Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one puzzle before we start. I have three equations in x, y and z, with one unknown number k sitting inside them. And the determinant of this system is zero, not for one value of k, but for every value of k. So, for every k, the system has no solution? Many students write exactly that, and lose the marks. The truth is more interesting, and we will find it by the end of this class. This is Part A of our unit problem bank, at JEE Main level. So, let's begin.

0:37Part A: a 40-minute test

So, here is what Part A holds. Twenty two solved problems, all at JEE Main level. Short, sharp questions, okay? And they are mixed: matrices, determinants, inverses, and systems of equations, all together. Nobody tells you which chapter a question comes from, just like the real exam. Right? The plan: finish Chapters one to five, then attempt Part A as one forty minute test. So today, two tools that Part A uses, and then two of its problems, an easy one and a harder one. With me?

1:11Tool 1: area of a triangle from a determinant

Tool number one, from Chapter two. Look at the figure: a triangle with vertices capital A, B and C, okay? The small lowered numbers in x one, y one, x two and so on are subscripts: they just tell you which vertex. Each vertex gives one row of a determinant: its x, its y, and then a one. Clear? Now look at the formula in the figure. Capital delta, this triangle-shaped Greek letter, stands for the area here. It is one half, times the determinant, but the determinant sits inside two pairs of vertical bars. The inner bars mean determinant. So why the outer bars? Because a determinant can come out negative, and an area never can! The outer bars are the modulus, the absolute value. So the area is one half of the modulus of the determinant. Got it? Now the sign. The figure tells us that too. The sign is plus when you go from A to B to C anticlockwise; the little arrows on screen show that order. List the same vertices clockwise, and the determinant comes out negative. Only the order changed, the triangle did not. Right? One more line from the figure. The double-headed arrow means, if and only if. The determinant is zero exactly when the three points are collinear, all on one straight line. Makes sense: no triangle, no area. Okay?

2:37Tool 2: when the determinant is zero

Tool number two, from Chapter four: three linear equations in three unknowns. Now capital delta means something new: the determinant of capital A, the coefficient matrix. If delta is not equal to zero, that is the crossed equals sign, then the double arrow gives us exactly one solution. With me? But now suppose delta is zero. Does that mean the system has no solution? Not always! When delta is zero, there are two possibilities: no solution, or infinitely many. Never exactly one. So delta alone cannot decide. This is a very common trap, okay? So, to decide, here is the topper's shortcut. Call the left sides of the three equations L one, L two and L three. If delta is zero, and the left sides L one and L two are not proportional, then L three is p times L one, plus q times L two, for some numbers p and q. Find p and q by matching two of the coefficients. Clear? Now the right sides: d one, d two and d three. If they follow the same rule, d three equals p times d one, plus q times d two, the system has infinitely many solutions. Otherwise, there is no solution at all. Right?

3:54Example 1: area of the triangle with three given vertices

Alright, example one, the easy one. Find the area of the triangle with vertices: minus two, minus three; three, two; and minus one, minus eight. Pause the video, and try it. Done? Okay, let's check. Tool one: each vertex gives a row. Minus two, minus three, one. Then three, two, one. Then minus one, minus eight, one. Got it? Now expand along the first row, with the signs plus, minus, plus. Minus two, times two plus eight. Then minus, times minus three, times three plus one. Then plus one, times minus twenty four plus two. Okay? Look at the middle term: a minus sign, times minus three. So what sign does that term get? Plus, yes! Minus times minus is plus. So we get minus two times ten, plus three times four, plus minus twenty two. That is minus twenty, plus twelve, minus twenty two: minus thirty. Clear? So the area is half of minus thirty, that is, minus fifteen, right? Oh no, careful! An area can never be negative. Remember tool one: take the modulus first. The modulus of minus thirty is thirty, and one half times thirty, the dot means times, is fifteen. And that minus sign only tells us the vertices were listed clockwise. Perfect!

5:20Example 2: discuss a system for all values of k

Now the harder one, and it is our puzzle from the start. Three equations: x plus y plus z equals three; x minus y plus z equals one; and two x plus k y plus two z equals four. Discuss the system for all values of k. Pause the video, and try it. Hint: start with delta. Done? Okay, let's check. Before you expand anything, compare the x coefficients and the z coefficients in each equation. Notice anything? They are equal in every equation: one and one, one and one, two and two! So in delta, column one equals column three; C one and C three on screen are those columns. Two equal columns, so delta is zero, for every k. Exactly one solution is impossible, whatever k is. Clear? So the system has no solution, for every k, right? No, no, wait! Tool two: delta zero means none, or infinitely many. Now, L one is x plus y plus z, and L two is x minus y plus z; they are not proportional. So write L three as p L one plus q L two. Match the x coefficients: p plus q equals two. Match the y coefficients: p minus q equals k. With me? Add and subtract these two equations. So p is two plus k, over two, and q is two minus k, over two. And the z coefficients: again p plus q equals two, so they match automatically. Okay? Now the right sides. d one is three and d two is one, so the rule needs d three to be three p plus q. Three p plus q is six plus three k, plus two minus k, all over two, which is four plus k. But the right side of the third equation is four. So four plus k must equal four. Got it? So k equals zero! Then the right sides follow the same rule as the left sides, and the system has infinitely many solutions. Right? And for any k not equal to zero, the right sides break the rule: no solution. Puzzle solved! Delta was zero for every k, yet k equals zero gives infinitely many solutions, and every other k gives none. So never stop at delta equals zero. Clear?

7:43Quick revision: half the modulus, zero determinant, L3 test

So, quick revision. One: the area of a triangle is half the modulus of the determinant, so never a negative area. Two: delta equals zero means no solution or infinitely many, never exactly one. Three: we decide by writing L three as p L one plus q L two, and then checking the right sides. Now go and attempt all twenty two problems on this page as one forty minute test. You will do well, I promise. All the best!