Consistent or inconsistent: what a vanishing determinant does and does not decide

0:00Before you compute anything at all, every system of equations belongs to one of exactly two kinds. Either it has at least one answer, or it has none. A system with at least one answer is called consistent. A system with none is called inconsistent. Notice what that split does not care about. It does not care whether the answer is easy to find. It does not care whether there is one answer or a thousand of them.

0:31A system with infinitely many answers is consistent, in exactly the same sense as one with a single answer. So the word is about existence, and nothing else. Hold on to that, because the machinery we are about to build answers a narrower question than the word does, and the gap between the two is the whole of this video. Pack the system the usual way. The coefficients go into a square array, the unknowns into a column, and the right-hand sides into a second column.

1:08Now compute one number: the determinant of that square array. If it is not nought, you are finished before you have started. The array has an inverse, you multiply the right-hand column by it, and out comes the answer. One answer, and never a second. So the entire non-vanishing branch is settled, and it is settled with a formula. This video is about the other branch. The determinant comes out nought, and everything you were about to do stops working.

1:43Look carefully at what a vanishing determinant actually costs you. The inverse is built by dividing the adjoint by the determinant. That division is the only place the determinant appears, and it is the only step that fails. So the inverse is gone. The adjoint is not gone. You can still build it, entry by entry, exactly as before; nothing in its construction divides by anything. And here is the trap the whole topic exists to disarm.

2:18Losing the inverse is losing a tool. It is not learning an answer. A vanishing determinant tells you that the system does not have exactly one answer. It does not tell you that the system has none. Those are different statements, and treating the first as the second is the single most common mistake made here. So we need a second test, and there is one. You already have the adjoint sitting there, unused.

2:50Multiply it into the column of right-hand sides. The adjoint on the left, the right-hand column on the right, in that order. Out comes a column of numbers. Compare it against the column of noughts. If the result is not the nought column, the system has no answer at all. If the result is the nought column, the test has nothing to say. Those are the two outcomes, and they are not symmetric.

3:19One of them is a verdict and the other is a shrug, and a great deal depends on not confusing them. Take the first outcome and turn it from a rule into a reason. Suppose, for the sake of argument, that an answer did exist. Then the packed equation holds: the array times the column of unknowns equals the right-hand column. Multiply both sides on the left by the adjoint.

3:49On the left you now have the adjoint times the array, times the unknowns. And the adjoint times the array is the determinant times the identity. The determinant is nought. So the whole left side collapses to the nought column, whatever the unknowns happened to be. Which forces the right side to be the nought column too. So if the adjoint pushes the right-hand column onto something that is not nought, no answer could have existed.

4:21Three lines, and they rest on one identity you already have. The rule is now a consequence. Here is that argument doing real work. Two unknowns, and a coefficient array with one and three across the top, two and six beneath. Its determinant is six minus six, which is nought. Build the adjoint: six and minus three on top, minus two and one below. The right-hand column is five and eight.

4:53Push it through: six and minus two. That is not the nought column, so this system has no answer, and we have just proved why. Now look at the equations themselves and see it without any machinery. The second row of the array is exactly twice the first row. So the left side of the second equation is twice the left side of the first. But eight is not twice five.

5:21The two equations demand different things of the same quantity, and no pair of numbers can satisfy both. In two unknowns you can draw this. Each equation is a straight line, and an answer is a point on both lines. Two lines that cross once give exactly one answer. Two lines lying on top of one another give infinitely many. Two parallel lines that never meet give none. Sweep every two-unknown system whose coefficients and right-hand sides are minus one, nought or one -- seven hundred and twenty-nine of them.

6:02Four hundred and thirty-two are a crossing pair. Ninety-seven are lines lying on top of one another. Two hundred are lines that never meet. The determinant vanishes on the last two groups together, and the last two groups have opposite answers. That is the difficulty, drawn. Go back to the array with one and three on top, two and six beneath, and leave it alone. Change only the right-hand column, from five and eight to five and ten.

6:36The determinant has not moved; it is still nought. The adjoint has not moved either. But push the new column through: nought and nought. The test has gone silent. And this system does have answers -- infinitely many of them, because now ten is twice five and the second equation says nothing the first did not already say. So the same coefficient array, with the same vanishing determinant, sits on both sides of the question.

7:08The array alone was never going to be enough. You might hope the silent outcome at least means the system is consistent. It does not. Here are three equations in three unknowns whose coefficient array has every entry three times its own first row pattern -- one, two, three on top, then two, four, six, then three, six, nine. Every two-by-two minor of that array vanishes, so every cofactor is nought, so the adjoint is the nought array.

7:44Which means the adjoint pushes every right-hand column onto the nought column. The test is silent on this array no matter what you feed it. Feed it one, one, one, and there is no answer at all. Feed it one, two, three, and there are infinitely many. One array, one test result, two opposite verdicts. The silence is not weak evidence for consistency. It is no evidence at all. So how often does the test actually speak?

8:17Take four collections of systems small enough that every candidate answer can be tried one at a time, and keep only the ones whose determinant vanishes. Fifteen thousand eight hundred systems. For each one, the truth is settled by trying every possible column of unknowns -- no determinant, no adjoint, no formula. The test speaks on nine thousand four hundred and thirty-two of them. It goes silent on six thousand three hundred and sixty-eight.

8:51That is two systems in five where all this machinery returns nothing. And of those six thousand three hundred and sixty-eight, one thousand five hundred and thirty-five have no answer while four thousand eight hundred and thirty-three have many. Both outcomes are well populated, in every one of the four collections separately. So the silence is not a rare corner case waiting to be tidied away. Now the other column of that table, and it is the one that matters most.

9:27Of the nine thousand four hundred and thirty-two systems the test declared to have no answer, how many actually had one? None. Not one, in any of the four collections. When the test speaks it is right, and that is what makes it worth having despite the silence. But a claim like never wrong is worth almost nothing until you have watched something else be wrong under identical conditions. So change one thing.

9:58Push the right-hand column through the ARRAY instead of through its adjoint, and keep every other line of the argument the same. That version declares no answer correctly five thousand five hundred and eighty-six times in the largest collection. It is right far more often than it is wrong, which is exactly how much a spot check is worth. And it is wrong two thousand two hundred and eight times.

10:27The counting was never in doubt. The adjoint was. Push that further. Six versions of the second test, each asked the same three questions over the same systems. How many does it declare to have no answer, how many of those declarations are wrong, and how many systems with no answer does it let through. The adjoint declares five thousand one hundred and eighty-four, is wrong about none, and lets one thousand two hundred and two through.

11:01The cofactors without the transpose declare exactly five thousand one hundred and eighty-four as well. The same number. And they are wrong about one thousand five hundred and twelve of them. An equal count is not the same test. One version declares every singular system to have no answer. It lets nothing through -- it misses nothing, ever -- and it is wrong two thousand four hundred and two times.

11:31A test that never misses can still be useless, and only the second column tells you. Three unknowns now, and the same procedure without a change. The coefficient array is one, two, three on top, then two, four, six, then one, one, one. Its determinant is nought, because the second row is twice the first. Its adjoint is minus two, one, nought; four, minus two, nought; minus two, one, nought.

12:03Take the right-hand column one, three, one. Push it through, and you get one, minus two, one. Not the nought column, so there is no answer, and the reason is again visible in the rows: the second equation asks for three where twice the first asks for two. Now keep the array and take six, twelve, three instead. Push that through and every entry comes out nought. Silent. And this one does have answers, infinitely many of them.

12:37One more, and it is the one that catches people. Same shape, but the last coefficient is a letter rather than a number: one, one, one; one, two, three; one, three, and the letter. Compute the determinant symbolically and it is the letter minus five. So there is not one answer to this question. There are two, and which one applies depends on the letter. For every value except five, the determinant is not nought and the system has exactly one answer.

13:11At exactly five, the determinant vanishes and you have to run the second test. With right-hand sides one, two and four, the adjoint pushes that column onto one, minus two, one, which is not nought, so at that one value there is no answer at all. A letter in a coefficient is not another number. It is a case split, and reporting a single verdict here is teaching somebody to walk past it.

13:43It is worth saying plainly why the silent branch cannot simply be patched. A vanishing determinant says the equations are not independent of one another -- one of them carries no information the others did not already carry. That fact is about the left-hand sides alone. Whether the dependent equation agrees with its neighbours or contradicts them is a fact about the right-hand sides. The determinant never looked at the right-hand sides.

14:15The second test does look at them, which is why it can catch a contradiction. But when it goes quiet, it has told you the right-hand sides are consistent with that one dependency, and there may be others it cannot see. Separating the two remaining cases needs a different tool altogether, and this material does not carry one. Naming the limit honestly is better mathematics than pretending the flowchart has two branches.

14:48So here is the whole decision, on one board. Compute the determinant of the coefficient array. If it is not nought: exactly one answer, and the inverse hands it to you. If it is nought, build the adjoint and push the right-hand column through it. If that result is not the nought column: no answer at all, and you can prove it in three lines. If that result is the nought column: either no answer or infinitely many, and nothing on this board can tell you which.

15:23Three branches, and the third one ends in an honest gap rather than a leaf. Most of what you will be asked lands on the first two. Recognising the third is what stops you from asserting something you have not established. You can sort any system into consistent or inconsistent by existence alone, before computing anything. You can compute the coefficient determinant and say precisely what it settles: not whether an answer exists, but whether exactly one does.

15:58You can build the adjoint when the determinant vanishes, push the right-hand column through it, and read the two outcomes correctly. You can prove the first outcome instead of quoting it. You can recognise the second outcome as a genuine stopping point and name both possibilities that remain. You can handle a coefficient that is a letter by splitting into cases rather than reporting one verdict. And you can hold the two words apart from the machinery: consistent means an answer exists, and the machinery on this board decides that question completely on one branch and not at all on the other.