 In this video we're going to present the solution to question 13 from the practice, from question 13 from the practice exam number one for math 2270. We are given a matrix A which you can see on the screen this is four by four matrix. We are going to determine with proof whether the column vectors of A are linearly independent or not. That comes down to us row reducing the matrix A which I'm just going to copy it down here 1 1 1 0 1 2 negative 1 0 1 3 1 0 and 1 4 negative 1 1. We want to determine if there are pivots in each and every column of the matrix. If every column is a pivot column the vectors will be linearly independent. So we get our first pivot position in the 1 1 spot. So we're going to compute r2 minus r1 r3 minus r1 and r4 minus r1 so we get minus 1 minus 1 minus 1 into 0 minus 1 minus 1 minus 1 into 0 minus 1 minus 1 minus 1 into 0. So the first row stays the same 1 1 1 0 then we're going to get 0 1 negative 2 0 0 2 0 0 and then 0 3 negative 2 1. This would then be the next matrix in this calculation here. So we have a pivot position in the first column. We then see that since the next column is non-zero we get a pivot position there as well. And so we want to get zeros below that pivot position. We're going to take row three and subtract from it two times row two and then for row four we're going to subtract from it three times row two. So we get minus two plus four and then a zero. We're going to get minus three plus six and then a zero. Copy down the next matrix. We have the first row. We then have the second row. Again these didn't change. The third row is going to look like zero zero four zero. And then the last row would look like zero zero. We're then going to get a four and then a one like so. And I don't need this matrix to be in row reduced echelon form. Any echelon form will work. So I don't need the pivots to be one. Moving on to the third column I'm going to get a pivot in the three three position. And then notice if I just take if I just take row four minus from that row three you're going to get a minus four and then minus zero right that doesn't do much. We then are going to get a matrix that looks like one one one zero zero one negative two zero zero zero four zero. If you want to scale that to a one that's okay but we didn't need to do it zero zero zero and then a one like so. And so now this matrix is in fact echelon form and then notice the pivots we have a pivot we have a pivot we have a pivot and we have a pivot and this matrix is in fact an echelon form. It's not row reduced but that's not necessary here. And so then we can conclude therefore the vectors the column vectors of A right the vectors are linearly you know here's the kicker right here they're linearly independent because we found there was a pivot in each and every column. There's no way of writing one vector as a combination of the others.