 One important use of the cofactors for matrix is that we can find the determinants of a matrix fairly quickly provided we have a good fast way of finding the determinants of a cofactor. So for example, this shows up in the following theorem, which is the adjoint method. If I want to find the inverse of an invertible matrix A, then that inverse could be found as follows. I can multiply 1 over the determinant of A times the transpose of some new matrix. And, well, where is this new matrix coming from? Well, this new matrix M is going to have entries that consist of the cofactors of each entry in my matrix A. And a concise way of expressing this, as we can say, is that the inverse is the transpose of the matrix of cofactors scaled by the determinant. And this gives us a method of calculating inverses directly, although it's limited by the requirement that we find this matrix of cofactors. So, for example, let's say I want to find the inverse of a 3 by 3 matrix. So first off, I do want to check to make sure that the matrix is actually invertible, so I need to find the determinant, which we're going to need in any case. So here's my matrix, and I kind of like this third column because it has a couple of zeros in it. So when I find the determinant by expanding along that third column, two of the terms in the expansion are going to be zeroed out. So here's my expansion for purposes of writing out the formula. I'll go ahead and write this down, term zero times the corresponding cofactor. That's my smaller matrix 2 negative 1, 1 negative 1, and my coefficient of the cofactor negative 1 to the power of 1 plus 3. Again, you know that's going to be positive because, again, our checkerboard pattern of pluses and minuses plus minus plus. This three will be minus. This zero will be positive. So my second term in the expansion, the term 3, negative 1 to the power of 2 plus 3, again, that's row 2, column 3. And again, with my checkerboard pattern, I also can predict that's going to be negative times eliminate row, eliminate column, take a look at the leftover matrix, and find its determinant. Last term zero, eliminate column, eliminate row, use the leftover matrix, find that determinant, and that's going to be plus, plus, that'll be plus. Or again, this is the row 3, column 3 entry, so that coefficient will be multiplied by negative 1 to the power of 3 plus 3. And after all the dust settles, I find my determinant is 6, and since my determinant is not equal to zero, then I know my matrix is invertible. Now, we know we need the determinant of A, so we should record it someplace we can find it again. So we'll go ahead and write that up here. We found that the determinant of A is equal to 6. Let's go find those cofactors. So our matrix of cofactors. So our first cofactor, this 1, 1 entry, so again, the cofactor is going to be the negative 1 raised to the power of row plus column, that's going to be negative 1 to the power of 1 plus 1, and multiplied by the determinant of the minor. So that's going to eliminate row, eliminate row and column, what's left over, that's going to be our minor, and that's a 2 by 2 matrix. So after all the dust settles, we find the determinant is equal to 3. So we'll drop that in our 1, 1 entry. Note the key difference here between calculating the determinant for the matrix and calculating this matrix of cofactors. I don't have the matrix entry entering into this expression here. So even though the 1, 1 entry is 1, I don't actually need that piece of information to calculate the cofactor. All right, let's take a look at the next entry, that's going to be our row 1 column 2 entry. So again, that row 1 column 2 entry, negative 1 to the power of 1 plus 2, eliminate row, eliminate column, my matrix 2, 3, 1, 0, find the determinant, multiply and get the value of that cofactor also going to be 3. Again, my next entry, my row 1 column 3 entry, that's negative 1 to the power of 1 plus 3, eliminate column, eliminate row, and what's left over 2, negative 1, 1, negative 1, find the determinant of that, after all the dust settles, I have that last entry in that first row is going to be negative 1. Next row, second row, first column entry, that's going to be negative 1 to the power of 2 plus 1, eliminate column, eliminate row, what's left over 1, 0, negative 1, 0, that's my minor and the determinant of that, I can calculate and after all the dust settles, that's going to be 0, that will be dropped into the 2, 1 position. Second row, second column, eliminate column, eliminate row, matrix is going to be 1, 0, 1, 0, and my cofactor is going to be 0 once again. And second row, third column entry, negative 1 to the power of 2 plus 3, second row, third column, eliminate column, eliminate row, what's left over 1, 1, 1, negative 1, there's our minor, we'll calculate the determinant and drop that into the appropriate spot. Third row, first column entry, negative 1 to the power of 3 plus 1, eliminate row, eliminate column, the minor, 1, 0, negative 1, 3, there's our minor and find the determinant and drop it into the appropriate spot. Third row, second column, negative 1 to the power of 3 plus 2, third row, second column, eliminate column, eliminate row, the minor, 1, 0, 2, 3, there's our matrix, find the determinant and find the cofactor. And finally our third row, third column entry, negative 1 to the power of 3 plus 3, eliminate column, eliminate row, what's left over is our minor and there's my last entry. Now the adjoint theorem tells us that I can find the inverse by multiplying the transpose of the matrix of cofactors by 1 over the determinant. I know what the determinant is, so I can find the inverse directly, A inverse, 1, 6 of the matrix M transpose. I'm going to rewrite my rows as columns and my columns as rows. So there's my first column becomes my first row, my second column 3 is here, negative 3 becomes my second row, my third column, negative 1 to negative 3 becomes my third row and there is what we believe to be our inverse and it's always a good idea to check. Now here I can check by multiplying this, by this, and seeing if I get the identity matrix. And that's a good thing to do because it gives you some nice practice finding the product of two matrices. But again just as a reminder, don't focus on the step-by-step algorithm because those are easily replaceable. So let's say I really had to find the inverse of this matrix. Again, not a paid spokesman, but Wolfram Alpha, I can find that inverse. So what do I want to do? I want to find the inverse and of the matrix 1, 1, 0, 2, negative 1, 3, 1, negative 1, 0. And Wolfram Alpha, oh Wolfram Alpha tell me what this is and let's see, verify here's something you should do anytime using technology. It thinks I want to find the inverse of this matrix. Well, I'll check and that is actually the same matrix that I started with. And so if it thinks that I want to find the inverse of this matrix, it says that the inverse is 1, 6, 3, 0, 3, 3, negative 3, and so on. And here's what Wolfram Alpha says is the answer, here's the answer that I got and the two are the same. So either we're both right or we're both wrong. And again, I'll emphasize the point. The point is not that when you need to find the inverse of a matrix, you should rely on technology, but it's the technology that the technology exists and that if you're going to focus on the algorithm that gets you here, the thing to realize is that if you focus only on the algorithm, you're going to be replaced by the machine sooner or later. And there's a much deeper structure to where this comes from and that's the important thing to learn.