Evaluating Determinants

Transcript

0:00Introduction

Hello everyone! Okay, a small challenge before we start. I give you a three by three determinant: nine numbers between two straight bars. Most students expand it the long way: three small two by two determinants, lots of chances to slip, right? What if today's example needs only one small two by two determinant? Magic? No. Just the right choice of row or column. Determinants come up often in the exam, and the adjoint and the inverse are built on them. So, let's begin.

0:36Every square matrix has a number

So, what is a determinant? Every square matrix A has a number attached to it: its determinant. The double arrow on screen means gives us. We write it with two straight vertical bars around A, or as det A, for short. And note the word square: only a square matrix has a determinant. Okay? Now, a quick question. These two bars look exactly like the modulus sign from your earlier classes. So is a determinant always positive, or zero? No, no, careful! These bars do not mean absolute value; a determinant can be negative. Look at the two by two case, entries a, b, c, d. Multiply down the main diagonal, a times d, then subtract the other diagonal, b times c. So the determinant is a d minus b c. If b c is bigger than a d, it is negative. Clear?

1:29Minors and cofactors

For bigger determinants, we need two new words. First, the minor. Capital M, with a small i j below it as a subscript, is the minor of the entry in row i, column j, okay? Delete row i and column j; the smaller determinant left over is the minor. With me? Next, the cofactor, capital C sub i j: the minor with a sign attached. It is minus one, to the power i plus j, times M i j. The i plus j up top is a superscript, a power. Got it? So tell me. For the entry in row one, column two, is the sign plus or minus? Minus! Yes, exactly. One plus two is three, an odd number, and minus one to the power three is minus one. Even sum, plus. Odd sum, minus. So the signs alternate like a chessboard: plus, minus, plus; then minus, plus, minus; then plus, minus, plus. The top left is always plus. Remember this picture, okay?

2:26Expanding a 3×3 determinant

Now, the three by three determinant; the cross in the heading is read as by. The first row is a one, b one, c one; the small numbers below are subscripts for the row. Expand along the first row. Take a one, times its minor: delete its row and column, and b two, c two, b three, c three is left. Then b one times its minor, and c one times its minor. With me? But look carefully at the signs: plus, minus, plus. Why is there a minus sign in front of b one? The chessboard, exactly! Here b one sits in row one, column two, a minus place. So each entry is really multiplied by its cofactor. And the most common slip? Forgetting this middle minus sign. Don't do that! Now the big result. You can expand along any row. For row i, the determinant of A equals a i one times C i one, plus a i two times C i two, plus a i three times C i three. Each entry times its own cofactor. Any column works too, and the answer is always the same. Right? So, if you are free to pick any row or any column, which one would you choose? The one with the most zeros! Correct! We call a row or a column a line. Every zero kills its whole term, because zero times anything is zero. So scan for zeros before you start. Clear?

3:52Two more facts

Two more facts. First, a quick question. What if I take the entries of row one, but multiply them by the cofactors of row two, and add? Always zero! a one one times C two one, plus a one two times C two two, plus a one three times C two three, equals zero. One row's entries with a different row's cofactors: the sum is zero. We do not prove it here, but this is the key to the adjoint in Chapter three. Got it? Next, a triangular determinant, like this one: only zeros below the main diagonal. And a diagonal one, with zeros above it too, is a special case. Expand down the first column. How many terms survive? Only one! Yes, exactly. Column one is a one one, zero, zero, so only a one one survives, times its minor: a two two, a two three, zero, a three three. That minor is a two two times a three three. So the answer is the product of the diagonal entries. Next time, just multiply the diagonal; no long expansion needed!

4:55Example 1: 3x3 determinant along the column with zeros

Now let's solve our challenge from the start. Evaluate this determinant: first row two, zero, one; second row three, zero, minus two; third row five, four, one. Expand along the best line, okay? Pause the video, and try it first. Done? Okay, let's check. First, which line should we expand along? Column two! Yes. Column two is zero, zero, four. Two zeros, so only one term survives: four, times its cofactor. One small two by two determinant, just as I promised! Now the sign. The four sits in row three, column two; the round brackets three, two give that position. Three plus two is five, odd, so minus one to the power five is minus one: a minus place. Students slip right here! Now the minor. Delete row three and column two: two, one, three, minus two is left. Two times minus two, minus one times three: minus four minus three, minus seven. With the minus sign in front, the cofactor C three two is plus seven. Right? So the determinant is four times C three two, that is four times seven: twenty eight. The cross on screen means times. Forgot the sign? Then minus twenty eight. Oh no! One small determinant, and the question is done. That is the power of zeros.

6:22Example 2: solve a 3x3 determinant equation in x

Now a harder one. Solve for x: first row x, one, one; second row one, x, one; third row one, one, x; the determinant equals zero. See the pattern: x on the diagonal, one everywhere else. Pause the video and try it. Done? Okay, let's check. No zeros here, so expand along row one. First, x times its minor, x squared minus one; the small two up top means squared. Then minus one, times its minor, x minus one. That minus is the chessboard sign, and the dot means times. Then plus one, times its minor, one minus x. With me? Open the brackets: x cubed minus x, then minus x plus one, then plus one minus x. Collect the terms: x cubed minus three x plus two equals zero. Okay? Now we need the roots of this cubic. Can you guess one root quickly? x equals one! Very good. Put one in: one minus three plus two is zero. So x minus one is a factor. Right? Dividing, the cubic becomes x minus one, squared, times x plus two. The little two outside the bracket means x minus one appears twice. Clear? So x equals one, one, and minus two. Careful: one is a repeated root, so it is listed twice. Later in this chapter, you will see a faster way. But expansion always works. Well done!

7:58Quick revision: cofactor signs, best line, triangular

So, quick revision. Two by two: a d minus b c. Three by three: entries along one line, each times its cofactor, then add. The cofactor is the minor with the chessboard sign, top left plus. Expand along the line with the most zeros; for a triangular determinant, multiply the diagonal. Now go and try the practice problems on this page. You will find them easy. All the best!