Properties of Determinants

Transcript

0:00Introduction

Hello everyone! Okay, a small challenge first. Suppose I give you a three by three determinant whose top row is one hundred and two, eighteen, thirty six. Big numbers, six products to expand, right? How long will it take you? Now, a topper looks at it for two seconds and says: zero. How? With the properties of determinants. They let us create zeros and pull out factors before expanding, and questions built on them come up often in JEE, so let's begin.

0:32Swap rows, equal rows

First, one promise. The two vertical bars mean determinant, and the small capital T at the top right means transpose. So, determinant of A transpose equals determinant of A, okay? Transpose turns rows into columns, so every property of rows holds for columns too. Property one. Capital R with a small one below means row one, and the double-headed arrow means swap. Swap any two rows, and the determinant changes its sign. Right? Now think. Suppose two rows are exactly equal, and you swap them. The matrix does not change, so the determinant should not change. But the swap flips the sign! So what can the determinant be? Zero! Yes, exactly. Only zero equals its own negative. So two equal rows give zero, and so do proportional rows, where one row is a multiple of another. Clear? Now, our challenge from the start. One hundred and two, eighteen, thirty six; then one, three, four; then seventeen, three, six. Don't expand! Compare row one and row three. Notice something? Yes! One hundred and two is six times seventeen, eighteen is six times three, and thirty six is six times six; the dot means times. So R one equals six times R three: proportional rows. The answer is zero, with no expansion. That's it!

2:02Factors out, rows split

Property two: factors. If every entry of one row has a common factor k, take it outside, just once. Here the first row is k a, k b, and the second row is c, d. So we get k times the determinant with rows a, b and c, d. Okay? Property three: splitting. If each entry of a row is a sum of two terms, like a plus p and b plus q here, the determinant splits into two: one with a, b, and one with p, q. The other rows stay as they are. Got it? Now a quick question. Suppose matrix A is three by three; the cross is read as by. What is the determinant of two A? Just two times the determinant of A? No, no, careful! In two A, every entry is doubled, so all three rows carry a factor two. Each row gives one two: two cubed, so the determinant of two A is eight times the determinant of A, okay? In general, for an n by n matrix, the determinant of k A is k to the power n, times the determinant of A, and forgetting this is a common trap. With me?

3:08The most useful move: Ri to Ri + kRj

Now the most useful property. The arrow means replaced by: row i becomes row i plus k times row j, where j is not equal to i; the slashed equals sign means not equal. And this changes nothing! This is how we create zeros. Okay? But why does it change nothing? Hint: use the splitting property we just learnt. Correct, split the new row! The first determinant is just our old one. In the second, row i is k times row j: proportional rows, so it is zero. Determinant of A plus zero. Simple, right? You can even add multiples of two other rows at once: R i plus k times R j, plus l times R m, as long as row i itself is kept with coefficient one. Clear? Careful, though: R one going to two times R one is different. That puts a factor of two in one row, so the determinant gets multiplied by two. Got it? One more. Can I replace R one by R one plus R two, and R two by R two plus R one, both at the same time? Oh no, careful! Done together, both new rows are R one plus R two: two equal rows, so every determinant would become zero. Wrong! Do them one after another. Remember this one!

4:28Example 1: determinant after row operations, given 5

Alright, an easy one first. The determinant with rows a, b, c; x, y, z; p, q, r equals five. Find two new ones. The first has rows two a, two b, two c; then x plus p, y plus q, z plus r; then p, q, r. The second has rows p, q, r; then a, b, c; then x, y, z. Pause the video, and try it first. Done? Okay, let's check. The first one has twos in it, and we just saw the k to the power n rule. So two cubed times five, forty, right? No, no, wait! Only row one has the factor two, not the whole matrix. So take two out of R one, just once. Then replace R two by R two minus R three. Okay? Subtract p, q, r from row two, and it becomes x, y, z: the given determinant again. Two times five: ten. That's it! Now the second one. The last row, p, q, r, has moved to the top. How many row swaps is that? Two swaps, yes! First swap R two and R three, then R one and R two, giving p q r, then a b c, then x y z. Count only one swap, and you write minus five. Careful! Each swap brings a minus sign: minus one, in brackets, squared, times five, is five. So the answers are ten and five.

6:00Example 2: prove the determinant equals 2(a+b+c)^3

Clear? Now a harder one. Show that this determinant equals two times a plus b plus c, the whole cubed. Row one: a plus b plus two c, a, b. Row two: c, b plus c plus two a, b. Row three: c, a, c plus a plus two b. Pause and try. Hint: add all three columns. Done? Okay, let's check. Here is a topper's trick: before expanding, add up each row. Row one: a plus b plus two c, plus a, plus b. What do you get? Two a plus two b plus two c: two times a plus b plus c! Rows two and three give the same total. So, with capital C for column, replace C one by C one plus C two plus C three: our safe move, on columns. Every entry of column one is now two times a plus b plus c; take it out, leaving one, one, one. With me? Now create zeros: replace R two by R two minus R one, and R three by R three minus R one. Both use row one, which does not change, so doing them together is fine. Row two becomes zero, a plus b plus c, zero, and row three becomes zero, zero, a plus b plus c. Right? Now look at this determinant: everything below the main diagonal is zero. How do you finish it in one line? Yes, exactly! It is triangular, so it equals the product of its diagonal entries: one, times a plus b plus c, squared. With the factor outside, that is two times a plus b plus c, cubed. Proved, without any messy expansion!

7:40Quick revision: row swaps, equal rows, row operations, kA

So, quick revision. Swap two rows, and the sign changes; equal or proportional rows give zero. Adding k times another row changes nothing, and a common factor of one row comes out just once. The determinant of k A is k to the power n, times the determinant of A, and since transposing keeps the determinant, all of this works for columns too. Now go and try the practice problems on this page. All the best!