Orders one, two and three, and expanding along a chosen row

0:00A determinant is one number belonging to a square array of numbers. Today: how you actually get that number, at the three sizes you will meet. Start at the bottom, where it looks like a joke. A one by one array holds a single entry, and its determinant is that entry. Take the five arrays holding minus two, minus one, zero, one and two. All five give back exactly what they hold.

0:30That line is not a joke. It is the bottom rung of a ladder, and without a bottom rung the whole thing has nothing to stand on. Because here is the shape of what is coming. Order two is going to be defined by being taken apart into order one. Order three is going to be defined by being taken apart into order two. There is no separate formula for the three by three. There is a reduction, applied twice.

1:01Almost everyone learns this as a formula and loses the reduction, and with it loses a freedom worth having. So let us build it the other way. Order two. Four entries: a and b along the top, c and d below. Two tracks run across the array. One goes from the top left entry down to the bottom right. The other goes from the top right down to the bottom left.

1:31Each track gives one product. a times d along the first. b times c along the second. And the value is the first product minus the second. a d minus b c. Which one gets subtracted is the whole content of the rule, and it is exactly the part a picture of two crossing arrows does not tell you. Draw the tracks by all means. Say the sign out loud.

2:00Two multiplications and one subtraction, and no rearranging of anything. Two evaluations, side by side, because they land in very different places. First, with numbers: two and four on top, minus one and two below. Two times two is four. Four times minus one is minus four. And four minus minus four is eight. Now with letters: x and x plus one on top, x minus one and x below.

2:34The first product is x times x. The second is x plus one times x minus one, which is x squared minus one. Subtract, and the x squared cancels. The value is one. Every trace of the letter is gone. Run that in letters and read the answer as an expression rather than a number: it carries no term holding the letter at all. Not for some values of x. For every value.

3:04Which is worth saying, because a student who reaches an answer with no letters left in it usually assumes a mistake. It is not a mistake. It is a determinant being an identity in disguise. Order three. Nine entries, and the question is what number they carry. The move is this. Pick one line of the array -- one row, or one column. Walk along it. Each entry you pass is going to contribute one term.

3:37The term is that entry, multiplied by a smaller determinant, with a sign attached. Three terms, added. That is the whole of it. Notice what has happened: the order three question has become three order two questions, and each of those becomes two order one questions, and order one is where we started. So there are exactly two things left to pin down. Which smaller determinant belongs to each entry, and which sign it carries.

4:12The smaller determinant first, because there is a misreading here that costs people marks for years. Choose an entry. Now delete the whole row it sits in, and the whole column it sits in. Not the entry. The row and the column. Nine entries go in. Five come out -- three in the row, three in the column, and the entry itself counted once, not twice. Four survive, and they close up into a two by two.

4:43That two by two is the smaller determinant belonging to that entry. Compare it with the misreading. If you delete only the entry, eight survive. And there is no whole number at all whose square is eight, so eight entries cannot be arranged into a square array. That reading does not give a wrong answer. It gives no answer. Asked for a value on all five hundred and twelve of the arrays we are about to sweep, it refuses all five hundred and twelve.

5:17A misreading that cannot even be carried out is the friendliest kind. The dangerous ones are coming next. Now the sign. Every entry sits at a position, described by two subscripts: which row, then which column. Add those two numbers. If the total is even the sign is plus. If it is odd the sign is minus. Write out the nine totals. Along the top row, two, three, four. Along the second, three, four, five. Along the third, four, five, six.

5:54And the signs that come off them: plus, minus, plus. Then minus, plus, minus. Then plus, minus, plus again. Five positions carry a plus and four carry a minus, and they alternate like a chessboard. Two of the three rows read plus, minus, plus -- and one does not. That one row is where almost every sign error in this subject lives. So say it precisely. The sign belongs to the POSITION. It does not belong to the step you are on.

6:29Along the top row those two descriptions happen to agree. Along the second row they are exact opposites. Let us run it once, slowly, on a real array. Take the array holding two, one, zero along the top; three, minus one, zero in the middle; and one, four, five at the bottom. Step one. Take the first entry of the top row, which is two. Its subscripts are one and one, and one plus one is two, which is even, so the sign is plus.

7:04Delete its row and its column. What is left is minus one, zero above four, five. That smaller determinant is minus one times five, minus zero times four: minus five. So the first term is plus two times minus five, which is minus ten. Step two. The second entry is one, at position one, two. One plus two is three, which is odd, so the sign is minus. Delete its row and column: three, zero above one, five, which comes to fifteen. The term is minus one times fifteen, or minus fifteen.

7:47Step three. The third entry of the top row is zero. Its term is zero, whatever its sign and whatever its smaller determinant. Step four. Add them. Minus ten, minus fifteen, plus nothing. Minus twenty-five. Now the claim that makes this worth learning properly. Take the same array and expand along its second row instead. Different entries, different smaller determinants, and different signs -- minus, plus, minus this time. Same answer. Minus twenty-five.

8:25Or go down a column instead of across a row. Same answer. A square array of order three has six lines you could expand along: three rows and three columns. A two by two has four. And the claim is that all six lines give the same number. That is a strong claim, and it is exactly the sort of claim that is easy to check dishonestly. If I expand along one line, then expand along another line with the same rule, and announce that they agree, I have compared a procedure with itself.

9:03So let us do it properly. Two things get taken away from me here. The first: nothing below decides an answer by expanding. Every array gets a value from a completely separate route -- a signed sum over the arrangements of the column indices, which never deletes a row, never deletes a column and never calls anything that does. Only then is an expansion compared against it. The second: the expansion rule itself becomes a parameter. Seven readings of take it apart along a line, run side by side.

9:44The real one: the sign from the sum of the two subscripts. No sign at all -- every term simply added. Plus, minus, plus along every line, whichever line it is. The sign turned over: minus where the real rule says plus. The sign from the product of the two subscripts instead of the sum. The sign from the row subscript alone, ignoring the column. And the one that deletes only the entry.

10:14Each of the seven is asked the same four questions over the same arrays: how often is it right along the first row, how often is it right along all six lines, how often does it agree with ITSELF along all six lines, and how often does it refuse to answer. That third question is the one people forget. Being consistent and being right are different properties, and a rule can have either without the other.

10:46The sweep is every three by three array built from zero and one: five hundred and twelve of them, all six lines each. The real reading: five hundred and twelve right along the first row, five hundred and twelve right along all six lines, five hundred and twelve self-agreeing, none refused. Perfect on every question. No sign at all: right on three hundred and forty-three. But look at its third number -- it agrees with itself on all five hundred and twelve.

11:23Read that again. A rule can give you the same answer along every one of the six lines and still be the wrong rule. Consistency is not correctness. If your only check is that two expansions agree, this rule passes your check and hands you the wrong number a hundred and sixty-nine times. The sign from the product of the subscripts: right along the first row on all five hundred and twelve, and right along all six on four hundred and sixteen.

11:56The sign from the row alone: three hundred and forty-three, then two hundred and sixty-five. And deleting only the entry: nothing, nothing, nothing, and five hundred and twelve refusals. Two of the seven deserve a scene each, and this is the first. Plus, minus, plus along every line. It is the commonest wrong rule in this whole topic, and it is wrong in the most inconvenient possible way. Along the first row it is right on all five hundred and twelve. Every single one.

12:35Along all six lines it is right on three hundred and thirty-eight. So if you only ever expand along the top row -- which is what most worked examples do, and what most students do -- this rule is indistinguishable from the correct one. It will never fail you. It cannot fail you. The moment you take the freedom this topic is offering, and expand along the second row to use its zeros, it hands you the answer with its sign turned over.

13:07In letters, that is exactly what happens: run this reading along the second row of nine distinct symbols and you get the correct expression with every sign reversed. Not sometimes. Always. Which is why the sign is derived from the subscripts every time rather than quoted from a pattern. The pattern is right for the row it was learned on and wrong for the next one down. The second one is stranger, and it is the reason the sweep is run at more than one size.

13:43Turn every sign over. Where the real rule says plus, say minus. At order three, this reading is right on all five hundred and twelve arrays, along all six lines. It is not approximately right. It is exactly the real rule. At order two it is right on thirty-three of eighty-one. The reason is worth a moment. Turning the sign over at the top level multiplies the answer by minus one. But the smaller determinants inside are computed by the same reading, so each of them is also multiplied by minus one.

14:24At order three the two reversals cancel and the answer is untouched. At order two there is only one reversal, and nothing cancels it. So a checker that only ever looked at three by threes would certify this rule as correct. The defect is not visible at that size. That is the general lesson, and it is bigger than determinants. If a rule is defined by recursion, testing it at one depth tests one depth.

14:58Back to the real rule, and to the claim that all six lines agree. Counting is evidence. Nineteen thousand six hundred and eighty-three three by threes built from minus one, zero and one, expanded along all six lines each: right every time, self-agreeing every time, never refused. But a sweep over a finite collection is not a proof about all arrays. So run the expansion once on nine distinct symbols, and compare the expressions rather than the numbers.

15:35All six lines give ONE expression. Not six expressions that happen to agree on the arrays we tried -- one expression. And look at what that expression is. Six terms. Three carry a plus and three carry a minus. Every term is a product of exactly three entries. In every term, the three entries come from three different rows, and from three different columns. So the six terms are the six ways of picking one entry from each row and each column, half of them added and half subtracted. That is what the number IS, and every one of the six routes is just a different order of assembling it.

16:20Which is the honest reason the routes agree, and it is a better reason than a table of agreeing answers. Now cash the freedom in. Every zero entry on the line you expand along kills a term before you compute anything. A zero entry does not make the value zero -- it makes one term of three disappear. So choose the line with the most zeros. Back to that array: two, one, zero on top; three, minus one, zero; one, four, five. Its last column holds zero, zero, five.

17:00Expand along that column and only one entry survives. Its position is three, three -- six is even, so the sign is plus. Its smaller determinant is minus five. Five times minus five is minus twenty-five. One smaller determinant instead of two, and the same answer we got the long way. Over all nineteen thousand six hundred and eighty-three arrays: always taking the top row costs thirty-nine thousand three hundred and sixty-six smaller determinants. Always taking the emptiest line costs twenty-two thousand and eighty.

17:42Twelve thousand four hundred and forty-two of those arrays cost strictly more if you always take the top row. And that saving only exists because all six lines are allowed. Learn one row's expansion as a formula and you have thrown it away. Two loose ends, both of them things people get wrong confidently. First: a determinant with a zero in it is not zero. Of the nineteen thousand six hundred and eighty-three arrays, nineteen thousand one hundred and seventy-one hold at least one zero, and eleven thousand six hundred and sixteen of those have a value that is not zero.

18:27It runs the other way too. Three hundred and twenty of these arrays have no zero entry anywhere and a value of zero. There is a whole family like that. Put zeros down the diagonal and make every entry the negative of its mirror image across that diagonal. All one hundred and twenty-five such arrays built from minus two through two come out at zero. The matching symmetric family, built the same way but without the sign flip, gives a non-zero value sixty-four times.

19:04Second loose end: multiplying every entry by the same number. Not by two. By two to the power of the order. One and two above three and four has value minus two. Double every entry and the value is minus eight -- and two squared is four, and four times minus two is minus eight. At order three the exponent is three. Over all five hundred and twelve arrays and five scale factors -- two thousand five hundred and sixty cases -- the order as exponent is right in all two thousand five hundred and sixty.

19:44Just multiplying the value by the scale factor is right in two thousand and thirty-eight of them. Using two as the exponent, which is correct at order two, is right in one thousand eight hundred and sixty-four. So the exponent is the order. It is not two, and it is not one. So take stock. You have a ladder. Order one is an entry. Order two is two products and a subtraction. Order three is three terms, each an entry times a smaller determinant times a sign read off the position.

20:25You have six routes through order three and permission to take whichever one has the most zeros in it, which is not a convenience but a saving you can count. You have a sign rule that survives leaving the top row, which is more than can be said for the pattern most people carry. And you have one small skill worth naming. If two determinants are set equal and one of them holds an unknown squared, undoing the square gives back two answers, not one. A determinant equal to twenty-seven with an unknown down its diagonal and three in both corners gives an unknown squared equal to thirty-six -- and that is plus six and minus six.

21:12What you do not have yet is any reason why the six routes agree beyond the six-term picture, and you do not have the shortcuts that would let you see a value is zero without computing it. Those arrive later. What arrives today is the reduction, and the freedom it hands you.