Question

I'm trying to use cofactor expansions to find the determinant of a 3x3 matrix. I'm pretty sure I understand the approach, but I keep missing the answer in my book and calculator by 4. Would appreciate an extra set of eyes. I've attached a screenshot of my work.

Answer 1

Ah, I think I may have jumped the gun on a technique. I'm going to try a different approach and see where I net out. :P

Answer 2

The second method is correct, i.e. 8 is the correct answer.
In the first method, the second and fifth expansions are duplicated.
I usually expand it by row or column, not the whole matrix.

Answer 3

UGH. :) I don't know how many times I looked at that top row and couldn't figure out why it wasn't adding up. Thank you Here goes.

Answer 4

I do for example:
3 4 1
2 4 2
1 2 3
=
3(4*3-2*2)-4(2*3-2*1)+1(2*2-4*1)=24-16+0=8

Answer 5

Ah, that's the way I was just about to try.

Answer 6

Also, the first matrix is not the same as the second one.

Answer 7

To avoid having to worry about the sign of the cofactor, I also extend the first column to the right so I always add:
3 4 1
2 4 2 2
1 2 3 1
so I have
instead of the second element of first row being -4(2*3-2*1), I do
+4(2*1-2*3)

Answer 8

The correct answer for the first one is 4.
BTW, how did you make the writing with a black background? With a special software?

Answer 9

It's one of my favorite math tools, Photoshop. I draw a lot, too. http://imgur.com/a/rxkCM#37 Or at least I used to, before physics. :)

Answer 10

Wow, did you draw all that? and your writing is beautiful as well! congrats!

Answer 11

I did, thank you. :)

Answer 12

I wish I were as patient with the manual bookkeeping of linear algebra.

Having been a programmer for the last 15 or so years, it's hard to want to do something out long when even my old HP calculator can hash it out in seconds.

Answer 13

"How I wonder what you are" Were those chocolates, or meant to be a twinkling star? :)

Answer 14

"Twinkie, Twinkie, little star..." :D

Answer 15

Your HP cannot do a 3x3? I guess HP25, but that's more than 15 years!

Answer 16

It *can*, it's just that my professor is going to want to see more work than an answer.

Answer 17

I have an HP50G (my 48g died).

Answer 18

Yes, linear algebra takes a little patience, but after a while, and with a little discipline, expanding a 3x3 won't be any challenge to a 15-year programmer! :)

Answer 19

It's much easier than linked lists or binary trees!

Answer 20

Once you know it. ;)

Answer 21

yep, the secret is "practice, practice, and practice", just like programming... Hands on, not a spectator sport! :)

Answer 22

Im following allllll the way up to the last step, det(A). What's going on there?

Answer 23

Well, after watching a video on Khan, I *thought* that's the step where I arrive at the determinant. I was trying to go from matrix of minors to the cofactor matrix, and from there, the sum of the orange arrows minus the sum of the blue ones.

Answer 24

Im trying to understand why 4(1)(1) = -4....etc.

Answer 25

That first screenshot is off. I'm trying to work through it now.

Answer 26

Ohh, ok.

Answer 27

\(\color{blue}{\text{Originally Posted by}}\) @mathmate
The correct answer for the first one is 4.
BTW, how did you make the writing with a black background? With a special software?
\(\color{blue}{\text{End of Quote}}\)

I just saw this. Hahahah. Sorry, I didn't bother reading through the other posts.

Answer 28

Funny, that is the biggest question I get here. :)

Answer 29

Im off, good luck!

Answer 30

Thanks.

Answer 31

Argh. The arithmetic is killing me.

Answer 32

Even with the 0, 1 and 2's? It's probably the bookkeeping that bothers you!

Answer 33

That + ADHD, yes.

Answer 34

What I usually try to do is to blank mentally (if you can) the row and column that we cross-out, and do the calculation of the 2x2 mentally. This will reduce a lot of your bookkeeping.
Khan used all the bookkeeping to explain, but surprises me he didn't show what people actually do.

Answer 35

So when I get to the 2x2, let's say it's this in the top-left corner:
21
12
That position should =3, right?

Answer 36

yep!

Answer 37

Pray continue!

Answer 38

Something's just not coming out in the wash.

Answer 39

Which method are you doing, the cofactors?
It's a little longer than expansion by row/columns.

Answer 40

Minors and cofactors and a little something extra probably.

Answer 41

What you can do, instead of writing out a whole matrix, you write out the cofactors, and alternate the signs,

2 1
1 2
=3

1 1
0 2
=(-) 2

1 2
0 1
= 1

for the first row, but you don't have to write out the cofactors, so it simplifies to
3-2+1

Answer 42

Basically a cross-product?

Answer 43

Yes!

Answer 44

Saves you a lot of writing.

Answer 45

Aw man, they could have just said *that* in the book. :|

Answer 46

Even better, you don't even have to expand all nine of them!
In fact, if you multiply each of the cofactor by the value of the element you're expanding on, you get the determinant:
2(3)+1(-2)+0(1)=4

Answer 47

So when they say "use cofactor expansions to evaluate", that's what I'm doing?

Answer 48

* determinant of each cofactor

Answer 49

And does that work for any square matrix?

Answer 50

That's good for any matrix (must be square), and any degree.
It's just that finding the determinant of a 4x4 means expanding the cofactor of a 3x3 16 times!

Answer 51

Well now, *that* actually makes sense! Maybe *you* should be writing textbooks. :)

Answer 52

As I said, Khan is very strong in explaining, but I am surprised they didn't show these "normal practices" at the end.

Answer 53

So want to try a 3x3?
2 0 1
1 2 0
0 1 2
You can check by the method, perhaps even mentally.

Answer 54

I think 9 is the answer to yours?

That's the first time I've been lead down a *more* difficult path on that site.

It seems like every weekend, I get right up to the point where I think I can't possibly decipher the next concept. Fortunately kind, patient people like yourself manage to get me to the next exam. I don't know what I'd do without this site.

Answer 55

Thank you.

Answer 56

Going to try a 4x4 now.

Answer 57

Yes, nine is correct.
You can solve it by the short method:
2(4-0)-0(2-0)+1(1-0)
or your second method : 2*2*2+1*1*1*1+0*0*0 -(1*2*0+0*1*2+0*1*2)=8+1+0-(0+0+0)=9

Answer 58

Great!
Do you need one to work on?

Answer 59

I think I'm good for now. Just going to see if I can get back on track with the rest of my homework.

Answer 60

Fortunately, my book has *all* the answers in the back, and I can make problems up with the calculator if I need more. Thanks again.

Answer 61

Yeah, that's good!
If you want, you can try
2 1 3 0
0 2 1 3
3 0 2 1
1 3 0 2
The answer should be 48, but it will take A LOT OF patience!
Catch up with you some other time!

Answer 62

Excellent. :)

Answer 63

In fact, it's not that bad, if you use the short method.
The 3x3 can be solved mentally using your second method.

Answer 64

Bye and thanks for the interesting conversation!

Answer 65

Thank you, as well. Take care.

Answer 66

you too!