Mixed Problems: Set 2 (Determinants, Adjoint and Inverse)

Transcript

0:00Introduction

Hello everyone! Okay, a quick puzzle before we start. I have a three by three matrix, call it A, and I will not tell you a single entry. I only tell you its determinant: three. Now find the determinant of adjoint of two A inverse. Scalar, inverse and adjoint, all stacked together. Is it even possible? Yes, and in under a minute! Questions like this come up often in JEE Main. So today, in Set two of our mixed problems, you will learn to open them layer by layer. Let's go.

0:35Tool 1: three determinant rules

So, before we open Set two, let's sharpen our tools. Take a three by three matrix, and double every entry in it. What happens to its determinant? It becomes eight times, not two times! Each of the three rows gives out one factor of two, and two times two times two is eight. So, for a square matrix A of order n, and a scalar k, determinant of k A equals k to the power n, times determinant of A, okay? The two vertical bars on screen mean determinant. Next, the inverse matrix, written A with a small minus one at the top. It exists only when determinant of A is not zero; the crossed equals sign means not equal to. And then, determinant of A inverse equals one over determinant of A, simple! Why? Because A times A inverse is the identity matrix, whose determinant is one. Right? Now the adjoint, written a d j. Determinant of adjoint of A is determinant of A, to the power n minus one. For a three by three matrix, n minus one is two, so determinant of adjoint of A is determinant of A, squared. Clear? And here is the topper's trick for nested questions. Work from the inside out, and keep every scalar in front. Peel one layer at a time, and never drop a scalar on the way. Got it?

2:01Tool 2: the 2x2 quadratic gives the inverse

Now, the second tool. First, one word: t r of A is the trace of A, the sum of the diagonal entries. Top left plus bottom right, for a two by two matrix. Okay? Every two by two matrix A satisfies this quadratic; the little cross is read as by. It reads: A squared, minus trace of A times A, plus determinant of A times I, equals capital O, the zero matrix. And I here is the identity matrix. With me? Now think. This is a quadratic in the matrix A, right? So how can it give you A inverse, without finding any adjoint? Take A out as a common factor! For example, say A squared minus five A plus seven I is the zero matrix. Move the A terms to the other side, and take A out. The double arrow means gives: A times the round bracket, five I minus A, equals seven I, okay? Now divide by seven. So A inverse is one seventh of five I minus A, because A times that matrix gives the identity. This works only because the constant term, seven, is not zero. Remember this move, we need it soon. Right?

3:17Example 1: |adj(2A inverse)| when |A| = 3

Now, remember our puzzle from the start? Let's solve it. It is the very first problem of Set two, and it is marked easy. Matrix A is three by three, and its determinant is three. Find the determinant of adjoint of two A inverse. Start your clock, pause the video, and try it. Done? Okay, let's check. Inside out, remember? The innermost piece is A inverse. It exists, because the determinant, three, is not zero. Its determinant is one over determinant of A, that is, one third. Okay? Next, the scalar two sits in front of A inverse. So the determinant of two A inverse is two times one third, that is, two thirds, right? No, no, careful! That is exactly the trap from our first tool. The order is three, so the two comes out as two cubed, eight; the small dot means times. Eight times one third gives eight thirds. Got it? Now the outer layer, the adjoint. Call the matrix inside M, so M is two A inverse. For order three, determinant of adjoint of M is determinant of M, squared. Clear? So the answer is eight thirds, squared, in the big round brackets: sixty four over nine! And see, we never needed a single entry of the matrix. Puzzle solved, in under a minute. Perfect!

4:46Example 2: A inverse = alpha I + beta A, find alpha + beta

Now a medium one from the same set. Matrix A, in square brackets, has first row one, two, and second row minus one, four. We are told that A inverse equals alpha times I, plus beta times A, where alpha and beta, these two Greek letters, are just numbers. Find alpha plus beta. Pause the video, and try it. Done? Okay, let's check. The answer has A inverse written using I and A, so our second tool, the quadratic, is the fast way. Trace first: one plus four, five. Okay? Now the determinant: one times four, minus two times minus one. So four minus two, that is two, right? Oh no, careful! Two times minus one is minus two, and subtracting minus two means adding two. So the determinant is four plus two, that is, six. Watch the signs inside a determinant, always. Clear? Now put trace five and determinant six into the quadratic. So we get: A squared, minus five A, plus six I, equals the zero matrix. Right? Same move as before. Move the A terms across: six I equals five A minus A squared. Then take A out, and we get A times the bracket, five I minus A, okay? Divide by six, which is allowed, because the determinant, six, is not zero. So A inverse is one sixth of the bracket, five I minus A, right? Open the bracket: five sixths I, minus one sixth A, got it? Now compare this with alpha I plus beta A, the form we were given. What is alpha, and what is beta? Alpha is five sixths, and beta is minus one sixth. Yes, exactly, the minus sign goes with beta! So alpha plus beta is four sixths, that is, two thirds. And we never wrote out the inverse entry by entry. That's it!

6:54Quick revision: |kA|, |A inverse|, |adj A|, the 2x2 rule

So, quick revision. For order n, determinant of k A is k to the power n, times determinant of A, never just k times determinant of A, remember. Determinant of A inverse is one over determinant of A, when that is not zero, and for order three, determinant of adjoint of A is determinant of A, squared. And for two by two matrices, the quadratic with the trace and the determinant gives A inverse fast. Now go and try the practice problems on this page, and then the whole of Set two with a clock running. You will do well, I promise. All the best!