Consistency: One, Infinitely Many or No Solutions

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Two equations, two unknowns: one answer, right? Take x plus two y equals three, and two x plus four y equals six. Now change just one number: make that six a seven. Still one answer each, right? Not so fast! One of these pairs has endless answers, and the other has no answer at all. Which is which, and why? Today we learn the three possible cases: exactly one solution, infinitely many, or none, and how to tell them apart, fast. This comes up often in JEE, so let's begin.

0:41Three cases, decided by Delta

So, three equations in three unknowns. We start with four determinants from the last video. Delta, the Greek capital letter like a little triangle, is the determinant of A; the straight bars mean determinant. Delta one, two and three, with small subscripts: the column B, the right-hand sides, replaces the first, second or third column of A, okay? Quick question from the last video. If delta is not zero, how many solutions? Exactly one! Yes! The slashed equals sign means not equal. The matrix method or Cramer's rule gives one solution, and only one. Clear? Now the interesting part. If delta is zero, does that always mean no solution? Not always! If delta is zero, and at least one of delta one, delta two, delta three is not zero, then yes, no solution. The system is inconsistent. Right? But if all four are zero, there are infinitely many solutions, or none. The determinants alone cannot decide; we must eliminate and see. Keep this in mind. Got it? The same idea in matrix language: if the determinant of A is zero, and adjoint of A times B is not O, the zero matrix, there is no solution. Here, adj is short for adjoint, from chapter three. With me?

2:01The trap: all four determinants zero

Here is the whole story as a flowchart. Start at the top, the system, A X equals B, and follow the arrows. Delta not zero: the green box, unique solution, with x equals delta one over delta, and so on. Delta zero: find delta one, two and three. In the red box, delta i, with a small i at the bottom, means any one of them; if some delta i is not zero, no solution. And if all are zero: eliminate, infinitely many or none. See, it even says check, with an exclamation mark! And the grey line at the bottom: when B is the zero matrix, the system is homogeneous. It is always consistent, since x, y and z all zero always works: only the zero solution if delta is not zero, infinitely many if delta is zero. Okay? Now, why that check? Take x plus y plus z equals one, then equals two, then equals three. What are the four determinants here? All zero! Yes, exactly! In delta, every column is one, one, one. In delta one, two and three, one column changes, but the other two are still equal columns, so each determinant is zero, right? So all four are zero. That means infinitely many solutions, right? No, no, careful! That is the trap. x plus y plus z cannot be one and two at the same time! So there is no solution at all. Never conclude infinitely many from the determinants alone. Eliminate and check, just as the flowchart says. Got it?

3:33Topper's trick: the row-combination shortcut

Now a topper's trick that saves a lot of time. Say delta is zero, and the left sides of the first two equations are not proportional: one left side is not just a multiple of the other, okay? Then the third left side is a combination of the first two. L one, L two and L three are the three left sides, and p and q are numbers. L three equals p times L one, plus q times L two. Find p and q by matching two coefficients. Right? Now think. If the left sides obey this relation, what must the right sides do? The same relation, exactly! With d one, d two and d three as the right sides, the system is consistent exactly when d three equals p times d one, plus q times d two, and then it has infinitely many solutions. Clear? And notice: no delta one, delta two, delta three needed. Much faster! We will use it in Example two.

4:27Example 1: same line or parallel lines, x + 2y = 3

Now, remember our puzzle? Let's solve it. First system: x plus two y equals three, and two x plus four y equals six. Second system: the same first equation, and two x plus four y equals seven. One solution, infinitely many, or none? Pause the video, and try it first. Done? Okay, let's check. Same left sides, so the same delta. Between the vertical bars: one, two on top, two, four below. One times four, minus two times two: four minus four, zero. So neither system has exactly one solution, okay? Now, how is two x plus four y related to x plus two y? Exactly twice! Very good. Two x plus four y equals two times the bracket, x plus two y. Right? In the first system, the right side also doubles: two times three is six; the little cross means times. So the second equation is just the first one times two: the same line, written twice. Infinitely many solutions. With me? But in the second system, seven is not two times three. Same left side, different right side: parallel lines that never meet. No solution at all. Puzzle solved! The right side decided everything.

5:49Example 2: a and b for unique, infinite or no solution

Now a harder one, JEE level. x plus y plus z equals two, x plus two y plus three z equals five, and x plus three y plus a times z equals b. For which values of a and b is there a unique solution, infinitely many, or none? The small roman numerals number the three parts. Pause the video, and try it. Hint: use the trick. Done? Okay, let's check. First, delta, with rows one, one, one; one, two, three; and one, three, a. Expand along the first row: one times, two a minus nine, minus one times, a minus three, plus one times, three minus two. So delta equals a minus five. Okay? Part one is quick: the solution is unique exactly when delta is not zero, that is, when a is not five. Clear? Now a equals five, where delta is zero. The first two left sides are not proportional, so use the trick: L three equals p times L one, plus q times L two. Find p and q. x coefficients: p plus q equals one. y coefficients: p plus two q equals three. Subtract: q equals two, so p equals minus one. The double arrow means this gives that. Right? So L three equals two L two, minus L one. Check all three coefficients, listed in the brackets: two minus one is one, four minus one is three, six minus one is five. Perfect! Now the right sides must follow the same relation. So what must b be? Two times five, minus two: eight. Correct! So at a equals five, the system is consistent only when b equals eight. With me? So, part two: a equals five and b equals eight, infinitely many solutions. Part three: a equals five, but b not eight, no solution. And here is the trap: some students see a equals five and write infinitely many at once. No, no! The value of b decides. See how fast the trick was?

8:00Quick revision: one, infinitely many or no solution

So, quick revision. One: delta not zero means exactly one solution. Two: delta zero, and some delta i not zero, means no solution. Three: all four zero means infinitely many or none, so eliminate, or use the row-combination trick. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!