Linear Algebra

[hans030390]

I'm supposed to find the formula for the inverse of a matrix. Basically the steps are as follows:

1. Find the matrix of cofactors
2. Take transpose of that and divide by the determinant of the original matrix.

Here's the matrix:

| 1 x y |
| 0 1 z |
| 0 0 1 |

Now, I've followed all the steps and have obtained the following:

| 1 x xz-y |
| 0 1 z |
| 0 0 1 |

However, the solution my professor gave says that the x and z values (not xz-y) should be negative...but mine aren't.

What did I do wrong/what am I missing?

 

[theflyingpig]

You know, It'd be really, really nice if you posted all of the steps you took. That might help.

 

[paulney]

If you are doing a matrix inverse, it's just this
At least, that's how I remember doing it in college.

 

[hans030390]

Originally posted by: theflyingpig
You know, It'd be really, really nice if you posted all of the steps you took. That might help.

I got the same answer as the solution, except the fact that the x and z are apparently supposed to be positive. I don't see why I'd need to post every step I did if someone knows how to quickly check those two things.

 

[hans030390]

Originally posted by: paulney
If you are doing a matrix inverse, it's just this
At least, that's how I remember doing it in college.

Yeah, that's basically what I did, then I had to transpose it and divide by the original matrix determinant (which was 1).

 

[Born2bwire]

Originally posted by: hans030390

Originally posted by: theflyingpig
You know, It'd be really, really nice if you posted all of the steps you took. That might help.

I got the same answer as the solution, except the fact that the x and z are apparently supposed to be positive. I don't see why I'd need to post every step I did if someone knows how to quickly check those two things.

Well, what are supposed to tell you then? It's pretty obvious from a quick check that you need the negative signs and you claim that you are doing the appropriate steps. We can't psychically tell you what you did wrong.

 

[Molondo]

Don't the signs alternate?

 

[hans030390]

Originally posted by: Molondo
Don't the signs alternate?

I don't know? I really don't understand this much. My professor only goes over the lesson one day a week, has a review/discussion the next class day, then a test over it the next class day. It's all very quick and he doesn't spend much time explaining...he also has a heavy accent and has horrible handwriting.

 

[Born2bwire]

Well when I do it I get:

1 -x xz-y
0 1 -z
0 0 1

 

[hans030390]

Originally posted by: Born2bwire
Well when I do it I get:

1 -x xz-y
0 1 -z
0 0 1

Ok...can you show me the steps on how you got -x and -z? Here's how I, for example, got x instead of -x:

In the matrix of cofactors I made, row 2, col 1 is going to be found by making that same point the cofactor value in the original matrix. Based on that, I find the determinant of the following matrix:

| x y |
| 0 1 |

So, that's (x * 1) - (y * 0) = x

So that value in the matrix of cofactors is x...and then the transpose...? Then I end up where I'm at, and apparently not where I'm supposed to be.

 

[Born2bwire]

Originally posted by: hans030390

Originally posted by: Born2bwire
Well when I do it I get:

1 -x xz-y
0 1 -z
0 0 1

Ok...can you show me the steps on how you got -x and -z? Here's how I, for example, got x instead of -x:

In the matrix of cofactors I made, row 2, col 1 is going to be found by making that same point the cofactor value in the original matrix. Based on that, I find the determinant of the following matrix:

| x y |
| 0 1 |

So, that's (x * 1) - (y * 0) = x

So that value in the matrix of cofactors is x...and then the transpose...? Then I end up where I'm at, and apparently not where I'm supposed to be.

It should be
-| x y |
| 0 1 |
You are taking the cofactor using the minor M_{12}. The sum of the indices is odd, hence you take the negative of the determinant (minor) for the cofactor value. You'll notice that this was done in the image linked from Mathworks in a previous post.

 

[hans030390]

Originally posted by: Born2bwire

Originally posted by: hans030390

Originally posted by: Born2bwire
Well when I do it I get:

1 -x xz-y
0 1 -z
0 0 1

Ok...can you show me the steps on how you got -x and -z? Here's how I, for example, got x instead of -x:

In the matrix of cofactors I made, row 2, col 1 is going to be found by making that same point the cofactor value in the original matrix. Based on that, I find the determinant of the following matrix:

| x y |
| 0 1 |

So, that's (x * 1) - (y * 0) = x

So that value in the matrix of cofactors is x...and then the transpose...? Then I end up where I'm at, and apparently not where I'm supposed to be.

It should be
-| x y |
| 0 1 |
You are taking the cofactor using the minor M_{12}. The sum of the indices is odd, hence you taken the negative of the determinant (minor) for the cofactor value. You'll notice that this was done in the image linked from Mathworks in a previous post.

...I really am not sure what you're talking about. Like I said, the professor goes over everything really quickly in one day and then expects us to be able to take a full test over it after our "review" class period (where he teaches nothing new). I've bolded parts I'm particularly stuck on or don't know what you're talking about.

 

[Born2bwire]

Originally posted by: hans030390

Originally posted by: Born2bwire

Originally posted by: hans030390

Originally posted by: Born2bwire
Well when I do it I get:

1 -x xz-y
0 1 -z
0 0 1

Ok...can you show me the steps on how you got -x and -z? Here's how I, for example, got x instead of -x:

In the matrix of cofactors I made, row 2, col 1 is going to be found by making that same point the cofactor value in the original matrix. Based on that, I find the determinant of the following matrix:

| x y |
| 0 1 |

So, that's (x * 1) - (y * 0) = x

So that value in the matrix of cofactors is x...and then the transpose...? Then I end up where I'm at, and apparently not where I'm supposed to be.

It should be
-| x y |
| 0 1 |
You are taking the cofactor using the minor M_{12}. The sum of the indices is odd, hence you taken the negative of the determinant (minor) for the cofactor value. You'll notice that this was done in the image linked from Mathworks in a previous post.

...I really am not sure what you're talking about. Like I said, the professor goes over everything really quickly in one day and then expects us to be able to take a full test over it after our "review" class period (where he teaches nothing new). I've bolded parts I'm particularly stuck on or don't know what you're talking about.

http://en.wikipedia.org/wiki/Cofactor_(linear_algebra)

The cofactor, denoted by C, is dependent upon the minor, denoted as M. The two are related as
C_{i,j} = (-1)^{i+j}M_{i,j}
Whenever the sum of the indices of the cofactor that you wish to find is odd, you take the negative of the associated minor.

 

[drinkmorejava]

Other than what paulney said, there isn't really a generalized formula for an inverse past a 2x2 matrix.

 

[Born2bwire]

Originally posted by: drinkmorejava
Other than what paulney said, there isn't really a generalized formula for an inverse past a 2x2 matrix.

There are algorithms though, like LU decomposition or Gaussian elimination.

 

[hypn0tik]

The cofactor also alternates in sign like the determinant.

|+ - +|
| - + -|
|+ - +|

So since you are looking at Row 2 Column 1, attach a negative sign in front.

 

[hans030390]

Originally posted by: Born2bwire

Originally posted by: hans030390

Originally posted by: Born2bwire

Originally posted by: hans030390

Originally posted by: Born2bwire
Well when I do it I get:

1 -x xz-y
0 1 -z
0 0 1

Ok...can you show me the steps on how you got -x and -z? Here's how I, for example, got x instead of -x:

In the matrix of cofactors I made, row 2, col 1 is going to be found by making that same point the cofactor value in the original matrix. Based on that, I find the determinant of the following matrix:

| x y |
| 0 1 |

So, that's (x * 1) - (y * 0) = x

So that value in the matrix of cofactors is x...and then the transpose...? Then I end up where I'm at, and apparently not where I'm supposed to be.

It should be
-| x y |
| 0 1 |
You are taking the cofactor using the minor M_{12}. The sum of the indices is odd, hence you taken the negative of the determinant (minor) for the cofactor value. You'll notice that this was done in the image linked from Mathworks in a previous post.

...I really am not sure what you're talking about. Like I said, the professor goes over everything really quickly in one day and then expects us to be able to take a full test over it after our "review" class period (where he teaches nothing new). I've bolded parts I'm particularly stuck on or don't know what you're talking about.

http://en.wikipedia.org/wiki/Cofactor_(linear_algebra)

The cofactor, denoted by C, is dependent upon the minor, denoted as M. The two are related as
C_{i,j} = (-1)^{i+j}M_{i,j}
Whenever the sum of the indices of the cofactor that you wish to find is odd, you take the negative of the associated minor.

Ah...got it. Looking back, some other problems now make sense...he just did a poor job explaining that to us (or expected us to know it). I mean, this guy decided we didn't need a book...no homework...just weekly tests, and crappy notes he found from some other guy online.

Edit: Actually, he did go over this...well, not really. I have a drawing showing basically what you just stated in words...but I was able to make no sense out of it during the lecture or after the class. Oh well, thanks a bunch.

 

[oboeguy]

Originally posted by: hans030390
Ah...got it. Looking back, some other problems now make sense...he just did a poor job explaining that to us (or expected us to know it). I mean, this guy decided we didn't need a book...no homework...just weekly tests, and crappy notes he found from some other guy online.

Edit: Actually, he did go over this...well, not really. I have a drawing showing basically what you just stated in words...but I was able to make no sense out of it during the lecture or after the class. Oh well, thanks a bunch.

He's probably trying to be a nice guy by not having a textbook requirement (saving you $$$). BTW, did you check your answer by multiplying it against the original matrix? You should get an identity matrix if the inverse is correct.

Anybody have a suggestion for a decent undergrad LA textbook for the OP to track down at a library? (or buy)

 

[BlahBlahYouToo]

it's orthogonal.

 

[drinkmorejava]

Originally posted by: Born2bwire

Originally posted by: drinkmorejava
Other than what paulney said, there isn't really a generalized formula for an inverse past a 2x2 matrix.

There are algorithms though, like LU decomposition or Gaussian elimination.

Err what, those are for solving systems by creating identity matrices or allowing backwards/forward substitution, not finding an inverse.