 An important property of a matrix is its determinant. To find the determinant, we'll have to introduce a couple of terms. First, we'll introduce the term minor. Let A be an N by N matrix. The ijth minor of A is the determinant of the matrix formed when the i-th row and j-th column of A are removed. And the i-th cofactor of A, written A ij, is the product of this determinant by negative one to the power of i plus j. This factor of negative one to the power i plus j is actually pretty easy to remember. Since negative one to an even power is equal to positive one and negative one to an odd power is equal to negative one, then every time we shift over one column or move down one row, we change the sign. And this means these powers form a checkerboard grid of plus and minus. The important thing to remember about these signs is these signs are in addition to the sign of the matrix entry. And this leads to the following method of calculating the determinant of a matrix A. The determinant of A, written debt A or bar A, is defined recursively as follows. If I have a one by one matrix, the determinant of A is just the single entry that's in A. If I have a larger matrix, then I can choose any row or column. I'll first of all find the cofactors of each entry in the row or column, and then I'll sum the products of the cofactors with the corresponding entry. So let's find the determinant of a two by two matrix. So the first thing that's useful to do is to set down that checkerboard grid of plus and minus. We can expand along any row or column, so let's choose the first row. We'll take the first row, first column entry, along with its sign. The minor is going to be the sub matrix we formed by crossing out the row and column that it was in. And since this is a one by one matrix, the determinant of that minor is just going to be the entry itself. Then we'll take the second entry in that first row along with its sign, cross out the row and column it was in to form the minor, and again as a one by one matrix the determinant is just the value of the entry itself. And then we can evaluate our products and sums to find the value of the determinant. And if we think about what we've just done, this leads to a general result. If I have a two by two matrix, the determinant is going to be expressed as follows, which is the product of the terms along one diagonal minus the product of the terms along the other diagonal. How about a three by three matrix? Again, we'll set up our grid of plus and minus signs. And if we start with the first row again, we'll take that first entry, cross out its row and column to form the minor, and then multiply that minor by the entry along with its sign. We'll take the second entry, cross out row and column, which gives us a minor, and then we'll multiply that minor by the entry along with its sign. And we'll do the same thing for the third entry, cross out row and column to form the minor, and then multiply the minor by the signed entry. Since all of our minors are two by two determinants, we can find their value by multiplying along one diagonal and subtracting the product along the other diagonal. So for our first determinant, we'll find the product of 0 and 2 and we'll subtract the product of 1 and 3. Likewise, for our second determinant, we'll multiply 0 and 2 and subtract the product of 1 and 5. And finally, our last determinant will need to multiply 0 and 3 and subtract 0 times 5. And we'll evaluate the arithmetic expression. Now a useful feature is that the determinant of a matrix doesn't depend on which row or column we happen to use to find it. And in this particular case, we notice that our second row consists of a bunch of 0s and a 1, so it might be easier to expand along the second row. So as before, we'll set up our grid of plus and minus. Starting with the first entry of the second row, we'll form the minor by crossing out its row and column, and then multiply the minor by the signed entry. We'll do the same thing for the second entry of the second row, cross out row and column, and then multiply the minor by the signed entry and the same thing for the third entry. And what's useful about this approach is that while these first two terms seem to require the evaluation of a 2 by 2 determinant, they're going to be multiplied by 0, so we don't actually care what they are equal to. The only thing we need is the value of this last determinant, which we can find by multiplying along one diagonal and subtracting the product of the terms along the other diagonal. And we find, again, that our determinant of the matrix is equal to 17. So the fact that the determinant of a matrix is independent of which row or column we use to find it is one useful result about determinants. Here's another important theorem regarding determinants. Suppose a and b are n by n matrices with the determinant of a equal to a and the determinant of b equal to b. Then the determinant of the product is equal to the product of the determinants. And this leads to several important results that should be fairly easy to prove. If it exists, the determinant of the inverse of a is equal to 1 over the determinant of a. And this is something you should be able to prove. Conversely, if the determinant of a is 0, then a inverse does not exist. And again, this is a nice simple result you should be able to prove on your own.