Algebra 2 Determinants

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In today's lesson, we are going to cover determinants.0002

And determinants are associated with matrices, and they are useful, because they can help you solve problems such as systems of equations.0004

OK, first of all, defining a determinant--what is a determinant?0014

Every square matrix has a number associated with it, called the determinant of the matrix.0017

And notice that this is limited to square matrices--matrices with the same number of rows and columns.0022

And just talking about some convention: when we write a determinant, we use vertical lines, like absolute value bars.0030

For example, if I had a square matrix that was 2x2, and I wanted to indicate0041

that I am talking about the determinant of that matrix, I would write it as such.0046

So again, this is for square matrices only; and you use these vertical lines to denote that you are working with the determinant,0054

rather than the matrix (this is the determinant), because if I just used these brackets, then that would actually be the 2x2 square matrix.0062

All right, starting out, we are finding determinants of 2x2 matrices.0081

Given the matrix with the elements a, b, c, d, the determinant written with these vertical bars, a, b, c, d, can be found using this formula.0087

And a determinant that is associated with a 2x2 matrix is called a second-order determinant.0097

So, this formula, ad - bc (you multiply these two elements and then subtract the product of these two elements) is pretty straightforward.0105

Let's try an example, using the 2x2 matrix.0118

To find this determinant, I am going to multiply 2 times 3; and from that product, I will subtract this product, BC.0127

So, I have ab, and I am going to then find bc, which is 0, -1; this equals 6 - -1, which is a negative and a negative; that gives me a positive 7.0140

Therefore, this second-order determinant (for a 2x2 matrix) is 7.0156

Finding the determinant of a 3x3 matrix is more complex, and there are a couple of methods.0171

In certain cases, one method works better; in others, the other works better.0177

Or you may just have a preference for one method over the other; and I am going to show you both of these methods.0181

And this first one is called expansion by minors: so, I am looking here, and it is asking me for a third-order determinant.0187

So, it is for a 3x3 matrix, so the determinant here is a third-order determinant.0198

And one way to find it is to do what is called expansion by minors, and break this down into three second-order determinants.0203

Once you get these second-order determinants...we already talked about finding a second-order determinant.0212

If you had one that was a, b, c, d, you would find that by taking ad, and then subtracting the product bc from it.0226

So, if I can get these into second-order determinants, then it is pretty easy to work with.0239

Now, looking at it, actually what is happening here is: what we did is something called expanding along the first row (expanding along row 1).0246

You can expand along any row; but right here, what they did is expanded along row 1.0258

And the reason I know that is: I look here, and I see a (that is from row 1), b, c.0263

The way you find these minors is: you find what you are working with (which is a), and you eliminate the column that a is in, and the row that a is in.0270

And that leaves you with e, f, h, i.0286

OK, looking at what happened here with b: for b, I eliminated the column that b is in, which would be this one,0291

and the row that b is in, which would leave me d, g, f, i, right here.0304

For c, eliminate the column that c is in and the row that c is in; and that is going to leave me with this: d, e, g, h.0312

Now, you notice that there are alternating positive and negative signs.0325

And it is important to understand how this works, because you can expand along other rows, and then you need to know what to make the sign.0329

So, it starts out positive; a is positive; b is negative; here, c is going to be positive; and then, you just alternate.0337

Since I was working with a right here, the associated sign is positive; for b, it is negative; and so on.0348

You can actually expand along any row; and you will see, in a little while, why you may want to choose another row.0355

You want to pick the row that will give you the situation that is easiest to work with, in terms of finding these determinants and multiplying.0363

If I were to expand along row 2, then I would take whatever is here first,0371

and I would eliminate this column and this row; and then I would have these four things left.0377

Then, I would take this second element; I would cross out this column and this row and take what is left; and so on.0386

Let's do an example of expansion by minors.0395

All right, the first example: 0, 2, 1, 2, 3, 2, 1, 2, 0; I need to find this third-order determinant using expansion by minors.0401

And I am going to expand along the first row, just to keep it simple, like we did up here.0416

So, I start out with this 0, and I know that there is a positive sign associated with that.0422

And I am going to end up with looking for this second-order determinant.0428

So, 0 is in this first column (so I cross it out) and this first row; that is going to leave me with 3, 2, 2, 0.0434

Now, next I am rewriting this and erasing this, so I can cross out again: 0, 2, 1...0450

The second section right here is going to be associated with this 2, and it is going to be negative: -2 times this second-order determinant.0467

2 is in the first row, second column; cross that out; that is going to leave me with 2, 2, 1, 0.0481

Next, I have a 1 here, and there is a positive sign associated with that; so it is + 1, and that is a second-order determinant.0496

And I am going to find that (let's reconstruct this: 0, 2, 3, 2)...so, my 1 is right here in this first row, third column.0508

Cross out the first row and the third column to leave me 2, 3, 1, 2.0527

Now, this is much easier to work with: and one thing that is really nice is that there is a 0 here.0535

What I need to do is find each of these determinants, and then multiply them by the scalar, the number that is right out here.0542

OK, well, no matter what this turns out to be, I am multiplying it by 0; so it is just going to drop out.0549

So, I am going to have 0 times something; and it doesn't really matter what it is that is going to drop out.0558

Then, I am going to have -2 times this determinant; so I am going to find this determinant, using ad - bc.0563

ad is 2 times 0, minus bc--that is minus 2 times 1; plus...here is 1, times ad (which is 2 times 2), minus bc.0571

I had ad minus bc (minus 3 times 1).0611

Simplifying this: this is just 0 times whatever was in here--it didn't matter, because 0 times anything is 0;0620

minus 2 times...2 times 0 is 0, minus 2 times 1, plus 1 times...2 times 2 is 4, minus 3 times 1, which is 3.0628

This equals 0 minus 2 times -2, plus 1 times 4 - 3, which is 1.0645

Now, just coming up here to finish this out: get rid of this 0 at this point: this is -2 times -2, which is 4; 1 times 1 is 1; so this is 5.0665

So, at last, we found this third-order determinant for this, and it is 5.0686

Just taking it step-by-step, I expanded along row 1; I found the minor associated with each of these numbers0694

by crossing out the column and the row that that number was in, and writing out the four numbers (elements) that were left.0702

Then, I multiplied by 0 for this first one; for the second one, I crossed out row 1, column 2.0711

I was left with 2, 2, 1, 0, and multiplied that by 2, but I made it a -2, because of the negative signs associated with this.0720

For the third one, I had 1; I crossed out row 1, column 3; I was left with 2, 3, 1, 2.0729

I multiplied by a positive 1 (because it has a positive sign associated with that).0736

Now, I mentioned that you might want to expand along a different row.0739

The best rows to expand along are ones with 0's; and if there had been a row with two 0's, I would have expanded along that row.0743

And you see why: I didn't have to do any work here, because this is going to drop out.0751

If the second row, for example, would have had two 0's, I would have chosen that row.0754

I would have done the same method and expanded along row 2.0758

And if I had had two 0's there, great--this drops out; that drops out; I would only have had to work with this one.0762

So, you choose the row to expand out that has the most 0's, because it will be the easiest one to work with.0768

This is one method of finding a third-order determinant; and once I get these,0775

I find the second-order determinants using what we talked about before, this formula ad - bc.0779

I knew the first one was just going to drop out; the second one is represented right here; it is -2 times ad - bc.0785

And the third one is 1 times ad - bc.0795

Then, I just did the math to get 5.0801

Now, there is another method we are going to learn today for working with 3x3 matrices when you are finding the determinant.0803

And it is called the diagonal method: using an example to illustrate this, if you need to find0812

this third-order determinant, and you decide you want to use the diagonal method;0829

the first step is to rewrite the first two columns of numbers, right next to the determinant that you are looking for.0833

So, I am starting out with this; and I am going to just copy these two columns over.0854

It is 3, -1, 0, 1, 2, -2: that is the first step.0858

Next, you begin at the upper left, and you go along the diagonals, going from upper-left to down.0866

And you find the sum of the product of the diagonals.0877

Start at upper left, and find the sum of the products of the diagonals.0886

Now, what does that mean? It is actually simpler than it sounds.0903

Here, I have this first diagonal, 3, 2, 3; I am going to find that product, 3 times 2 times 3.0907

And then, I want to add it--I want to find the sum with the next diagonal, starting at the upper left: 1, 4, 0...1 times 4 times 0.0914

Then, I am going to add that to the next diagonal, 0, -1, times -2.0928

Then, you start at the lower left, and you find the products of those diagonals and subtract those.0939

Start at lower left, and subtract products of the diagonals.0948

Starting at the lower left, I have 0, 2, 0 on this diagonal--so minus 0 times 2 times 0.0973

On the next diagonal, I am going to subtract (just to make this clearer, let's go like this): minus this next diagonal, which is -2, 4, and 3.0986

Then, I am going to subtract this next diagonal: 3 times -1 times 1.1005

OK, so all you are going to do is find the products of the diagonals, starting up here and going down; find those products: 1, 2, 3; add them.1013

Find the products of these diagonals going up; subtract them.1024

All right, just doing the math: 3 times 2 times 3 is 9 times 2; that is 18; plus 1 times 4 times 0...0, plus 0 times -1 times -2...0,1030

minus 0 times 2 times 0 is 0; minus -2 times 4 (is 8) times 3 (is -24); minus 3 times -1 times 1 (so that is just -3).1046

I am simplifying things a bit: that is 18; let's get rid of these 0's--we don't need those; and this is minus -24, so I am going to write that as + 24.1067

Minus -3 is plus 3; 18 + 24 + 3 is 45.1077

If you would have used the other method, expansion by minors, you would have gotten the same result of 45 as this third-order determinant.1086

But you need to decide which method you like better and, given the situation, which method seems easier.1094

So again, in order to use diagonals to find this third-order determinant, we have to first rewrite the first two columns of numbers right next to the determinant.1100

That allowed us to form sets of diagonals; and we took the downward-going diagonals from here,1110

starting up at the upper left, and going down; and we found the products of each of those and added those.1117

Then, we started at the lower left, and found the products of those diagonals, and subtracted them.1123

Working out that math gave me 45, so that is the third-order determinant.1130

In this first example, we are going to find the second-order determinant.1142

If we have a second-order determinant with elements a, b, c, d, the determinant can be found1151

by multiplying ad--finding that product--and subtracting bc.1157

All right, so doing that right here is going to give me 1 times 4, minus b times c, which is -1.1162

OK, this equals 4 minus -3; if I have a negative and a negative, that gives me a positive: 4 plus 3, so it is 7.1181

So, the determinant here is 7, using this formula.1193

So, the second-order determinant...you just need to follow this formula--it is pretty straightforward.1197

A third-order determinant: I have two choices--I can either use diagonals, or I can use expansion of the minors.1204

And I am actually going to expand; and I am going to choose which row I should expand along.1211

And recall that I said, if you have 0's, to go to that row, because then things cancel out, and it makes your life a lot easier.1218

So, looking right here, I have a good situation, because I actually have two 0's up here in this row.1225

So, I am going to choose to expand along row 1.1232

Now, recall that, before we even do anything else, the signs associated would be positive, negative, positive;1236

negative, positive, negative...just so we make sure we are using the right signs for the row we are expanding along.1247

All right, so first, I am going to work with the 0.1252

And it doesn't matter if it is positive or negative, because it is 0; but it would have been positive.1255

Now, to find that associated minor, I am going to cross out the row that it is in and the column that it is in, so row 1 and column 1 are gone.1260

That leaves me with 3, 4, 2, 5; OK, I am rewriting this, just to make it really clear what is happening.1270

Now, next I am working with this 0; it is a 0, so it doesn't matter; but had it been another number,1285

it would have mattered that this would have been a negative sign associated with it.1291

So, right here, I am going to write the negative 0; and then, I am going to find the minor associated with it.1295

OK, this 0 is in row 1, column 2; that leaves me with 2, 4, -1, 5.1308

Next, I am writing this again, just so you can see each step.1322

The next number I am working with is 1; and that is going to be associated with a plus sign, because there is a positive sign right there in that position.1332

It's times this minor, which is going to be...I am going to cross out row 1, because that is where the 1 is, and column 3, to leave me with 2, 3, -1, 2.1340

OK, so as you can see, these 0's make the math much easier, because 0 times whatever this turns out to be is...1355

if I had bothered to find this determinant, it doesn't matter what it is, because it is just going to drop out.1367

The same with this one--I don't know what it is, and it doesn't actually matter.1373

Now, I need to find this determinant; and remember that the way I do it is: given that these are the positions,1377

a, b, c, d, I am going to find the product of a and d; and from that, I will subtract the product of b and c.1385

So, this is plus 1 times this; 2 times 2, minus 3 times -1.1394

This is 0; it drops out; 0 drops out; all I am working with is this.1417

1 times 2 times 2 is 4, minus 3 times -1 (that is -3), equals 1 times 4; a negative and a negative gives me a positive, so that is 1 times 7, which equals 7.1421

OK, so in Example 2, here we found the third-order determinant to be 7.1438

And I did that by expanding the minors along row 1; and I chose row 1 because I saw 0's.1444

So, I expanded by eliminating the column and row that that first 0 was in, leaving me 3, 4, 2, 5; and that was associated with a plus.1450

For this next number, it was a 0, and I saw that there was a negative associated with that, and a 0.1466

And then, I crossed out the row and the column that 0 was in, and was left with 2, 4, -1, 5; I wrote those there.1473

For the third element here, which was 1, I saw a positive associated with that, so I put + 1.1484

I crossed out the first row and the third column, and was left with 2, 3, -1, 2.1493

Then, once I had all this, I had to find these determinants.1499

And then, I realized that, since I am multiplying this by 0, I don't need to waste time even finding that; it is just going to drop out.1504

The same with this one; for this one, I used the formula to find this second-order determinant, which is ad - bc.1510

So, I have 0, 0, plus 1, times ad, which is 2 times 2, minus 3 times -1.1519

I calculated that out to find out that it is 7; the result is 7.1530

For this third-order determinant, I am going to use the diagonals method.1537

Recall that, with the diagonal method, the first thing you need to do is copy over those first two columns of numbers right next to the determinant.1542

You copy those over to expand the array; and then, you are going to find the products of diagonals,1555

beginning at the upper left: this diagonal, this diagonal, and this diagonal.1565

And you are going to add those products.1570

OK, so first, I have 2 times 4 times 1; then, I am going to add that to the product of this next diagonal, which is 3 times 6 times 0.1572

For the next diagonal, I have 1 times -1 times 8; OK, so that took care of those diagonals.1589

Then, I am going to start at the lower left, down here; and I am going to find the product of those diagonals; and this time, I am going to subtract.1599

So, minus 0 times 4 times 1; then, I am going to subtract the product of this next diagonal, and the product is the product of 8, 6, and 2.1606

Then, the next diagonal: subtract that product, which is the product of 1 times -1 times 3.1630

OK, then it is just working out all the arithmetic and making sure that you don't make any errors.1640

2 times 4 times 1 gives me 8, plus 3 times 6 (that is 18); then, this is...let's see...1 times -1 is -1, times 8--that gives me -8.1645

Actually, it is 3 times 6 times 0; correction--that is 0: 3 times 6 times 0--that is 0.1671

This is 1 times -1, is -1; times 8--that is -8.1678

Here, again, I have a 0; so this is just going to become 0.1683

Here, I have minus 8 times 6, which is 48, times 2--that is going to give me 96, minus 1 times -1 is -1, times 3 is -3.1690

Simplifying this a little further, this is 8; eliminate the 0; minus 8; eliminate the 0; minus 96; and this is minus -3, so it is plus 3.1710

8 minus 8 is 0, so that leaves me with -96 plus 3, to give me -93; so, this determinant is -93.1722

And I used the diagonal method, first copying these first two columns of numbers right here,1733

then finding the products of the diagonals, going from upper left to lower right, and adding those products.1740

Then, I went to the lower left, and along those diagonals, I found the products; and I subtracted those, did my arithmetic, and got -93 as the determinant.1749

OK, in Example 4, again, I am looking for a third-order determinant.1763

And you could use either method; I am going to actually expand along row 1.1770

That has a 0 in it, so I am going to use expansion of the minors.1779

You could have also used row 2 or 3; they all have 0's in them; but these also have smaller numbers, so I am going to go with row 1.1783

Now, with the expansion of the minors, I want to keep track of my positive and negative signs.1790

I am going to start with 0, and that is associated with a positive.1800

And when I expand that, I am going to eliminate this first column and this first row; and it is going to leave me with 3, 0, 6, 3.1805

Then, next, I am going to rewrite this...2, 3, 4...so that you can see very clearly.1819

The next number I am working with is 2, and that is associated with a negative sign; so I am going to write -2.1827

It is in row 1, column 2; get rid of those, and see that I am left with 4, 0, 0, 3.1832

Next, I am rewriting this original; I am now working with the 3, and that has a positive sign associated with it, so plus 3.1844

Eliminate the row it is in and the column it is in; 4, 3, 0, 6.1859

Now, I have 0 times this determinant; that is just going to drop out, so I am not even going to waste my time figuring out that determinant.1868

These other ones...I use ad - bc to find them.1878

So, I have minus 2, times...this is 4 times 3, minus 0 times 0, plus (so I have this taken care of) 3 times ad - bc.1887

So, that is 4 times 6, minus 0 times 3; OK, do the math now.1906

This is just 0; it is going to drop out; minus 2, times 12 minus 0, plus 3 times 24 minus 0.1914

This becomes 0 minus 2, times 12, plus 3 times...this is just 24.1925

Let's get rid of the 0 at this point: -2 times 12 is -24; 3 times 24 is 72; -24 + 72 = 48.1935

OK, so we found this third-order determinant by using expansion of the minors along row 1,1949

and being careful to keep the signs straight on the numbers associated with those,1955

and crossing out the row and column for each number to find the associated minor.1961

Then, once I had those, I saw that this first one was just going to drop out.1969

For the second and third, I used ad - bc to figure this out, and multiplied this one by -2 and this one by 3; working that out gave me 48.1973

That concludes this session of Educator.com on determinants; I will see you next lesson!1988