Mixed Problems: Set 1 (Matrix Algebra and Powers)

Transcript

0:00Introduction

Hello everyone! Okay, a quick challenge to start. Take a two by two matrix with first row two, minus one, and second row one, zero. Call it A, okay? I want A to the power fifty, and the sum of all its entries. So, will you sit and multiply A by itself forty nine times? No way, right? That would eat up half your exam! Today we learn three quick moves that bring a huge power down to one or two lines. Problems like this come up often in JEE Main, and each one should take you only a minute or two. So, let's begin.

0:36Move 1: when A squared is a multiple of A

So, here is move one. Our matrix A must be square, otherwise A times A is not even defined, right? Always start by finding A squared; the small two up top is the power. Sometimes A squared turns out to be a number times A, and we call that number k. So, A squared equals k times A, as you see on screen. Okay? Now tell me. If A squared is k times A, what is A cubed? Let's see. We write A cubed as A times A squared; the dot means multiply. Now replace A squared by k times A, and the number k comes out in front: k times A squared. Use the same fact again, and we get k squared times A, just like that. Clear? Now look at the pattern. So, A to the power n is k to the power n, times A, right? No, no, careful! Count again: A squared has k to the power one, and A cubed has k squared. The power of k is always one less than the power of A, okay? So, A to the power n equals k to the power n minus one, times A; that n minus one sits up top as the power of k. This slip is very common, so watch it. Got it?

1:52Move 2: a power that equals I

Now, move two. Sometimes a power of A is the identity matrix, capital I: ones on the diagonal, zeros everywhere else. If A to the power m equals I, then after every m steps the powers come back to the start. They simply go round in a cycle. With me? For example, one matrix in this set has its cube equal to I, okay? And one hundred is three times thirty three, plus one. So, what is A to the power one hundred? Yes, exactly! The answer is just the matrix A itself! Write A to the power one hundred as A cubed, in brackets, to the power thirty three, times A once more. Now A cubed is I, and I to any power is still the identity. So we get I times A, which is simply our matrix A again. Only the remainder, after dividing by three, matters.

2:47Topper's trick: write A=I+N

Clear? And now move three, a topper's trick. Write A as I plus N, where N squared is capital O, the zero matrix, all zeros. Such a matrix N is called nilpotent. Okay? Now, A to the power n is I plus N, in brackets, to the power n. Can we expand it with the binomial theorem, just like one plus x, to the power n? Yes, but only for one reason! For matrices, the binomial theorem needs the two matrices to commute, that is, I times N must equal N times I, as on screen. The identity commutes with every matrix, so here we are safe. But never expand a sum of two matrices like this unless they commute. That is a classic trap. Right? So, expand: I, plus n times N, plus n choose two times N squared, and so on. The bracket with n above two is n choose two, the binomial coefficient, and the three dots mean the remaining terms. Okay? But N squared is zero, so N cubed, and every higher power, is zero too. Everything after n times N just vanishes! So, A to the power n is just I plus n times N, nothing more. Two terms, for any power. Beautiful, isn't it?

4:08Example 1: sum of all entries of A^10 using A^2 = 2A

Okay, example one, an easy one. Here, A is the two by two matrix with every entry one; the big square brackets hold the matrix. Find the sum of all the entries of A to the power ten. Which move fits? Pause the video, and try it first. Done? Okay, let's check. First, A squared, as always. Each entry is one times one, plus one times one: two. So every entry of A squared is two; that is just two times A, you see? So this is move one, with k equal to two. Clear? Now, A to the power ten. Is it two to the power ten, times A? Oh no, careful! The power of k is always one less. So, A to the power ten equals two to the power nine, times A, and two to the power nine is five hundred and twelve. Got it? Now every entry of A is one, so every entry of A to the power ten is five hundred and twelve. Four entries, so the sum is four times five hundred and twelve: two thousand and forty eight. One small pattern, and the whole question is done!

5:22Example 2: sum of all entries of A^50 using A = I + N

Now, remember our challenge from the start? Here it is. This time, A has first row two, minus one, and second row one, zero. Find the sum of all the entries of A to the power fifty. Pause the video and try it. Hint: look at A minus I, okay? Done? Okay, let's check. Let N be A minus I, the identity. We just subtract one from each diagonal entry. So N has first row one, minus one, and second row one, minus one. With me? Now, what is N squared? Multiply row by column. Zero! Correct. First row times first column: one minus one, zero. First row times second column: minus one plus one, zero. The second row of N is the same as the first, so it gives the same zeros. So N squared is capital O, the zero matrix, and our topper's trick applies. Right? So, A to the power fifty is I plus fifty times N, by our trick. On the diagonal: one plus fifty is fifty one, and one minus fifty is minus forty nine. Off the diagonal: minus fifty, and fifty. So we get fifty one, minus fifty, in the first row, and fifty, minus forty nine, in the second. Okay? Now add all four entries: fifty one minus fifty, plus fifty minus forty nine. Just two! See? The fiftieth power of a matrix, and the answer is two, without a single long multiplication. Our challenge from the start, solved in four lines!

7:03Quick revision: A^2 = kA, A^m = I and A = I + N

So, quick revision. One: if A squared is k times A, then A to the power n is k to the power n minus one, times A, okay? Two: if A to the power m is the identity, only the remainder of the power on dividing by m matters. Three: if A is I plus N, with N squared zero, then A to the power n is I plus n times N, nothing more. Now go and try the practice problems on this page. Give each one a minute or two, and you will find them easy. All the best!