27;m going to convert it into Watts. And what that does is it turns this three into a nine. Since that mega won is a lot of watts. In fact, it&#x27;s 1,000,000 once what finds out gives us 2.16 times 10 to the 14 Jules and to figure out the mass of the coal used. What weaken use is the jewels for kilogram that&#x27;s given in the problem. And so if we take this amount here and divide by 3.65 times 10 to the seven Jules per kilogram, this is given the problem, and it converts the energy into a mass that we need to use. You&#x27;ll see the jewels would cancel, the coherence will flip up and we get 8.15 times 10 to the six kilograms. Thats how much coal we need to supply to get this amount of energy, and that&#x27;s the amount of energy that we&#x27;re going to get in one day. So it&#x27;s the Parsi to figure out the power going into the cold reservoir. We need to use this equation. Actually, it&#x27;s supposed to be in a tch. Q H is equal to W plus QC, and so, you see, is equal to Kyu age minus of you and now just taking the dirt of respects time or converting it into power like we did in party we get PC is equal to P. H minus PW and then now we can plug in for P H and P. W. PH, we found was 2.5 times 10 to the third megawatts. It&#x27;s three in the exposure and POW is 1,000. Maybe it wants. And so this ends up being 1.5 times 10 to the third. Mayor wants This is the amount of power going into the cold reservoir. And so is the answer to Marcy Snow in the party. We just found what PC was, and this is the amount of power going into the river. And so, just to reiterate that he employed into the river is 1.50 times 10 to the nine jewels for second and recall Jules per second isn&#x27;t want. So this year is why we&#x27;re in here. And that&#x27;s the heat and put into the river. And so now we&#x27;re going to use this equation. Q is equal to Elm Sea Delta. T C is the specific e It was the mass, and lt is a change in temperature are going to use. Lt is equal to 0.5 degrees Celsius. This has given the problem and we&#x27;re going to solve this for elm. When we do that, we get you over. See Delta T we know CIA&#x27;s we know adult in tears and so we can plug in what this is for one second if we take one second here than this just turns into 151.50 times 10 to 9 jewels. That&#x27;s the he entering the river in one second. And this will be equal to our cue in this equation here. And so playing this this and see for water in which is a known constant 4,190 Jules per kilogram. Kelvin, we can solve this and we get seven point 16 times 10 to the fifth kilograms. This is the kilograms of water that is heated by this amount. In one second, we can convert. Convert this into a volume by using the density of water supply musicals of mass over the density, and the density of water is well known. And so it&#x27;s just a number. You go well, if you want. The volume is of being 716 meters tube and so this is the mass of the water being heated by this much in one second This is the volume of the water being heated by this much in one second. And so the river floor. It must be seven sixteen 16 meters Q and A one second. So says the Florey. So now we have to figure out the entropy change. And since the temperature of the river is nearly constant, we can imply the so called ice, a thermal expanse of ice, a thermal approximation. And this is discussed in 16 8 of the books of Section 16 8 And this is Thie Isil thermal X approximation for finding entropy. And so in one second, 7.16 times 10 to the fifth kilograms of water goes from 2 91.1 killing 2 to 91.6 kelvin. All right. It changes by half a degree, and the amount of heat added is 1.50 times 10 to the nine jewels. Heat added to get this change in temperature to happen. And so the interview change is just equal to 1.50 times 10 to the nine. Where is the heat added over the 2 91.6 and the reason we used to 91.6 is the Isil thermal approximation. If you want more information, c section 16.8 In the textbook, when you do the sound, you get 5.1 times 10 to the six jewels for killing and that.

Vipender Yadav · Answer

so here in the situation, we&#x27;re given that the net power output from the stream power plant it is 300 megawatt on the overall tumble efficiency of the problem is 32%. That wins 0.32 on the air fuel ratio. It is given us proof on getting value for the oil. For the call, it is given us 28 28,000 College Aleppo, if you can. What this thing to make a jewel. So we&#x27;ll get 28 Mega Jewell package e Onda if we have to calculate in a part of migration, amount off cold consumed in 24 hour time. No, first of all will be calculating the heat input so we know that the heat input it is includes two the work net output divide by the efficiency. So working it were given as a 300 megawatt on the deficiency. We&#x27;re going us a 0.3 toe. So from here cure input. It will be good. It&#x27;s true. A 9 37.5 megawatt Next will be calculating the rate off. Mosler it off the oil. So that means don&#x27;t must loaded off the fuel. So it will be equals. Two Que dort in divide by the heating value secure envy have calculated it does 9 37 0.5 megawatt Divide by hitting value is 28 mega Jewell Park itchy and make up What we can read it sting us mega Jewell per second from here The master It comes out to be us 33 point 48 kg per second if you have to. If you are looking calculators in the mass for 24 hour period so it will be equals two the m dot off fuel James Delta T and hurt No dirty is 24 hours And if you convert a 24 hour and do second the village it 24 times toward this extended second so you can plug a level is here. We know this value This is 43.48 kg per second. James, this torture dame, this is 24 hour. That means times 2602nd. So it comes out to be as nearly two point 893 Indo 10 to the power six Cagey so we can conclude that the moss consumed in 24 hour period would be 2.893 and pretended about six k. G. So this is a solution for the a part. Next, we have to calculate, in the be part the rate off air flowing to the foreign. It&#x27;s so it will be girls, too. Must flow rate for air. It will be close to air fuel ratio times the mass Laurette off fuel air fuel ratio were given us 12th names The mask mustard mass Florida fuel. We have calculated as 43.48 Did you per second. So mass flow rate for air comes out to be us nearly 401.8 kg per second. So this is the answer for the Beeb artists. I hope you were the problem. Thank you.

Anjana Saravanan · Answer

All right, so this problem gets a little bit complicated. So, um, to start out, I think I&#x27;m just going to right out all the information. We have to do it in a ah, more visual way. Just because think that will help us understand what&#x27;s going on. So there&#x27;s two separates. Steam engines, um, dropped him here. Then we&#x27;re given temperature information for both. Um, so I was gonna fill these in another piece of interest. Interesting information that we&#x27;re given is that these two are working in tandem where, um, call this demon. And one This could be steam engine or two, um, so Q l one equal to Q h two. Um and that&#x27;s what the problem means when it says that the operating temperature or I&#x27;m sorry the output of heat from the 1st 1 is Q A one is the same as the input of to the 2nd 1 which is qh two. So the first thing we&#x27;re gonna do is to solve for the efficiency. Um, and we have we know our equation. That e guy equals one minus t c or T h. He can apply that to both and find e. I for both. Um, and this is your simple plugging in, so I&#x27;m not gonna walk through it, But I&#x27;m going to, um, I&#x27;ll let you guys know what I got, which is that e I? The first is about 0.2349 Yeah, I for a second is your point 1991 But of course, we&#x27;re also told that the efficiency is 60% of the ideal efficiency. So we multiply each of these values times 0.6, then we&#x27;ll get the actual efficiencies. I&#x27;m cynical. Have not subscript here so we can keep track. But let&#x27;s put in the efficiencies here. So you want. So this times 0.6, you wanted the same pretty too, Um, again, that&#x27;s just plugging in. But you get is or oh, night. Four pretty one e to ISS 2.1195 All right, so next we want to somehow find a way to, um so for the, um, the rate that the cool must be burned, which might seem like a quantity that were not really sure what to make up. But the rate of coal being burned is actually just the heat input for the first, um, engine. So the big question mark here is Q H one equals What? So in order saw for that, um, one more piece of information that we&#x27;re given is that the was actually two more. But the heat of combustion of coal use at the very end. Um, you the plant puts out 1100 megawatts of power. Um, which means that the work done by the first engine, plus the work done by the 2nd 100 I&#x27;m sent to the six because megawatts, the prefix mega is tend to the sixth. Um, so that&#x27;s something we can also keep in mind. So now what I&#x27;m going to do is just a bunch of rearranging, um, of equations so that we can kind of manipulate our way into working with the information you have. We can find a way to find qh one. So the way I went about doing this waas first, I worked with the efficiency equation for the 1st 100 saying that e one is the work done by the first engine or acute h of the first century Cures. Qh is what we&#x27;re selling for. But we don&#x27;t have enough information because we don&#x27;t know what w oneness. But we do know this. So if we can somehow, um, relate w one and w two, then that is that would be a deal. So what I&#x27;m gonna do here is to, um another equation that we have is right here that q no one was q h to. So if I can, I&#x27;m solved in terms of Q l one. I can just change up to Cuba age, too. And that will be a convenient way off, uh, switching information about the first engine to the second. So what I&#x27;ll do here is just plug in by conservation of energy. We know that, um, the Q hot Q. What should I write? This, um, maybe over fired up Q r equals the work. Done. Plus Q. Go. Um, so let&#x27;s just go ahead and plug levin right here so we can Sahlins we can get something in terms off l one, which is what we have here. So that means that e one equals work one over work one plus Q Oh, sorry. That should be a subscript 01 All right. And now we&#x27;re just gonna rearrange to solve in terms of Q l one. And and of getting that work Times one over the efficiency minus one. All right, so now we&#x27;re going to keep this equation mind and turned to second at you. So this is all, um, engine one. Me, actually, in difficulty. Have you read? Now? We&#x27;re gonna work with engine two. You say that second efficiency is equal to make the equal sign not look like the less than sign, Uh, the work done by the second engine over Q h two. Um, now, something very important that we can use is that if this is equal to Cuba one, we have our formula here that tells the cute L one and huge two of the same thing. Which means that these two are interchangeable, and I can just plug in, put that right into there. So let&#x27;s go ahead and do that. Um, Major is green to keep this uniform. This is for clean up. A little plunge into, um, so rewriting you get, it&#x27;s gonna right over here. Work to over work. One one over. E one, minus one. All right. Beautiful. Now we want to rearrange, and in doing so, we you should end up with an equation. Um, where does he too is by itself on that side? Um, it&#x27;s on left side. You&#x27;ll have e to one over e one minus one times w one equals W two. So now you have a relationship between w one and W two right here. And you also know this equation, um, and e two and e one we already solved for We have those right here. So what we do, it&#x27;s great. Seles lines, um, it&#x27;s plug in or e two and e one values. And that should give us, um, plug. Ian. Simple flying have to give us 0.72 a create times. The work done with the 1st 100 equals the work done by the second, and it Now, we go ahead and plug that into this one. Lots of errors here. Um, but I hope you see that each step is giving us information necessary to kind of complete this system of equations here. Um And then if we put into their, we can solve for w one because well, essentially get that 1.72838 w one equals 1100 times 10 to the sixth. Um, solving we will then get that w one equals about 6.4 times 10 to the me Jules. Every second. Um, And now we&#x27;re almost there because we have W one. And again we&#x27;re solving for Q H one and we have our equation again. Ah, very important question that e equals the worked on over Q h rearranging u H was the work done over the efficiency. And if we add in our sub scripts here, we know our e one. And we know our work once and how we can solve off Rick. You actually, um so doing that plugging in plugging this year? E one is the value from up here. Plug it in to eat one. Then you should get that Q H one equals about 4.5 times 10 to the ninth jewels per second. And now we&#x27;re almost done. We can go back to the piece of information that the heat of combustion of cool. It&#x27;s 2.8 times 10 to the seventh jewels per kilogram coming around and were asked the rate of coal that would be burned. Right now we have the rate of energy that must be inputted in order to convert coal. We just use a conversion factor that they give us. Which is that, um, for every 2.8 times 10 to the seventh. Jules Energy equal to a one kilogram of cool. Now, of course, the jewels were canceled and will be left with a unit of kilograms per second. Um, and if you do this division, you should get note 1.6 times 10 to the second kilograms of coal per second, which have to be inputted to satisfy this system.

Problem

Let $\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}$ be p…

Problem 28 Easy Difficulty

Answer

$x 1 = 3.9$ tons of nathracite conl nold $x 2 = 1.8$ tons of hituminous coal.

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Catherine R.

Alayna H.

Maria G.

Michael J.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

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

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$x 1 = 3.9$ tons of nathracite conl nold $x 2 = 1.8$ tons of hituminous coal.

Linear Algebra and Its Applications

Catherine R.

Alayna H.

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.

Alayna H.

Maria G.

Michael J.

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

Linear Algebra and Its Applications