Building the adjoint, and why it multiplies back to a scalar times the identity

0:00There is a question sitting underneath everything we are about to do, and it is worth saying out loud before we start. You already know that some square arrays can be undone. There is a second array that multiplies with them to give the identity, and we call it the inverse. What nobody has told you yet is WHICH arrays have one. Not how to compute it. Which ones even possess it.

0:27That is the hunt. And the first move towards it is a construction that, the first time you meet it, looks completely arbitrary. It is called the adjoint. Today we build it, and then we find out why it is built that way and not some other way. Here is the recipe. Two moves, no more. Move one. Take your square array and replace every single entry by that entry's cofactor. The minor of that position, with the sign of the position already welded on.

1:04You now have a second array of the same size. Call it the cofactor array. Move two, and this is the strange one. Flip that array about its main diagonal. Rows become columns. Whatever sat in the second place of the first row now sits in the first place of the second row. That is the adjoint. Cofactor everything, then flip. And the flip is where every reasonable person stops and asks why. Why would you go to the trouble of computing nine cofactors and then move them all around?

1:43Watch one entry make the journey, because the bookkeeping is the whole difficulty here. Take the cofactor belonging to row one, column two. In the cofactor array it sits exactly where its own entry sat: top row, second place. After the flip it is in row two, column one. Its two subscripts have swapped places on the page while the label on the cofactor itself has not changed at all.

2:14That is the sentence to hold on to. The cofactor keeps its name. The position changes. And the commonest single mistake in this whole topic is doing move one, admiring the array of cofactors, and delivering it as the answer. You might wonder how much that mistake actually costs. Sometimes the cofactor array is already symmetric, and then flipping it changes nothing. So we measured it. Take every three by three array you can build out of noughts and ones. There are five hundred and twelve of them.

2:53For how many does the unflipped cofactor array behave the way the real adjoint does - multiply back to the value times the identity? One hundred and eighteen. Out of five hundred and twelve. So the shortcut is not never right. It is right about one time in four, on arrays as simple as these, and that is exactly what makes it dangerous. It will let you through a homework problem and then fail you on the one that counts.

3:26Let us do a small one entirely by hand, so the recipe is concrete before we ask what it is for. Here is a two by two. Two and three across the top, one and four underneath. Four cofactors. For the entry in row one column one, cross out that row and that column, and what remains is four. The sign of that position is positive, so the cofactor is four.

3:54Row one, column two. Cross out, and one remains. This position carries a minus, so the cofactor is minus one. Row two, column one: three is left, the position is negative, the cofactor is minus three. And row two, column two gives two, positive, so two. Cofactor array: four and minus one on top, minus three and two below. Now flip. Four and minus three on top. Minus one and two below. That is the adjoint.

4:31For a two by two there is a picture that gets you there in one move, and it is worth learning properly. Draw a loop around the two entries on the main diagonal - the one in the top left corner and the one in the bottom right. Those two swap places. Draw a second loop around the other two, the ones off the diagonal. Those two stay where they are and change sign.

4:57Swap the diagonal pair. Flip the sign of the other pair. Done. Try it on our array. The diagonal pair is two and four, so they swap: four in the corner, two in the other corner. The off-diagonal pair is three and one, so they keep their seats and become minus three and minus one. Four and minus three on top, minus one and two below. Which is what the long way gave us.

5:28Now, that picture is so satisfying that people reach for it at order three. It does not work there, and I want to show you how badly. Take the same five hundred and twelve arrays of noughts and ones, at order three, and build the two-arrow answer instead of the real adjoint. It multiplies back correctly on thirty-two of them. And on how many is it actually the same array as the real adjoint? Fourteen.

5:58At order two, over every two by two built from minus one, nought and one - eighty-one arrays - the picture is right on all eighty-one. So this is not a rule with exceptions. It is a rule about two by twos that happens to be stated in a way that tempts you to carry it upstairs. At order three there is no picture. Nine cofactors, then the flip. Here is the claim that makes the whole construction worth having.

6:31Multiply an array by its adjoint. What you get is the array's own determinant, multiplying the identity of the same size. Every entry on the diagonal is that one number. Every entry off the diagonal is nought. And the same is true the other way round. Adjoint first, array second. Same answer. That should make you sit up. Multiplying two arrays is not usually something you can do in either order and get away with. Here it is.

7:05One small thing about notation before we go on. We are handling arrays now, so they sit inside square brackets. The only vertical bars on the page are around the determinant on the right hand side, because that is a single number, not an array. So why is that true? Not verified on an example - forced. Look at the entry that ends up in row one, column one of the product. To get it, you take the first row of the array and run it into the first column of the adjoint.

7:43Now remember what the flip did. The first COLUMN of the adjoint is the first ROW of the cofactor array. So that entry is the first row's entries, paired with the first row's own cofactors, added up. You have seen that sum before. Entries of a line against the cofactors of that same line. It is the determinant. Run the same argument on the second diagonal entry, and the third. Each one is a different line paired with its own cofactors. Each one is the determinant again.

8:20That is the first half of the diagonal question answered. Now the harder half. Notice that those three diagonal entries were not the same sum. They came from three different rows. Row one against its own cofactors. Row two against its own. Row three against its own. Three genuinely different arithmetic expressions. And they come out equal. All three give the same number. That is not a coincidence and it is not new. It is exactly the fact that you can evaluate a determinant along whichever line you please and get the same answer.

9:04So the diagonal of the product is not merely full. It is uniform. One number, repeated. Now the off-diagonal entries, and this is where something you were shown earlier finally gets spent. Take the entry in row one, column two of the product. First row of the array, run into the second column of the adjoint. The second column of the adjoint is the second row of the cofactor array.

9:36So that entry is the first row's entries, paired with the SECOND row's cofactors. Mismatched. Entries from one line, cofactors borrowed from another. And you know what a mismatched sum gives. Nought. Every time. Whatever the numbers are. Six off-diagonal positions. Six mismatched sums. Six noughts. Let us do that with no numbers at all, to be sure it is the argument doing the work. Take a three by three of nine letters. Build the adjoint symbolically. Multiply the two together and look at the nine entries of the answer.

10:18All three diagonal entries come out as the determinant. Not equal to it after simplifying - identical to it, term for term. All six off-diagonal entries come out as nothing at all. Every term cancels. Zero terms left standing. The same nine, done in the other order, give the same nine. And a numerical sweep agrees. Across nineteen thousand six hundred and eighty-three three by three arrays, the construction gives the value times the identity on every single one, in both orders.

10:59Now go back and ask the question we started with. Why the flip? Because without it, the first column of the second array would be the first column of the cofactor array - and running the first ROW of the original into that gives you a sum with no name. The flip is what puts the row of cofactors where the multiplication needs a column. It exists to line matched things up with matched things.

11:28It is not decoration. It is the entire mechanism. Here is a nice consequence. You can flip first and then take cofactors, instead of the other way round, and you get exactly the same adjoint. All five hundred and twelve times out of five hundred and twelve. But do something else - reflect about the other diagonal instead, or simply reverse the rows - and it collapses. Those give the right answer on one hundred and fifty-six and eighty-six of the five hundred and twelve.

12:06Let us check a three by three all the way through. Here is the array. Its determinant, expanded along the top row, is one times seven, minus three times one, plus three times minus one. Seven, minus three, minus three. One. Nine cofactors. Seven, minus one, minus one across the top. Minus three, one, nought in the middle. Minus three, nought, one at the bottom. Flip them. Seven, minus three, minus three. Minus one, one, nought. Minus one, nought, one.

12:43Multiply the original by that, and you get the identity. Now be careful here, because this example is quietly rigged. The determinant is one. So the value times the identity IS the identity, and the very factor the theorem is about has gone invisible. An example with determinant one cannot show you the theorem. It can only fail to contradict it. So take one where the determinant is nought. Two and three on top, minus four and minus six below. Its determinant is two times minus six, minus three times minus four. Minus twelve plus twelve. Nought.

13:31Does it have an adjoint? Of course it does. Nothing in the construction ever divides by anything. Minus six and minus three on top, four and two below. Multiply. Both orders. You get the array of all noughts. And that is the theorem behaving perfectly. The determinant is nought, so the determinant times the identity is nought times the identity, which is the zero array. This is the case that turns the statement from a coincidence into a theorem. It survives the situation where nothing is invertible at all.

14:13One more, with a determinant that is neither nought nor one, so you can see the number actually appear. Here is a three by three with two noughts down its middle column. Its determinant is eleven. Its adjoint, row by row: nought, three, two. Minus eleven, one, eight. Nought, minus one, three. Multiply either way round and every diagonal entry is eleven, every other entry is nought. Eleven times the identity, exactly as promised.

14:47There is the factor, sitting on the diagonal where the first example hid it. One last question, and it is the sort a paper likes to ask. The adjoint is an array. What is ITS determinant? You can get it from the theorem. Take determinants of both sides. The left becomes the determinant of the array times the determinant of the adjoint, because the determinant of a product is the product of the determinants.

15:19The right, for an array of order three, becomes the determinant cubed. Cancel one factor and the adjoint's determinant is the original's, squared. In general: raised to one less than the order. I want to be honest about what that rests on. Two statements were used and neither was proved here - that the product rule holds, and the theorem itself, which we argued at order three rather than proved in general.

15:49But the rule is testable, so we tested it. Across every three by three with a determinant that is not nought - one hundred and seventy-four of them - and every such two by two - forty-eight - the exponent one less than the order is right every time. The tempting alternatives are not. Saying the adjoint's determinant is the original squared works at order three, and fails at order two. Saying it is just the determinant does the reverse. Neither survives both.

16:26So where does this leave us? You can build the adjoint of any square array: cofactor every position, then flip. You can do a two by two in one move with the two loops, and you know not to try that at order three. You know what it multiplies back to, in either order, and you know why - matched sums on the diagonal, mismatched sums everywhere else. And now look at that theorem again with fresh eyes. The array, times the adjoint, equals the determinant times the identity.

17:02If the determinant is not nought, divide by it. The array, times the adjoint over the determinant, equals the identity. You just built an inverse. And you can see, written into that division, precisely which arrays have one and which do not. That is the answer to the question we opened with, and it is one line away.