Overview

Transcript

0:00Introduction

Hello everyone! Okay, a quick challenge before we start. Take the two by two matrix with first row two, three, and second row five, minus two. Its inverse is some number k, times the matrix itself. Find k. So, will you find the adjoint, then the determinant, then divide? That works, but it is slow. There is a two line route, and you will see it today. Look, in this chapter, problems from the whole unit come mixed together, just like a real JEE paper. So today, I will show you how to use this unit problem bank well. Let's go.

0:37How this chapter works

So, look at how this chapter is built. First, Part A, at JEE Main level: twenty two solved problems, all mixed. Okay? Now tell me, when should you open Part A? Right after Chapter one? No! Only after Chapters one to five, the whole unit. The little dash on screen means one to five. And attempt Part A as a forty minute test, with a clock running. Take it completely seriously, right? Next comes Part B: longer problems, at JEE Advanced level. We will look inside it in a minute. Then Part C, the concept checks. These are the statements most often tested in true or false form, or in which is correct form. Got it? And at the very end, a unit test of fourteen problems, with answers only, no worked solutions. So there, you check yourself honestly. Clear?

1:30Topper's trick: budget your 40 minutes

Now, a small trick that toppers use. Every problem carries a badge: easy, medium or hard. Count them before you start. Part A has thirteen easy and nine medium problems. With me? So, twenty two problems in forty minutes. On average, how many minutes does each problem get? Forty divided by twenty two: about one point eight minutes. That is the division sign, and the wavy equals sign means approximately, okay? So, a little under two minutes per problem. And that is only an average. The easy ones must go fast, so that the medium ones get the extra time they need. Right?

2:14Part B: JEE Advanced level

Alright, now Part B, the JEE Advanced level. Fourteen longer problems, and every single one is marked hard. So give each one real time, okay? What kind of problems? Determinant identities, where you prove that a determinant equals some expression. And systems of equations with two parameters, two unknown constants sitting inside the equations. Our second example today is one of these. Clear? And arguments with inverses and special matrices, where you must show that something is true, not just find a number. Got it?

2:47Part C: is the statement exactly right?

Now, let's try one item from Part C, okay? The transpose of the product A B, inside the round brackets, equals A transpose times B transpose. The small T up top means transpose, and the question mark asks: is it true? Looks perfectly natural. What do you say? Oh no, careful! It is false. The transpose of a product reverses the order: the transpose of A B equals B transpose times A transpose. That is the reversal law. One small slip in the order, and the answer is wrong. That is exactly what Part C catches. Right?

3:24Example 1: find k when A inverse = kA, using the trace

Now, remember our challenge from the start? Let's finish it. It is a Part A problem, marked easy. Here, A is the matrix inside the square brackets: first row two, three, and second row five, minus two. And A inverse, the A with minus one up top, equals k times A, for some number k. Find k. Start your clock, pause the video, and try it. Done? Okay, let's check. Before any adjoint, look at the diagonal: two, and minus two. What do they add up to? Zero, yes! That sum is called the trace of A, and on screen it is written t r. And the determinant of A, with the two vertical bars, is two times minus two, minus three times five: minus four minus fifteen, so minus nineteen. Not zero, so the inverse exists. The slashed equals sign means not equal, okay? Now the key fact. Every two by two matrix satisfies this: A squared, minus trace of A times A, plus determinant of A times I, equals O, the zero matrix. And here I is the identity matrix. Right? Put in trace zero, and determinant minus nineteen. So, A squared, minus zero times A, minus nineteen I, equals the zero matrix. The double arrow means implies: A squared equals nineteen times the identity. With me? Now move the nineteen across: A times, A over nineteen, equals the identity. So A inverse is one over nineteen, times A, and k is one over nineteen! No adjoint at all. That is the two line route I promised.

5:01Example 2: a linear system with two parameters p and q

Clear? Now a hard one from Part B, a system with two parameters. Two x plus p y plus six z equals eight. x plus two y plus q z equals five. x plus y plus three z equals four. Discuss the system for all values of p and q. Hard means more time. Pause the video, and try it. Done? Okay, let's check. Three equations, three unknowns. So what do we compute first? The determinant of the coefficients, yes, exactly! We call it delta, the Greek capital letter that looks like a little triangle. Expand along the first row: two times, six minus q, minus p times, three minus q, plus six times, one minus two. Simplify, and it factorises beautifully: p minus two, times q minus three. Okay? So, when p is not two and q is not three, delta is not zero, and there is a unique solution. Right? Now, delta is zero when p is two, or when q is three. So in both cases, infinitely many solutions, isn't it? No, no, wait! This is the JEE trap. Delta equal to zero only says there is no unique solution. It can be infinitely many, or none at all. So check each case. Got it? Take p equals two. The first equation becomes two x plus two y plus six z equals eight: exactly two times the third. So only two equations really remain, and there are infinitely many solutions, for every q, okay? Now q equals three, with p not two. Subtract the third equation from the second: the x and z terms cancel, and y equals one. Then the third equation gives x plus three z equals three. And the first equation? Two x plus six z is two times, x plus three z. So it reads: two times three, that is six, equals eight minus p. That needs p equals two. But p is not two! False, so no solution. So: unique when p is not two and q is not three; infinitely many when p is two; and none when q is three but p is not two. With me?

7:20Quick revision: how to use Parts A, B and C

So, quick revision. Part A: a forty minute test, after Chapters one to five. Part B: hard problems, and when delta is zero, check every case. Part C: is the statement exactly right? And then, the unit test. Now open Part A, start your clock, and begin. You have all the tools now, I promise. All the best!