 One of the quantities we can compute for square matrices is called the determinant, and that emerges as follows. This is a somewhat complex definition, but we define determinants recursively. So if I have an n by n matrix, the determinant of a, which we write using the vertical bar surrounding the matrix itself, we're going to define it recursively as follows. First of all, if a is a one by one matrix, in other words, if a consists of a single entry, then the determinant is just going to be the only entry that's in the matrix. If a is a two by two matrix, I'm going to define the determinant as being this difference of the cross products, this product of the main diagonal a11, a12, minus the product of what you can think about as a counter diagonal a12, a21. And so that gives us, for example, if I have this matrix three negative one, two, one, I'm going to form the cross product of the diagonal here, with the product of the diagonal here. So that's going to be the product three times one minus the product of the other diagonal minus one times. And after all the dust settles, we find that our determinant is equal to five. Now, if we have a larger matrix, we have to employ a little bit more notation. So we have to define one important idea here, which is the minor. So let's say I have an n by n matrix, the ijth minor of a is going to be the determinant of the matrix formed when I remove the i-th row and j-th column of a. And the ijth cofactor of a, written capital A ij, so it has the same symbol as the matrix, except it has the index that we associated with the entries of the matrix. It's going to be the product of this determinant of the minor by minus one to power i plus j. And so note that if i plus j is odd, this value here is going to be negative. If i plus j is even, the value here is going to be positive one. And so this produces what is sometimes described as a checkerboard pattern for the cofactors. So for example, let's take a look at this matrix A and let's say I want to find the ijth cofactor. So that's going to be the ijth cofactor. I'm going to proceed the determinant by minus one to power i plus j. That's i is two and three. So that's minus one to the power two plus three times the determinant of the matrix that I get when I eliminate the second row, third column. So that's going to be this matrix here and that's a two by two matrix. And I know how to calculate that determinant. It's going to be the difference of the cross products three times one minus one times two. And again, after all the dust settles, I find my determinant. Well, let's find the determinant of larger matrices. And so we can do this as follows. We're going to choose any row or column that we want. We're going to find the cofactors of each entry in the row or column. And we're going to sum the product of the cofactors with the corresponding entries. So we find those cofactors. We multiply them by the actual entry in the matrix. So for example, let's take this matrix here. It's a three by three matrix. So I'll pick any row or column. And so maybe I'll expand along this first row here. And so my determinant is going to be the row, the entry that I'm working with in this case one times the cofactor. That's going to be minus one. This is the one one entry. So that cofactor is going to have a coefficient negative one to power one plus one times the determinant of the matrix I get by eliminating first row, first column. And so now that I pick this entry here, I'm going to eliminate the things that are in the same row and the same column. And what I have left is going to be the matrix whose determinant I need to find. So there's my first term. Now I'm going to add in the second term. So that'll be negative one times the coefficient of that cofactor is going to be negative one to the power one plus two. Again, this is in the first row, second column. So that's going to be negative one to the power one plus two times the determinant of the matrix that I get by wiping out the same row and the same column as my entry. So that's going to get rid of this column. That's going to get rid of this row and my entry, my matrix will be three, negative one, one, three. And I have to find that determinant. And then finally I'm going to take the last entry in that row. This is the first row, third column entry. That's going to be zero times negative one to power one plus three times the determinant of the matrix I get by wiping out the same row and same column as my entry, zero. And that's going to be this matrix here, three, one, one, two. One thing to note is that the signs that I get for the cofactors form this nice checkerboard pattern. If the sum of the row and column is even, then this negative one is being raised to an even power. It's going to be one. If the sum is odd, this negative one will be raised to an odd power. I'll get an odd number. And so the signs that I get, you can remember those as plus, minus, plus, minus, plus, and so on. They form a checkerboard pattern of pluses and minuses. So now I have a whole bunch of things I can compute. Negative one to some power, that's easy. This is a two by two determinant and I do have a way of computing those. So we'll go ahead and fill in the blanks. This is one times negative one to the second. The determinant of the two by two will be the difference of the cross-products, one times three, that's three, negative one times two, that's whatever it is, and I'll be subtracting that. My next term I'll have negative one, negative one to power three, and the determinant of the two by two matrix, I'll find the cross-products, three times three is nine, negative one times one is negative one, so that determinant nine minus negative one. And my last term, zero times negative one to power four, three times two six, minus one times one is one. Now because that coefficient there is going to be zero, I don't actually need to compute this last determinant, but we'll leave it in there as part of our computation. And so now at this point it's a bunch of arithmetic and algebra, really arithmetic that we have to do, and we find our determinant is equal to 15. Now you might be a little bit uncomfortable with the fact that we can find the determinant by expanding along any row or column. Is it possible that we may make the wrong choice and expand along a row or column and not get the value of the determinant? And the answer is no. While there are better choices for which row or column to use, there's no actual wrong choice as we do our expansion correctly. So for example let's say I want to expand along the second column, so it's going to be along these entries here. So I'll start off with minus one times negative one to the power of, well that's going to be the first row, second column, one plus two. Again if you think about that checkerboard pattern, this is going to be plus, this will be minus plus and minus will be our values there, but my cofactor minus and I'll eliminate the same column, the same row, and my entry is three, negative one, one, three, and I'll find that determinant. My next entry, the zero, eliminate the row and column that it's in, I'll get the matrix one, zero, zero, three, and the sign of the cofactor is going to be positive, plus one, negative one to the power two plus two, and there's my smaller determinant. And then finally this last entry two, eliminate the row and column, the cofactor is going to have sign negative one to power three plus two, that's a row three column two entry, and again you can go by the sign, the checkerboard signs, this is a plus, plus, this will be minus, so again that'll be minus two times the determinant, wipe out the row, wipe out the column, and I have my one, zero, three, negative one as my last two by two matrix. And I'll go ahead and let the dust settle on that, and after all the dust settles we find the determinant is going to be fifteen. And so the determinant doesn't actually depend along, among which row or column we choose to use. Now is there a better choice? Well let's think about that. When we expanded along the second column I had to use my second column entries, negative one, positive one, two. So there was my second column expansion. When I expanded along the first row the coefficients ideas were one, negative one, zero. And while we're only multiplying by constants and it doesn't really make any great difference in the grand scheme of things, the fact that one of these terms was zero made our computations a little bit easier, the fact that the magnitudes of these numbers were both one also made our computations easier. So there is a preferred row or column that will make the computation easier, but there's no actual best row or column by any objective standards.