 In this video, we provide the solution to question number 13 from project exam number 2 for math 1050 We're given a three by three linear system three equation three unknowns. We have to solve this linear system We have to show all our steps What for which we can do this by substitution We can do this by elimination or we can use matrices that isn't do Gaussian elimination in general that last approach is actually going to be the most Effective the most accurate so that's the recommendation. I'm going to offer to you and that's what I'm going to present here if we write this in I should just say we write the system as a augmented matrix. We're going to get one negative two three draw your line seven And then we're going to get two one one and four and then negative three two Negative two and negative ten like so so looking at the one one position. I already have a one in that pivot position So that's great. So I'm going to utilize that to zero out everything below the one So I'm going to take row two and I'm going to replace it with row two minus two times row one and For row three, we're going to replace that with row three plus three times row one like so Now to help me avoid some arithmetic mistakes I'm going to take for the second row. I'm going to take the first one times of my negative two right as little super script So you get a negative two plus four Minus six in that situation and then minus 14 and then for the last row We're going to take the first row in times of my three. So we're going to get three minus six We're going to get plus nine You don't have to write the plus if you don't want to but the minus signs are pretty critical there And then you're going to get 21 right there So then we do do these row replacements Like so the first row stays the same nothing changed one negative two three and seven Just just copy it down the second third row of course will change you're going to get a zero below the one one plus four is five one minus six Is a negative five and then four minus fourteen is a negative ten right? I can see in that row I already have everything divisible by five so without even doing the third row I know my next thing I'm going to do is I'm going to take row two and I'm going to divide everything by five We'll get to back to that in a second. So then we get zero here negative three plus three Next we're going to get two minus six, which is negative four. We're going to get negative two plus In that case that was a nine I didn't I didn't write that very well, but you're going to get a seven Right there and then lastly twenty one take away ten is eleven like so So yeah, I'm sticking with what I'm what I said before my new pivot position is going to move to the two-two spot I would love a one there if I divide by five that'll give me exactly that so just copy down Your matrix again This time the second row becomes zero one negative one and negative two then zero negative four seven and eleven like so For which then with my pivot position still right here. We are going to take the We're going to take the third row and we're going to replace it with row three This time plus four row two if you want to you could take out this two right here right now That seems fine. Let's just do it We're going to take row one and replace it with two times row two like so We want to get zeros other than the one there So if we take row two and times it by two, we're going to get a two We're going to get a negative two and we're going to get a negative four Then in the bottom We're going to get a plus four We're going to get a minus four and then a minus eight like so. So then what is the next? Equivalent matrix here. We're going to get one zero one and three Row two didn't change so we have one negative one negative two and then last we get zero zero Three and then eleven take away eight is likewise going to be three right like so so again that worked out pretty nicely for us I'm going to divide row three by three itself This then gives us our next matrix Because after all our pivot moves now to the three three position there So if we divide by three we still get one zero one and three We're going to get zero one negative one and negative two and then last so we get zero zero one one And so using this final pivot, let's get rid of all the other entries here So in the second row, we're going to take row two just add to it row three and then for row one We're going to take row one minus row three like so. So we're going to add one add one We're gonna minus one minus one and then that should put us in our row reduced echelon form for which We get one zero zero two The next row we get zero one zero negative one and we get zero zero one one And so then we can record our final answer here that XY and Z are equal to Two negative one and one the final solution for which we did this using Gaussian elimination. Like I said, you could have also used some of the other techniques we developed in this class But I think this one works out pretty nicely in the end