Let's now compute the determinant using the previous formula of a 3 by 3 matrix. So there's my matrix A, let's say. And, once again, I'm allowed to expand along any row or column and I'll get the equivalent result when I compute this determinant. And then in conclusion, depending on whether that number is zero or not, that tells me whether or not the original matrix was invertible. I'm just going to use just that first general formula we wrote here for a determinant by expanding along the first row. So I take the number 5 and then I multiply by the determinant of the corresponding submatrix which is now 0, 2, -1, 3. And then I go minus, I alternate signs, so check this minus -3, goes +3 so we flip that and then I remove again row 1 column 2, and I just plug in the determinant here, 1,2,2,3, the corresponding submatrix. And then lastly I pluck out the 2, the last entry in that row, and then I multiply that by the corresponding submatrix here, that determinant here 1, 0, 2, -1. So there's my determinant formula. And again, just as a friendly reminder, the determinant of a 2x2 matrix is this sort of cross multiplication process, ad-bc. So let's go ahead and clear this up here. So we have 5, ad-bc, so this is 0-(-2) so this is going to be a +2, so 5(2). So plus 3 times ad, which is 3-(-4) is -1. And lastly we have +2, ad-bc, is -1. So we have that here. And let's add these numbers up. We get 10-3-2 and that results in 5. That means that my original matrix A here, is invertible. So there's some matrix out there I could find and multiply by that inverse matrix and then produce the identity. Since we haven't done a column yet, let's expand along column 2 here. And we will compute the determinant that way and we see that we also get 5, but we'll have to do little less work because of the 0 here. I just want to note right we have this checkerboard pattern with the signs. I'm just going to remind you for the determinant we have a negative with that term, a positive and then a negative, we alternate signs. Let's now compute the determinant of A one more time, just expanding along this time column 2. Ok, so I have, negative -3 which is +3 times the determinant of the submatrix, the same as it was before. So 1, 2, 2, 3. Ok, now alternate sign +0, ok I'll just leave this as a place holder so we can see it. Plus zero. Of course, that'll go away. But the submatrix associated with zero when I remove now column 2 and row 2 is going to be 5, 2, 2, 3. Ok so that just goes away. And then I go minus -1 which is plus 1 when I alternate sign, and then I multiply by the determinant of the submatrix corresponding with the -1 here, so I'm going to remove row 3 and column 2 and that leaves me with 5, 2 and 1, 2. Let's compute that determinant. So again, I use the formula for a 2x2 determinant ad-bc here. Right, so 3 times 1(3),ad, 3-4 is -1. Ok. plus 0, and then plus 1 times 10-2 is going to be 8. So what I have when the dust settles is -3+8 sure enough results in 5. Point being the determinant can be ascertained by expanding along any row or column we get a consistent result.