In this problem, we are given a lot of information about a saleswoman, and the different amounts of cell phones is she sells. Ah, and the commission that she makes for each week. So for every problem, access to a Let X, Y and Z represent the Commission on Standard Deluxe and Super Deluxe cell phones, and we want to write these right this information as a system of equations. Also, we know that the standard is going to correspond to X deluxe model, correspond to why and Super Deluxe will correspond to Z. So essentially each of our weeks is going to correspond to one of our equations in our system. I'm so for a week when we know that we have nine standards. So nine x plus 11 times our deluxe, which will make 11. Why plus eight times are Super Deluxe and this is going to give us $740 on me too. We're going to sell 12 standard models a 12 X plus 15 deluxe wife plus 16 Super Deluxe Easy, which is going to give us $1204. And finally, on our third week, we have three standard models of three X plus 70 look +71 plus 14 Super Deluxe See is going to be equal to $828. I'm so that's part a big problem for a party over problem. We want to write this system of equations as a matrix equation. I'm so essentially we're just going to be able to write this actually do that on the top of here as a co efficient matrix, which we know I's going to be. In our first equation, we have nine 11 eight. Our second equation has coefficients of 12 15 and 16 and our third equation as coefficients 37 and 14. And we're multiplying this by are variable matrix, which is X y Z, and this will be equal to the constants on the right hand side of the equation so equal to 740 1204 and 828 and finally, for party ever problem. We want to find the inverse of the co efficient matrix A and use it to solve the matrix equation in part beat. I'm so once we do that, it's going to answer the question How much commission does the sales saleswoman earn on each model of cell phone? I'm so we could go ahead and start working on finding our universe of our matrix here. Um, I'll go ahead and give us some extra space to work as we know that we're going to write this out, um, as a three by six matrix, where the left three by three Matrix is going to be our original matrix that were given 37 14 and then on the right inside, we're going to draw dash line and then write our identity matrix out. And so, essentially, all we're going to dio is ro reduce so that the left hand side of our equation is going to give us, uh, I knew I did in the Matrix. Um, so right away, we can see that, Uh, regardless of what we do, this is going to give us quite a few fractions here, and so I'm going to go ahead and begin by dividing my entire first road. Uh, fine, actually. Always And start with our third row, we can go ahead and compute row two minus four times our third row and row one finds three times or third route. And so, essentially, the reason that I chose to do this, um, was because I know it's going to give us some zeroes in our position there. Um, so our top row row one minus three, row three. Now we get nine minus nine, which is zero 11 minus three times seven. So I love in my eyes. 21 will give us negative 10 and then we have eight minus 14 Times Street. Um and so when we will buy that out, it's a tract that's going to leave us with negative 34. Well, then we have 10 negative. Three, our second row. We have ro to minus four times row three. So we have 12 minus 12 which is zero of 15 minus four times seven. Eso 15 minus 28 is going to be negative. 13 on. Then we have 16 minus four times 14 which is going to give us negative 40. And then we have 01 negative three and our third row is going to see is him 37 14 001 And from here we can see that we now have nothing that's going to cancel really nicely. Um, so I'm going to go ahead and start by dividing my top equation by 10 or my top row by Ted. So, row one over 10 this is going or negative time. This will give us 01 3.4 and then 0.10 negative 0.3. Our second row is still zero negative. 13 negative. 40 01 negative three. And her third row is 3 7/14 001 But actually, I'll go ahead and divide this. Divide everything by three here to change this, to be in the form that we want our equation to be in or our identity matrix to be in so again, dividing everything by 33 over three will give us one. I'm just going to leave this as 7/3 and 14 3rd and then we have 00 1/3. So from here, we can see that in order to get rid of the negative 13 in Row two, we're going to want to add a 13 row one to get to there. I'm so we'll go ahead and I'll actually replace our performer. Rose Swap on rose one and three so no one is not going to be one 7/3. 14 3rd 00 1/3 um, And Row three is not going to be 01 3.4 a negative 0.1 zero. Ah, and 0.3. And row two, we are going to be subtracting a 13 of our original row one from I'm so we get 00 a negative force yet? Negative. 40 minus 13 times 3.4 is going to give us negative. 84.2 Ah, in our next position. Now we have zero minus a 13 times 130.1. So that'll be negative. 1.3 ah one minus zero and negative three minus or should be plus, actually a negative three plus a 0.3 times 13. Essentially. Ah, that's just the same thing as three times a point. Three times 13 minus three actually. Give us negative 11.7. Um, And in our first position right here, I realize that I accidentally subtracted, but I should have actually added so that should have been negative. 40 plus 13 times, 3.4, which would actually be 4.2, huh? Um and so from here, we can go ahead and perform somewhere. Rose Wops. I'm going to go ahead and swap rose two and three here, huh? But And so instead of continuing Thio more finds out, I'll go ahead and just get to the end here because it's going to take quite a bit of time. But once we end up finishing this role reduction, we're going to be left with are inverse matrix as seven over four. Negative seven over four one in her first wrote our second rule being negative. 27 over 28. 31. Number 28. And then finally negative. Five over seven. And our third row will be negative. 29 over 56 25 over 56. Negative, one over seven. And so from here, now that we have are inverse matrix. We know that we're just going to multiply this by our matrix that we found, um, or are a constant matrix, which started with 740 from our first week 1204 from our second week and 828 from our third week. And once you go to buy these together, we find that are a standard model is going to give us $16 in commission. The deluxe model will be $28 and the Super Deluxe model is $36 in commission.