 In this video, we will provide the solution to question 3 for exam 1 in Math 2270. In this question, we are given a matrix, which is 3 by 3a. It's given us 301 first row, second row is negative 120, and third row is 001. And we're given a matrix, we're given a vector in R3, whose entries are negative 12 and 3. And we're asked to compute the matrix vector product, a times x. Now the matrix vector product by definition is we're going to take the first scalar in the vector x, and we times that by the first column vector of A. We add that to the second scalar in x times the second column vector of A. And then thirdly, we add that to the third scalar in x times the third column vector of A. Now the number of scalars inside the vector x needs to equal the number of rows inside, sorry, excuse me, the number of columns inside the matrix in order for this product to be defined. If those are not the same numbers, like this three columns and three numbers, that has to be the same. Otherwise it doesn't work. The number of columns in A needs to equal the number of rows in x, viewing x as a column vector. And so as such, we're looking for linear combinations of these three columns right here. And you can see that, you can see that choice A has, it's a linear combination of the columns of A. So is choice C and so is choice, I take that back, choice C, the third column is actually not the third column of A. So that one would rule it out automatically. I'm just kind of talking about elimination right here. If you look at choice E, also the columns do not coincide with the columns of A whatsoever. So C is not right, E is not right. If you look at choice B, that is not the first column of A, that's actually just x itself. These are acceptable columns though, so that one wouldn't work. If you look at choice D, let's see those, that's x again, that's not the second column of A. So D doesn't work. If you look at F, you're going to get, these are the correct columns of A. So just by looking at are these linear combinations of the columns of A or not? It turns out A and F are the only ones that actually work in that regard. Now the coefficients need to be the entries of x, right? So notice that you take the first entry of x right here, that's times the first column of A, that's good. The second one, you take the second entry of x times the second column of A, so far that's good. And then the third entry of x times the third column of A, that's good. So we see that A would be the correct choice for this one. I want to mention that with choice F, although the column vectors are right, the linear combinations are not. I mean, notice the coefficients 0, 3, 0 and 1, that is not the vector x. So F is an incorrect choice, A is the correct choice on this one.