The Factor Theorem and Standard Determinants

Transcript

0:00Introduction

Hello everyone! Okay, a small challenge before we start. I give you a three by three determinant full of letters, and I ask you to factorise it. Most students expand it first: six products, one long messy expression, and then they hunt for factors. What if you could find every factor without expanding at all? Magic? No. Just the factor theorem from polynomials, used inside a determinant. These standard determinants come up often in JEE problems. So, let's begin.

0:34The factor theorem inside a determinant

So, here is the key idea. If the entries of a determinant are polynomials in x, then expanding it only multiplies and adds them. So the determinant itself is a polynomial in x. We call it capital delta of x; the triangle is the Greek letter delta, and the double arrow means so. Okay? Now, you know this from the last lesson: two equal rows make a determinant zero. So if putting x equal to a makes two rows equal, then delta of a is zero. Two equal columns work the same way. Right? Now think back to polynomials. Delta of a is zero. What does that tell you about delta of x? x minus a is a factor! Correct! That is the factor theorem, and we did not expand anything. Clear? So, the full method. Find all the factors this way. Then compare degrees, to see what is still missing. And fix the constant with one easy substitution, or by comparing one coefficient. Got it?

1:33Vandermonde, without expanding

Let's use it on the most famous one: the Vandermonde determinant. The two straight bars mean determinant. Rows: one, one, one; then a, b, c; then a squared, b squared, c squared. Treat it as a polynomial in a, and put a equal to b. Then C one equals C two; C with a small one below means column one. Two equal columns, so delta is zero. With me? So a minus b is a factor. Which other factors can you find the same way? Very good! Put a equal to c: columns one and three match, so a minus c is a factor; that is c minus a, up to a sign. And b equal to c gives b minus c. Three factors. Okay? Now the degree. Each term of the expansion takes one entry from each row, with degrees zero, one and two: total degree three. The three factors already make degree three, so only a constant k is missing: delta is k times these three factors. Clear? Last move: find k. Compare the coefficient of b c squared. In the determinant, it comes only from the diagonal, one times b times c squared, so it is one. In the product, it comes only from minus b, times minus c, times c: also one. So k equals one. No expansion at all!

2:57More standard determinants

Now, these three factors turn up again and again. Make the last row cubes: a cubed, b cubed, c cubed. Equal columns still give zero, but now the degree is four. So one more factor appears: a plus b plus c. Okay? And this one: rows a, b, c; then their squares; then b c, c a, a b. Same three factors, times a b plus b c plus c a. One family, easy to remember. With me? Now the cyclic determinant: rows a, b, c; then b, c, a; then c, a, b. Each row is the row above, shifted one place. Its value is a cubed plus b cubed plus c cubed, minus three a b c, right? No, no, careful! There is a minus sign in front of the whole bracket. In factors: minus, a plus b plus c, times a squared plus b squared plus c squared, minus a b, minus b c, minus c a. Remember that factor a plus b plus c. Got it?

3:59Patterns with x on the diagonal

Next, x down the diagonal and a everywhere else. Its value is x plus two a, times x minus a, squared; the two outside the bracket is the square. Quick check: put x equal to a, and all three rows become a, a, a. Equal rows give zero, so x minus a is a factor. Right? And this one: rows x plus a, b, c; then a, x plus b, c; then a, b, x plus c. Take away the x on the diagonal, and every row is a, b, c. Its value is x squared, times x plus a plus b plus c. Keep this pattern in mind for Example two. Clear?

4:40Example 1: cyclic determinant of roots of x^3 + px + q = 0

Now, Example one. Alpha, beta and gamma, these three Greek letters, are the roots of x cubed plus p x plus q equals zero. Find the determinant with rows alpha, beta, gamma; then beta, gamma, alpha; then gamma, alpha, beta. Pause the video, and try it first. Done? Okay, let's check. Please don't expand! Which standard determinant is this? The cyclic one! Yes, exactly. So it has the factor alpha plus beta plus gamma. Right? Now look at the equation, x cubed plus p x plus q: there is no x squared term. The sum of the roots is minus the x squared coefficient over the x cubed coefficient, so it is zero. So the determinant is zero times something: zero! One line, and done.

5:34Example 2: trigonometric determinant equation in theta

Now a harder one, a JEE pattern. Find every theta, that is this Greek letter, between zero and pi, for which this determinant is zero. The symbol like a rounded E means belongs to, and the round brackets leave out the end points. Row one is one plus sine squared theta, cos squared theta, four sine two theta. Rows two and three are alike, with the extra one moving down the diagonal. Pause the video, and try it. Done? Okay, let's check. Take away the extra ones on the diagonal, and all three rows are the same. So the matrix is I, the identity matrix, plus a matrix with three equal rows: just like our x plus a, b, c pattern. Okay? So, same method. Add all three columns into column one. What does each entry become? Two plus four sine two theta, in every row! Very good. The arrow means replace, and C one is column one. Sine squared theta plus cos squared theta is one, so one plus one gives two. Right? Take that factor outside, so column one is all ones. Then subtract row one from rows two and three; R means row. The determinant left is triangular, with diagonal one, one, one, so its value is one. With me? So the whole determinant is two plus four sine two theta. Set it to zero, and sine two theta is minus one half. Clear? Now, careful. Theta lies between zero and pi. So where does two theta lie? Between zero and two pi! Yes, exactly. There, sine is minus one half at seven pi over six, and at eleven pi over six. Keep two theta between zero and pi by mistake, and you find no answer at all. That is the trap! Divide by two: theta is seven pi over twelve, or eleven pi over twelve. Both lie between zero and pi. One pattern spotted, and a scary determinant became one line. Well done!

7:36Quick revision: factor theorem, Vandermonde, cyclic

So, quick revision. If x equal to a makes two rows equal, then x minus a is a factor; then compare degrees, and fix the constant. Vandermonde is a minus b, times b minus c, times c minus a; and the cyclic determinant is minus, a cubed plus b cubed plus c cubed minus three a b c. For the x plus a, b, c pattern, add all three columns into column one, take out the factor, and make a triangle. Now go and try the practice problems on this page. All the best!