 In this video, we are going to present the solution to question number 13 from the practice exam number 2 for math 2270 We're given a basis for our for a non-standard basis. It contains four vectors 1 1 1 1 1 1 2 3 4 1 negative 1 1 negative 1 and 0 0 0 1 You don't need to prove this as a basis. You can just assume it's a basis and we have a vector b for negative 2 0 1 And we need to show that little b Well, we need to compute the coordinate vector of b So to do this, we're going to take our basis b and augment it with the vector. We're trying to find the coordinates for If this system turned out to be inconsistent, it would mean that little b cannot be expressed by the capital B vectors But as it's a basis, we know it can so we need to set up this matrix here 1 1 1 1 1 2 3 4 1 negative 1 1 negative 1 and 0 0 0 1 Augment it with 4 negative 2 0 and 1 Like so we need a row reduce this matrix the solution of this matrix is going to give us our coordinate vector We do have to show all of our steps here as well So look at the first the first pivot position. We need to get rid of what all the ones below it So we're going to take row two minus row one row three minus row one and row four minus row one So we get a minus one minus one minus one. I'm going to skip when there's zeros minus four And we just can do that again minus one minus one minus one minus four Minus one minus one minus one and minus four So then the next matrix in this process here the first row stays the same All the other rows change though. We're going to get zero one negative two zero and negative six third row becomes zero two zero zero negative four and then the third row becomes zero three negative two one and negative three So check to make sure that works out great The next thing to do is we look at the next pivot. That's going to be the two two spot That's already one which is great. I want to get rid of the numbers below it. So I'm going to take I'm going to take row Three it's a track from a two times row two. I'm also going to take row four. It's a track from a three times row Two as well. So if we do row three minus two times row two, we're going to get a minus two plus four That's a zero and then we're going to get plus 12 in that spot For row four, we're going to get minus three. We're going to get plus six There again, we add zero to one that doesn't do much and then you're going to get a plus Here because we're subtracting you're going to get plus 18 And so draw this down again. So the first row stays the same the second row Stay is the same as well The third row we're going to get zero four Zero in this case you're going to get an eight and then for the third row you're going to get zero zero You're going to get four one And then you should get a 15 like so So now my pivot moves down here like so I'm going to do two things at the same time. I'm going to take One fourth times row three because I notice everything in row three is divisible by four in fact And then I'm also going to take I'm going to take row four. So track the current row three So I'm going to get a minus four and a minus eight right there And so copying down what we did there first row second row stay the same Third row like I said divided by four. So I'm going to get a one a zero And then a two and then for the third row we're going to get zero zero zero One and then we'd have 15 take away Eight which is seven. So we're in a pretty good spot right now Um, we definitely are in echelon form now. We need to be working backwards. If you look at the last column This one's already perfect for a row reduced echelon form. So we need to go here to our third column We want to go to the numbers above that. So we're going to take row one minus row three We're going to take row two plus two times row three So we get minus one minus two And we're going to get plus two Two Plus four in that situation Copy down what you have one one zero zero And then you get four minus two, which is two you're going to get zero one zero zero negative six plus four, which is a negative two And then the next two rows are just identical to what they were before And so then the last thing to do is when we look at the second column We got to get rid of the one above it. So we're going to take row one minus row two So we get a minus one right there and we get a plus two right there And so then our ref we're going to get one zero zero zero zero one zero zero zero one zero zero zero zero one That is the identity matrix, which we should expect right here if we take two plus two That's a four we get a negative two Two and seven and so notice what we have here the first coefficient and the linear combination should be four The second coefficient should be negative two the third coefficient should be seven and the fourth coefficient should be The third coefficient should be two and then the fourth coefficient should be seven right there So then we have our answer what we're looking for the coordinate vector of b With respect to these b coordinates This is the vector the first coordinate was four the second coordinate was negative two The third coordinate was positive two and then the last coordinate was seven If we have any doubt on whether this was correct or not, we can actually verify it, right? So notice if we take 4 times the first vector, which remember that was 1, 1, 1, 1, next we should take negative 2 times the next vector, which was 1, 2, 3, 4, and then we get 2 times the third vector, which was, recall, 1, negative 1, 1, negative 1, and then lastly we get 7 times the fourth vector, which is 0, 0, 0, 1. When you try to combine these together, this looks like, and if you put it all together, we're going to get 4 minus 2 plus 2 plus 0, that's going to equal 4, just like it is supposed to with B. The next one we're going to get 4 minus 4 minus 2 plus 0, that turns out to be negative 2, just like B. The next one we get 4 minus 6 plus 2 plus 0, which gives us a 0, and then the last part we're going to get 4 negative 8 minus 2 plus 7, which that adds up to be 1, giving us the vector B.