Packing three equations into AX = B and reading the solution off the inverse

0:00Here is a page of algebra. Three equations, three unknowns, nine coefficients scattered across them, and three numbers on the right. Everything you have built so far has been leading here, and the first thing that happens is that the page gets shorter. Much shorter. Those three equations are about to become one equation. One array, times one column, equals one column. That is not an abbreviation. It is a true statement, and today we are going to check that it is true rather than agree that it looks tidy.

0:39And once it is one equation in one unknown, we undo it the way you undo any multiplication -- except that the thing being undone is an array, so the undoing is an inverse, and the side it goes on will turn out to matter enormously. Start by lifting three things out of the page. The nine coefficients, pulled out and left in their places, make a square array. That is the coefficient array, and it is the only square thing here.

1:12The three unknowns, written downwards instead of across, make a column. That column is the single object we are solving for. Not three unknowns any more -- one column. And the three numbers on the right make a second column. Three objects, and every one of them sits inside square brackets. Not a single upright bar anywhere. Hold onto that, because the bars are going to reappear in exactly one place, and when they do they will mean something completely different.

1:48Now the part almost everybody skips. Multiply the packing out. Row into column: the first row of the coefficient array runs across the column of unknowns, and what comes back is the first coefficient times the first unknown, plus the second times the second, plus the third times the third. That is the first equation. Not something like it -- it. Do the second row and you get the second equation, and the third row gives the third.

2:23I ran that expansion in letters, with nine named coefficients and three named unknowns, so that nothing could be quietly assumed. The product has one entry per equation. Each entry carries exactly three terms, one per unknown. Every term is exactly one coefficient multiplied by exactly one unknown -- nothing squared, nothing crossed. And the coefficients that show up in the first entry are the first row's three, and nobody else's.

2:57The packing is a translation, and it is exact. Treat it as notation and you will one day pack a system wrong and have no idea why the answer is nonsense. Before anything else happens, one number gets computed. Here is where the bars come back. Square brackets around an array mean the array. Upright bars mean its determinant -- a single number squeezed out of it. Compute the determinant of the coefficient array, and ask one question: is it nought?

3:33If it is not, everything below works. If it is, the method is unavailable and nothing you compute afterwards is worth anything. Check it first. Not at the end. I swept nineteen thousand six hundred and eighty-three coefficient arrays built from minus one, nought and one. Seven thousand eight hundred and seventy-five of them are singular. That is forty in every hundred. Forty in every hundred where a student who checks the determinant last has computed nine cofactors, assembled an adjoint, and thrown the whole lot away.

4:15Suppose the determinant is not nought. Then the coefficient array has an inverse, and here are the four lines. Line one: multiply both sides by that inverse, on the left. Say the side out loud every single time. Line two: regroup the two arrays on the left, which you are allowed to do, because multiplying arrays can be regrouped even though it cannot be reordered. Line three: the inner pair meet, and an array times its own inverse is the identity.

4:50Line four: the identity multiplied by anything leaves it alone, so the column of unknowns is standing there by itself. The column of unknowns equals the inverse, times the right-hand column. Four lines, one division, and the whole method. Now the sentence I want you to remember. The inverse goes on the LEFT, and it is not a convention. Look at the shapes. The unknowns are a column: three tall and one wide.

5:23The inverse is three by three. Put the inverse on the right of that column and the shapes do not meet. It is not wrong arithmetic. It is not arithmetic at all -- there is no such product. So I built a version where both sides ARE legal, by making the right-hand side square instead of a column, and asked both. Three thousand eight hundred and eighty-eight systems at order two.

5:52The inverse on the left solved all three thousand eight hundred and eighty-eight. The inverse on the right solved four hundred and ninety-six. At order three: eighty-nine thousand and eighty-eight systems, all of them solved on the left, one thousand six hundred and eighty on the right. Twelve in a hundred, then one in a hundred. So the wrong side is not always wrong. It is occasionally right, which is far more dangerous, because occasionally right is what gets past a spot check.

6:29And here is the part that shows the side is not arbitrary. Turn the packing round. Write the unknowns along a ROW instead of down a column, so the equation reads: unknowns, times coefficients, equals right-hand side. Run the same sweep. The inverse on the left now scores four hundred and ninety-six, and one thousand six hundred and eighty. The inverse on the right now scores three thousand eight hundred and eighty-eight, and eighty-nine thousand and eighty-eight.

7:04The two columns have swapped places exactly. The side is not something to memorise. It is decided by which side of the coefficient array the unknowns were written on, and if you can see the packing you can always work the side out again. That gives us an answer. It is also the ONLY answer, and that is a claim of a completely different kind. You cannot check it by producing an answer, because producing one tells you nothing about whether there is a second.

7:39So here is how I checked it instead. I moved to arithmetic where there are only finitely many columns to try, and then I tried all of them. Four sweeps. Thirty-four thousand two hundred and seventy-four systems, and for every single one, every candidate column that exists was substituted in and either worked or did not. Here is what came back. Where the coefficient array is not singular -- eighteen thousand four hundred and seventy-four of those systems -- the number of columns that work is one.

8:18Not usually one. One, every time, with no exceptions. Where it is singular, the number of columns that work is never one. It is nought, or it is many. Ten thousand nine hundred and sixty-seven of them have no answer at all, and four thousand eight hundred and thirty-three have more than one. So a singular system is not a harder system. It is a system that has stopped having a single answer, and no amount of care will produce one.

8:53The formula is the inverse, on the left, times the right-hand column. How do I know it is not one of the other things you could build out of the same pieces? I ran six constructions side by side over three sweeps, and asked each one whether it landed on the answer that elimination found. Not the answer the inverse found -- elimination. A completely separate route, which never forms a determinant or a cofactor at all.

9:25The inverse on the left: four hundred and thirty-two, one thousand three hundred and ninety-two, eleven thousand eight hundred and eight. Everything. The adjoint with no division: two hundred and eight, seven hundred and sixty-two, three thousand four hundred and eighty. The array itself over its determinant: eighty, two hundred and twenty-five, two hundred and eighty-eight. No inverse at all -- just the array times the column: one hundred and seventy-six, three hundred and six, four hundred and forty-four.

10:03The transpose's inverse: one hundred and ninety-two, four hundred and ninety-two, eight hundred and seventy-six. The right-hand column divided by the determinant: eighty, two hundred and twenty-eight, sixty-six. Exactly one of the six is right on every system of every sweep. Look harder at the second row, because that is the mistake people actually make. Forgetting to divide by the determinant. On the small sweep it is right forty-eight times in a hundred.

10:40Nearly half. Widen the sweep and it drops to twenty-nine. The construction did not change. The collection did. It is right exactly when the determinant happens to be one or minus one, and small collections of small numbers are full of those. Which is why a worked example whose determinant is one teaches you nothing about the division -- there is nothing to see. If you want to know whether you have understood the formula, test yourself on an array whose determinant is not one.

11:18Let us do one properly. Two equations, two unknowns. The coefficients are two and five on the top row, three and two underneath. First, the test: the determinant is four minus fifteen, which is minus eleven. Not nought, so we may proceed. The right-hand column is one and seven. The inverse is the adjoint divided by minus eleven. Multiply it into the right-hand column, on the left, and the answer comes out three and minus one.

11:53Now substitute back. Two threes plus five times minus one is six minus five, which is one. Three threes plus two times minus one is nine minus two, which is seven. Both equations hold. That substitution is not decoration -- it is the only step in the whole method that checks itself. Now three. Three, minus two, three across the top; two, one, minus one; four, minus three, two. The determinant is minus seventeen.

12:30Not nought. The right-hand column is eight, one, four. Nine cofactors, transposed into the adjoint, divided by minus seventeen -- that is the inverse. On the left of the right-hand column, and the answer is one, two and three. Substituted back into all three equations, all three hold. And I want to be plain about how that array got onto this board. I did not copy it from anywhere. I was told its nine cofactors and its determinant, and I reconstructed the array from them -- and then recomputed its determinant and all nine of its cofactors from the reconstruction, and got the same numbers back.

13:16Every entry came back a whole number, which it had no obligation to do. That is what it takes for a number on this board to be a measurement. Here is a worded one, because that is where the marks actually go. Three numbers. They add up to six. The second, plus three times the third, is eleven. And the first and the third together are twice the second. Two of those are already equations.

13:47The third one is not. It is a comparison, and it has to be moved into standard form before anything can be packed: the first, minus twice the second, plus the third, is nought. That rearrangement is the whole difficulty of worded problems, and it happens before the matrices ever appear. Now pack it. One, one, one on the top; nought, one, three; one, minus two, one. The determinant is nine.

14:19The answer is one, two and three. One plus two plus three is six. Two plus nine is eleven. One plus three is four, which is twice two. All three conditions, as stated in words. Now watch what one sign costs. In that array, the middle entry of the bottom row is minus two. Suppose it were written as plus two -- one missing minus sign, one character. The determinant is no longer nine.

14:51It is minus three. The system still has an answer, because minus three is not nought. The answer is nineteen thirds, minus six, and seventeen thirds. And here is the cruel part. Those three still add to six. The second plus three times the third is still eleven. Two of the three conditions are satisfied perfectly. The third one misses by twenty-four. One sign did not make the answer slightly wrong.

15:22It made it the answer to a different problem, and two thirds of your checks would not have noticed. One more, and it is the item that shows what the method is really about. Sometimes you are not asked to build an inverse. You are handed two arrays and told to multiply them. Here they are. Multiply, and the product is the identity. Multiply in the other order, and it is the identity again.

15:53So the second array IS the inverse of the first -- and no cofactor was computed, no adjoint assembled, no determinant divided by. Now solve a system whose coefficients are the first array. The right-hand column is eleven, minus five, minus three. Multiply the handed inverse into it, on the left, and the answer is one, two, three. Elimination, which knows nothing about any of this, agrees. The method needs an inverse.

16:27It does not need the recipe for one. Wherever an inverse comes from, the four lines are the same four lines. Last one, and at first it looks like it has nothing to do with any of this. Three equations, but every unknown is underneath a fraction bar. Two over the first, plus three over the second, plus ten over the third, equals four -- and two more like it.

16:54That is not a linear system. Doubling all three unknowns does not double the left-hand side; it halves it. But give the reciprocals their own names. Call one over the first unknown by a new letter, and the same for the other two. In the new letters, the system is completely ordinary. Coefficients two, three, ten; four, minus six, five; six, nine, minus twenty. The determinant is twelve hundred. The three new unknowns come out as a half, a third and a fifth.

17:29Which means the original three are two, three and five. And then you go back and put two, three and five into the original equations, with the fraction bars still in them, and check that all three hold. They do. The substitution is not a trick you either know or do not know. It is what you reach for whenever a system is linear in something other than its unknowns.

17:58So here is the whole of it. Pack the system: coefficients as a square array, unknowns as a column, right-hand sides as a second column. Multiply it out once to convince yourself the packing is true. Compute one determinant, and stop right there if it is nought. Otherwise multiply the right-hand column by the inverse, on the left, because that is the side the unknowns were written on. Substitute the answer back.

18:31You now know why the answer is the only answer -- because in thirty-four thousand systems where every candidate was tried, a non-singular one had exactly one, every time. You know that the wrong side is right about one time in a hundred, which is exactly often enough to fool you. You know that forgetting the division is right about half the time in easy examples and much less often in general.

19:01And you know that a fractional answer is not a mistake. Eleven in every hundred of the widest sweep's systems come out with a fraction in them, and every one of those was substituted back and held. Every count on this board was measured, not remembered.