Why cofactors borrowed from the wrong row always sum to zero

0:00Here is where we got to last time. Take any single line of a three by three - any row, any column - and pair each of its entries with its own cofactor. Multiply each pair. Add the three products. Out comes the value of the whole array. Six lines, three rows and three columns, and every one of them gives the same answer. We checked that on nineteen thousand six hundred and eighty-three arrays, and it held on all of them.

0:33So the sentence is short: entries of a line, times the cofactors of that same line, added. Notice how much work the word SAME is doing in that sentence. Everything today is about what happens when you take it away. Here is the mischief. Take the entries of the top row. And now borrow the cofactors of the BOTTOM row. Not the ones that belong to these entries. Somebody else's.

1:04Line them up anyway. First entry of the top row, times the first cofactor of the bottom row. Second times second. Third times third. Add. Stop and ask yourself what you expect. There is no obvious reason for this to be anything in particular. It is a sum of three products of numbers that were never meant to meet. You might expect junk. Some number that depends on the array, different every time.

1:35It is nought. Every time. For every array, for every choice of the two lines, whatever the nine numbers were. That ought to feel surprising, and if it does not, hold on to the surprise, because the reason it happens is genuinely lovely. Before the reason, the same discipline as always. The statement is true, which means a check that computes a mismatched sum and finds nought has found only its own arithmetic.

2:08So the PAIRING becomes the thing under test. Six ways to multiply the entries of a line by cofactors and add. One: its own line's cofactors. The correct pairing. Two: a different line's cofactors. The mismatched one. Three: its own line's MINORS, with no sign attached. Four: its own line's cofactors, but read backwards - first entry against the last cofactor. Five: a row's entries against the matching COLUMN's cofactors. That one is not invented; it is one of the options in a standard exam question.

2:49Six: a different line's cofactors, rotated by one place. And four questions asked of each, over all five hundred and twelve three by threes built from nought and one, along every line of every array. Does it give the value every time. Does it give nought every time. Does it at least give ONE answer whichever line you pick. And does it give the value on the arrays whose value is not nought.

3:20That last column is the one that matters, and I will say why when we get there. The correct pairing: five hundred and twelve, three hundred and thirty-eight, five hundred and twelve, one hundred and seventy-four. The mismatched pairing: three hundred and thirty-eight, five hundred and twelve, five hundred and twelve, nought. Read those two rows against each other, because they are a mirror. The correct one gives the value on all five hundred and twelve arrays. The mismatched one gives nought on all five hundred and twelve.

4:00But look at the mismatched row's first number. Three hundred and thirty-eight arrays where the mismatched sum DOES give the value. That is not a crack in the result. Of the five hundred and twelve arrays, exactly three hundred and thirty-eight have value nought, and one hundred and seventy-four do not. On those three hundred and thirty-eight, nought IS the value. Which is exactly why the last column exists. On the arrays whose value is not nought, the mismatched pairing lands on it nought times out of one hundred and seventy-four.

4:39The other four readings. Minors with no sign: three hundred and twenty-four, two hundred and seventy-four, three hundred and twenty-four, fifty. Cofactors read backwards: two hundred and two, four times over - and nought in the live column. It never once lands on a value that is not nought. A row against the matching column's cofactors: one hundred and seventy-eight, one hundred and twenty-six, one hundred and eighty-two, fifty-six. It is right sometimes. That is worse than being wrong, and it is why that exam option catches people.

5:21And rotated by one: fifty, fifty, fifty, nought. One reading of the six gives the value along every line of every array. One gives nought along every line of every array. And three of them never land on a live value at all. Now the reason. It takes three lines and it is worth every second. Write the mismatched sum out with the definitions put back in. Each borrowed cofactor is a signed minor: minus one to the power of its two subscripts added, times the determinant left when its row and its column are struck out.

6:04The entries stay where they are. They are the top row's. The signs and the minors all belong to the bottom row. So you get three terms. First entry, times plus, times a two by two. Second entry, times minus, times a two by two. Third entry, times plus, times a two by two. Look hard at that expression, because it is trying to tell you something. Three entries. Alternating signs. Each multiplying a two by two made by deleting a row and a column.

6:40That is an expansion. That is exactly the shape of a determinant being taken apart along a line. So the question stops being what does this sum equal, and becomes: an expansion of WHAT? The minors came from the bottom row, so the array is being expanded along its bottom row. Which means the bottom row's entries have been replaced by whatever is multiplying those minors. And what is multiplying them is the TOP row.

7:14So here is the array. Top row: the original top row. Middle row: untouched. Bottom row: the top row again. The same three numbers, written down twice. That is the whole argument, and it is invisible in the algebra and obvious in the picture. We did not take this on trust. For every one of the five hundred and twelve arrays, and for every ordered pair of two different lines, we computed the mismatched sum and separately computed the value of the array with the line written twice - by a route that never forms a cofactor at all.

7:55Six thousand one hundred and forty-four comparisons. Six thousand one hundred and forty-four agreements. Nought disagreements. Rows and columns alike. And the same six thousand one hundred and forty-four checks, run again with the cofactors' signs left off, disagree one thousand one hundred and fifty-two times. So the alternating signs are not decoration. They are what makes the expression an expansion. One step left, and it is the step to be careful about.

8:33The array has two rows the same, so its value is nought. That sentence is doing a lot of work. It is not an observation about this array. It is an appeal to a general fact: any determinant with two matching rows is nought. And I want to be straight with you. Nothing we have established so far gives us that fact. It is being brought in from outside. You will meet it properly later, with a proper proof. Today it is an import, and pretending otherwise would leave you with an argument you cannot actually reconstruct.

9:12So let us do two things instead. Measure the fact, and then prove the smallest case of it on screen. Measuring it first, because a fact you are importing deserves the same treatment as a fact you are proving. Seven readings of what makes a determinant nought, run over nineteen thousand six hundred and eighty-three three by threes built from minus one, nought and one. Two rows the same. Two rows proportional - one a multiple of the other. Two columns the same. A row of noughts.

9:50And three that sound just as plausible. Two rows with the same total. Two rows sharing their first entry. Some entry appearing twice anywhere in the array. For each, two numbers: how many arrays meet it, and how many of THOSE have value nought. Two rows the same: two thousand one hundred and thirty-three arrays meet it, and two thousand one hundred and thirty-three of them are nought. Every single one.

10:22Two rows proportional: five thousand nine hundred and fifty-five, and all five thousand nine hundred and fifty-five nought. A wider condition, still watertight. Two columns the same: two thousand one hundred and thirty-three, all nought. Same count as the row version, and it is not the same set - fifteen hundred arrays meet the column reading without meeting the row one, and fifteen hundred the other way about. A row of noughts: two thousand one hundred and seven, all nought.

10:58Now the three that fail. Two rows with the same total: nine thousand seven hundred and fifty-nine arrays meet it, and only four thousand seven hundred and ninety-one of those are nought. Two rows sharing their first entry: fifteen thousand three hundred and nine meet it, six thousand two hundred and eighty-five are nought. And some entry appearing twice: nineteen thousand six hundred and eighty-three meet it. Every array in the sweep. Nine entries drawn from three numbers - of course something repeats.

11:36Seven thousand eight hundred and seventy-five of those are nought, which is just the number of nought-valued arrays in the whole sweep. A condition that everything satisfies explains nothing at all. Four of the seven force nought. Three do not. So the fact the argument leans on is a real one and a sharp one, and the near misses beside it are not. And now prove it where it is cheap. Order two.

12:07A two by two with its two rows the same. Its value is the first entry times the last, minus the other two multiplied. But the rows are identical, so those two products are made of exactly the same pair of letters. One product minus the same product. Nothing. Not a small number - no terms at all. Done in one line. Written the same way with the second row's letters swapped, the same expression leaves two terms, so the machinery was not simply refusing to produce anything.

12:45And over the eighty-one two by twos built from minus one, nought and one, nine have their two rows the same, and all nine are nought. That does not prove the three by three case. It does show you the shape of why it is true, and it makes the import feel small instead of hand-waved. Numbers, so you can see it happen. An array with two, minus three, five on top; six, nought, four in the middle; one, five, minus seven at the bottom.

13:19Its nine cofactors: minus twenty, forty-six, thirty; four, minus nineteen, minus thirteen; minus twelve, twenty-two, eighteen. Take the top row - two, minus three, five - and the bottom row's cofactors: minus twelve, twenty-two, eighteen. Two times minus twelve is minus twenty-four. Minus three times twenty-two is minus sixty-six. Five times eighteen is ninety. Minus twenty-four, minus sixty-six, plus ninety. Nought. Now here is the number that stops you suspecting the array was rigged. Its own value, from the correct pairing, is minus twenty-eight.

14:06Not nought. Not close to nought. This array is nothing special, and all twelve of its mismatched pairings still come out nought while all six of its matched ones come out minus twenty-eight. The zero is a property of the mismatch. It is not a property of the array. One thing people quietly assume, so let us settle it. When you borrow a cofactor for a sum it was not built for, does it change?

14:36It does not. A cofactor belongs to a position in the original array and is computed once. Borrowing means using it somewhere else, not recomputing it. And there is a reason it cannot change, which is worth seeing. The cofactor of a position in the bottom row is built by deleting the bottom row. It never reads the bottom row's numbers at all. So you can overwrite the entire bottom row with anything you like and its own cofactors do not move.

15:11We ran that. Nine thousand two hundred and sixteen comparisons of a cofactor before and after the line it belongs to was rewritten. Nine thousand two hundred and sixteen unchanged. Nought moved. Which is also, quietly, the reason the whole argument works: the substituted array has the same bottom-row cofactors as the original, so the expansion really is an expansion of it. There is a standard question that tests exactly this, and it is worth doing properly because three of its four options are decided by today's result.

15:54It shows you four sums built from entries and cofactors and asks which one is the value. The first pairs the top row's entries with the BOTTOM row's cofactors. Different lines. Nought. The third pairs the second row's entries with the FIRST row's cofactors. Different lines again. Nought. The fourth pairs the first column's entries with the first column's cofactors. Same line. That is the value, and it is the answer.

16:28We checked all four in letters, over nine distinct symbols, comparing them as expressions rather than as pictures. The first and the third have no terms left at all. The fourth has six terms and is the value exactly. That leaves the second option, and it is the interesting one. It pairs the first ROW's entries with the first COLUMN's cofactors - matching neither pattern. It has six terms. Six, like a real determinant. But two of those six terms use one entry twice, and no term of a determinant ever does that. It shares only two of its six terms with the real value.

17:13And numerically, across the five hundred and twelve arrays, that second sum happens to equal the value on three hundred and forty-eight of them, happens to be nought on three hundred and four, and is neither on one hundred and ten. Which is the nastiest kind of wrong answer. It agrees with the truth often enough that a student who tries one example walks away convinced. Put the two results side by side, because together they make a shape.

17:47Nine sums. Each row's entries against each line's cofactors. Same line down the diagonal, different lines everywhere else. Down the diagonal: the value, three times. Everywhere else: nought. The value, nought, nought; nought, the value, nought; nought, nought, the value. That held on all five hundred and twelve arrays, with nought exceptions. That is not a curiosity. That grid is the value multiplying an array with ones down the diagonal and nothing else - and an array like that is the one that leaves everything it multiplies alone.

18:28Which is the beginning of undoing a matrix. The correct pairing supplies the diagonal. Today's zero supplies everything off it. Without the zero there would be no reason for that grid to be a diagonal at all. So this is half of a machine you have not been shown yet, and it is the half that is easy to mistake for a party trick. What you can do now. Given any square array, predict a mismatched sum before you compute it, and be right. Entries of one line, cofactors of another, added - nought, always, and you can say why: it is the expansion of the array with that line written twice.

19:14Run the argument yourself. Substitute the definitions, watch the alternating signs appear, name the array, and then say honestly which step you are importing. Prove the imported step at order two in one line, so it is not a magic word. Evaluate a determinant along whichever line you like, including one full of noughts, because all six give the same answer. And read a written sum of entries times cofactors and say what it is before you touch a number - the value if the lines match, nought if they do not, and neither if the subscripts have been shuffled some third way.

19:57One last thing. This looks like a curiosity and it is not one. It is the reason a whole construction you have not been shown yet comes out diagonal, and the reason you will be able to trust that construction instead of memorising it.