Equality as two demands, and why the orders have to agree before the entries are looked at

0:00Two rectangular arrays of numbers, and one equals sign between them. It looks like the most harmless statement in the whole subject. It is not. That single sign is making two separate demands, in a fixed order, and the second one is not even allowed to be asked until the first has been answered. Here are two arrays. Three, two, on the top row; zero, one, underneath. And beside them, the same again. Three, two; zero, one.

0:33Both are two by two. Every position holds what the position facing it holds. So the sign is honest: these are equal. Now watch how little has to change for that to stop being true. Keep the first array exactly as it was, and in the second one, swap the two entries on the top row. Two, three; zero, one. Nothing has been added and nothing has been taken away. Both arrays are still two by two. Gather up the four entries of each and you get the same collection: a zero, a one, a two and a three, on both sides.

1:15And they are not equal. Two of the four positions disagree - the two on the top row - and they disagree in the most annoying way possible, by holding each other's entries. The three is in the second array; it is simply not where the three has to be. So the entries are not the content of an array. Which entry sits where is the content, and equality is a statement about positions from beginning to end.

1:45That is easy to say and easy to nod at, so let me say how common the trap is. Take every two by two array you can build out of the three values zero, one and two, and put each one against every other. That is six thousand five hundred and sixty one ordered pairs. Eighty one of them are equal - each array with itself. Six hundred and thirty nine of them hold exactly the same collection of entries.

2:14Subtract, and five hundred and fifty eight pairs hold precisely the same values as each other and are still not equal. Nearly seven of those for every one genuine equality. And the closest of them differ at only two positions. Now the demand that comes first, and the one that gets skipped. Here is an array with two rows of three: one, two, three, over four, five, six. And here is one with three rows of two: one, two, over three, four, over five, six.

2:51Six entries each. The same six entries, gathered up: one, two, three, four, five, six on both sides. If entries were what mattered, we would be finished, and the answer would be yes. The answer is no, and not by a narrow margin. The first array's order is two by three. The second's is three by two. And once you write out the positions each of them actually has, four of them have no counterpart at all. The first array has a position in the third column; the second has no third column to answer it. The second has a third row; the first has no third row.

3:35So the demand about entries cannot even be stated here. It says: at every position, the two entries agree. On these two arrays there is no such thing as every position, because the two of them do not have the same positions to talk about. That is why the order of the two demands is not a matter of taste. The first one is a gate. It asks a question about the two shapes and nothing else, and it has to be answered before the second demand means anything.

4:11It is worth seeing exactly what goes wrong if you skip the gate, because the mistake is a quiet one. Suppose you compare two arrays by walking through the positions they have in common and checking those. It sounds reasonable. It is what a careless eye does. Put the two by three array from a moment ago against a two by two: one, two, three, over four, five, six, against one, two, over four, five.

4:42They share four positions - the top left two by two block - and on all four of them the entries agree. So the careless comparison returns yes. The two arrays are not equal. Two positions of the larger one have no counterpart, and a demand that quietly ignores whatever it cannot pair up will agree with almost every pair you show it and disagree with exactly the pairs this matters for.

5:12Notice also what the gate does not compare. It does not compare how many entries there are. Two by three and three by two both hold six. It compares two counts, in order: how many rows, then how many columns, and both have to match. So: gate first, entries second. Now the part that makes equality useful rather than merely careful. Once the gate has been passed, the second demand runs at every position at once. And that means one equals sign between two arrays is not one statement. It is one statement per position, all asserted together.

5:57Here is a three by two array of six letters - a, b, over c, d, over e, f - set equal to a three by two array of values: minus three halves, zero, over two, root six, over three, two. The gate passes. Three rows against three rows, two columns against two columns. So the second demand applies, at all six positions. Six positions, six statements. a is minus three halves. b is zero. c is two. d is root six. e is three. f is two.

6:34There is nothing to solve. Every entry on the left is a single letter standing alone, so each letter simply is the entry facing it, and the whole equality is read straight off. Two things worth noticing. Look at c and f: both take the value two. Equality has no objection whatever to that. Two letters may hold the same value; what they may not do is sit in the same position.

7:04And look at d. It takes root six, which is not a fraction and never will be. The demand never asked what kind of number was sitting there. It asked whether the two entries were the same entry, and that question makes perfect sense for a root. Reading straight off is a privilege, not a method, and it belongs to exactly one shape of question: the one where every entry on a side is a single letter with nothing done to it.

7:38Change one entry of that array from c to two c, and five of the six are still bare letters but the sixth is not, and the reading-off stops. Two c equals two is a small piece of work. It is still one statement per position; it is just no longer a statement you can take without doing anything. Here is a bigger one. Two three by three arrays. On the left: x plus three, z plus four, two y minus seven, on the top row. Minus six, a minus one, zero, on the middle. b minus three, minus twenty one, zero, on the bottom.

8:19On the right: zero, six, three y minus two. Minus six, minus three, two c plus two. Two b plus four, minus twenty one, zero. Both are three by three, so the gate passes and there are nine statements. Not one of the nine entries on the left is a bare letter, so nothing here is read off. Every one of them has to be turned into an equation and solved.

8:48And they solve, one at a time. x plus three is zero, so x is minus three. z plus four is six, so z is two. Two y minus seven is three y minus two, so y is minus five. a minus one is minus three, so a is minus two. Zero is two c plus two, so c is minus one. And b minus three is two b plus four, so b is minus seven.

9:15Six unknowns, six values, and every one of them came out of a single position. That is the thing to hold on to: each of the six informative positions mentions exactly one unknown, which is why this one never needs any elimination at all. Six values from nine positions. So what were the other three doing? Look at the middle left. Minus six, against minus six. Look at the bottom middle. Minus twenty one, against minus twenty one. Look at the bottom right. Zero, against zero.

9:55Those three positions are still making demands. They are just making demands that were already met before anybody arrived. Subtract one side from the other at each of them and you get nothing at all - not a small number, not an approximation. Zero, with no letters left over. That is how you tell. A position carries information exactly when the difference between its two entries is not identically zero. Three of the nine are identically zero, six are not, and the six are the whole content.

10:34This is worth being deliberate about, because there is a real temptation to count nine positions and go looking for nine facts. There are six. The other three are the equality confirming that it has nothing to add. Now here is a small thing that follows, and that catches people out. A position carries information when the difference is not zero. Notice what that definition does not say. It does not say the position has to hold a letter.

11:09Go back to the very first near miss - three, two, zero, one against two, three, zero, one. Two arrays of pure numbers, no unknowns anywhere. Run the same machinery on them. The gate passes, so there are four statements. Two of them are identically zero: the bottom row agrees with itself. The other two are not. Three equals two. And two equals three. Those are equations. They hold no unknowns between them - zero unknowns in the whole system - and they cannot be satisfied. So the near miss is not merely a pair of unequal arrays. It is a system of equations with no solution, and that is a stronger and more useful way to say the same thing.

12:00It also tells you what an equals sign between two arrays really is. It is an assertion. Sometimes it is true, sometimes it is false, and sometimes it holds letters, in which case asking whether it is true becomes asking what the letters would have to be. The nine-position example was gentle, because each position handed over one unknown by itself. Most of them do not. Here is a two by two: two a plus b, and a minus two b, on the top row. Five c minus d, and four c plus three d, underneath. Set equal to four, minus three, over eleven, twenty four.

12:47Four positions, four equations, and every single one of them carries information. Four unknowns between them: a, b, c and d. Not one of the four equations can be solved on its own, because every one of them names two unknowns. But look at which two. The top row's equations mention a and b, and never c or d. The bottom row's mention c and d, and never a or b. So this is not one problem in four unknowns. It is two problems in two unknowns, standing next to each other and sharing nothing.

13:28Two a plus b is four, and a minus two b is minus three. Take the first, double the second, and add: five a is ten, so a is one, and then b is two. Five c minus d is eleven, and four c plus three d is twenty four. Triple the first and add the second: nineteen c is fifty seven, so c is three, and d is four.

13:55a is one, b is two, c is three, d is four. And the important part is not the answers - it is that you were never obliged to hold four unknowns in your head at once. The equality told you it split, and it told you by which unknowns turned up in which positions. Now one that does not split, so you can see the difference. Two by two again. a minus b, and two a plus c, on the top. Two a minus b, and three c plus d, underneath. Against minus one, five, over zero, thirteen.

14:34Four equations again, and again every one of them names exactly two unknowns, so again not one of them can be solved alone. But this time they are all joined up: a is in three of them, c is in two, and following the shared unknowns from any equation reaches all four. One block, not two. So there is no half of the problem to peel off. You have to find a pair of positions that determine something between them, and there is exactly one such pair here: the top left and the bottom left.

15:12a minus b is minus one. Two a minus b is zero. Subtract the first from the second and the b disappears: a is one. Then b is two. And now the rest falls, because a is known. Two a plus c is five gives c equals three. Three c plus d is thirteen gives d equals four. Same four answers as before - one, two, three, four - reached in a completely different way, because the equations were wired together instead of sitting in two independent halves. Reading a matrix equation position by position, left to right, top to bottom, is a habit worth breaking. The positions are asserted all at once, and you may attack them in any order you like.

16:04So far every one of these has had exactly one answer. That is not guaranteed, and it is worth seeing both of the other outcomes. Take a two by two whose positions give three statements: x plus y is six. Five plus z is five. And x times y is eight. The fourth position agrees with itself and says nothing. The second one is easy: z is zero. The first one is not enough by itself - x plus y is six leaves a whole line of possibilities. Run elimination on the two linear statements and it will tell you so: it pins z, and it reports x and y as still loose.

16:51The third statement is what pins them, and it is not linear. x times y is eight. Elimination has nothing to say about a product; you have to go looking. Look, then. Which whole numbers add to six and multiply to eight? Two and four. And four and two. Two answers. x is two with y four, or x is four with y two. Both satisfy every position of the original equality, so both are correct, and if you were asked for the value of x, in the singular, the honest reply is that there are two.

17:30And this is a property of these particular numbers, not of the method. Ask for a sum of six and a product of nine and there is exactly one answer, three and three. Ask for a sum of six and a product of ten and there is none at all. Which brings us to the last outcome, and the sharpest one. A two by two. Three x plus seven, and five, on the top row. y plus one, and two minus three x, underneath. Set equal to zero, y minus two, over eight, four.

18:07Both are two by two, the gate passes, four statements. Start anywhere. Top right: five equals y minus two, so y is seven. Bottom left: y plus one equals eight, so y is seven. Both positions that mention y agree perfectly. If you had checked only y, you would have found nothing wrong and moved on satisfied. Now x. Top left: three x plus seven is zero, so x is minus seven thirds. Bottom right: two minus three x is four, so x is minus two thirds.

18:47Minus seven thirds and minus two thirds. Those are not nearly the same number and they are not a rounding difference; they are a whole five thirds apart, and no value of x can be both. So there is no assignment at all. Not a hard one, not an ugly one - none. And notice how that verdict was reached. It was not reached by trying things and failing to find one. It was reached by pointing at two specific positions, the top left and the bottom right, and showing that they demand different values of the same letter.

19:26That is the difference between not knowing an answer and knowing there is not one. And it is the reason it is worth thinking of a matrix equation as a system rather than as a puzzle: a system can be inconsistent, and inconsistency is something you exhibit, not something you give up and conclude. One last thing about this one. The consistent letter is not a lucky accident here; it is the trap. An equality can be perfectly well behaved in one unknown and impossible in another, and finding one letter that works tells you nothing whatever about the rest.

20:10So. One equals sign between two arrays. It makes two demands, and they are asked in order. First, the two orders must agree - both counts, rows then columns, and not the number of entries, because six entries can be laid out as two rows of three or three rows of two and those are different arrays. That demand is a gate. If it fails, the second demand has nothing to talk about, because the two arrays do not share their positions.

20:42Second, once through the gate, every position must agree with the position facing it. Not the collection of entries - the positions. Two arrays can hold precisely the same values and be unequal, and among small arrays that happens far more often than genuine equality does. And then the payoff. Because the second demand runs at every position at once, one equals sign between two arrays is one equation per position, all asserted simultaneously. Nine positions, nine equations. Four positions, four equations.

21:24Some of those equations carry nothing, and you can tell which by subtracting and getting an honest zero. Some are solvable on sight. Some split into independent halves that you can attack separately, and some are wired together so that you must find a pair that yields to nothing but each other. And the system has exactly three possible verdicts. One assignment. More than one, when something in it is not linear and elimination leaves letters loose. Or none at all, when two positions demand different values of the same letter.

22:03Every one of those outcomes is available to you only because the equals sign was a system in the first place. That is the whole reason a single sign between two rectangles of numbers can be solved. Two demands, in order. And then, at every position at once, an equation.