Mixed Problems: Set 3 (Linear Systems)

Transcript

0:00Introduction

Hello everyone! Okay, a fun puzzle to start. I throw a die twice. The first number we call a, the second we call b, and both go into a system of three equations in x, y and z. What is the chance that this system has no solution at all? Strange question, isn't it? But it is really a determinant question, and by the end of this video you will solve it in two minutes. Systems with a parameter like this come up often in JEE Main. So, let's begin.

0:32The determinant decides

So, every problem in this set starts the same way. Three equations in three unknowns become one matrix equation, A X equals B: A holds the coefficients, X the unknowns x, y, z, and B the right-hand sides. Then we find capital delta, the Greek letter that looks like a little triangle. It is the determinant of A; the two straight bars mean determinant. The map on screen shows the whole decision. Okay? Now, first question. The equals sign with a slash means not equal. So, if delta is not zero, how many solutions does the system have? Yes, exactly one solution! Cramer's rule finds it, as in the green box: x equals delta one over delta, and so on. Delta one, with a small one written below, is the determinant of A with its first column replaced by the right-hand sides; delta two and delta three replace the second and the third column. Clear? Now, the other branch: delta is zero. Then find delta one, delta two and delta three; the small i below delta stands for any of them. If even one is not zero, there is no solution. That is the red box. With me? And if all four determinants are zero? Then there are infinitely many solutions, right? No, no, careful! This is the classic trap. With all four zero, the system has either no solution or infinitely many. The determinants alone cannot decide; we must eliminate and check, as the last box says. And there is a fast way to check, coming next. Got it? One more line, for the homogeneous case. When the right-hand sides are all zero, B is capital O, the zero column. Then x, y and z all zero is always a solution, the trivial one. Non-trivial solutions exist exactly when delta is zero; the double-headed arrow means if and only if. This line is also written below the map. Right?

2:34Topper's trick: the row-combination shortcut

Now, a topper's trick. Call the left sides of the three equations L one, L two and L three; the small numbers below are just labels. If delta is zero, and the first two equations are not proportional, then L three equals p times L one, plus q times L two, for some numbers p and q. Okay? And to find p and q, just match two of the coefficients, say the x and the y coefficients. Clear? Now, the right-hand sides of the three equations. What must they do, for the system to have any solution? Yes, exactly! They must obey the same relation. Call them d one, d two and d three. If d three equals p times d one, plus q times d two, the third equation adds nothing new: infinitely many solutions. With me? And if d three is not equal to that, the third equation contradicts the first two: no solution. Much faster than three more determinants, isn't it?

3:35Example 1: for how many k does the system have no solution

Okay, example one, an easy one. k x plus two y equals three, and eight x plus k y equals six. For how many values of k does this system have no solution? The same idea works for two equations in two unknowns. Pause the video, and try it first. Done? Okay, let's check. Delta, between straight bars, has rows k, two and eight, k. So delta is k times k, minus two times eight: k squared minus sixteen. That is zero when k is plus or minus four; the double arrow means implies. Right? So, two values of k make delta zero. Then the answer is two, isn't it? No, no, careful! Put k equal to four: four x plus two y equals three, and eight x plus four y equals six. The second equation is just two times the first. The same line! So for k equal to four, infinitely many solutions, not none. Got it? Now k equal to minus four. Minus four x plus two y equals three, and eight x minus four y equals six. Divide the second by minus two; that is, minus four x plus two y equals minus three. Same left side, different right sides: parallel lines. No solution. With me? So only k equal to minus four gives no solution. The answer is one. See the trap? Delta zero only tells you the solution is not unique. Always check the right-hand sides!

5:15Example 2: dice probability that a system has no solution

Now, remember our die puzzle from the start? Here it is. The die is thrown twice, showing a and then b. The equations: x plus y plus z equals one; x plus two y plus a z equals two; and x plus four y plus a squared z equals b. Find the probability that the system has no solution. Pause the video and try it. Hint: find delta first. Done? Okay, let's check. Delta has rows one, one, one; then one, two, a; then one, four, a squared, the squares of the second row. That is the Vandermonde determinant, a standard determinant, in one, two and a. So delta is two minus one, times a minus one, times a minus two. Clear? The die can show any a from one to six. For which values of a does the system have exactly one solution? Correct! For a equal to three, four, five or six, delta is not zero: exactly one solution, whatever b is. Trouble is possible only when a is one or two. Right? So, our topper's trick. The first two equations are not proportional, so it applies. Take three times L two, minus two times L one. The x part is three minus two, one; the y part is six minus two, four. That matches L three. Now the z part: three a minus two must equal a squared. For a equal to one, one equals one; for a equal to two, four equals four. It works for both! Okay? Now, the right-hand sides must follow the same rule. So, what must b be, for the system to have solutions? Yes, exactly! b must be three times two, minus two times one: four; the dot means times. If b is four, infinitely many solutions; otherwise, none. With me? So, no solution exactly when a is one or two, and b is not four. The symbol like a rounded E means belongs to, and the curly brackets hold the set one, two. Two choices for a, times five choices for b: ten outcomes. Careful: b equal to four is out; it gives infinitely many solutions. Got it? And two throws of a die give thirty six equally likely outcomes. So capital P, the probability, is ten over thirty six, which is five over eighteen. There it is, our puzzle from the start, solved in a few lines!

7:54Quick revision: determinant test, L3 = pL1 + qL2, AX = O

So, quick revision. One: if delta is not zero, exactly one solution; if delta is zero, either none or infinitely many, so never stop there. Two: when delta is zero, write L three as p times L one, plus q times L two, and test the right-hand sides. Three: a homogeneous system has non-trivial solutions exactly when delta is zero. Now go and try the practice problems on this page. You will enjoy them. All the best!