What is the efficient way to find the determinant of a matrix?

[Tenshou]

What happened to the rest?

$x_1=2u-v \\x_2= u+v \\ x_3=v-u\\$

So the book has this arranged in a 2-by-3 matrix below
$\left[ \begin{array} { c c } 2 & -1 \\ 1 & 1 \\ -1 & 1\\ \end{array} \right]$
Then end up taking the determinant of this matrix, but the two by two part at the top which looks like the below
$det\left| \begin{array} { c c } 2 & -1 \\ 1 & 1 \\ \end{array} \right|$
The question is how did they do this (I see why it is to find the rank of this transformation) but why would they only use the top part, the bottom part has a determinant of 2 which is less than three... this is sooo confusing

 

[Simon Bridge]

So the book has this arranged in a 2-by-3 matrix below

... which book?
What was the section attempting to demonstrate?
What was the end result supposed to be?

You say you understand that it is to do with the rank of the transformation ... if you have an nxm matrix: what determines it's rank? What does the rank signify?

 

[lurflurf]

What book is that so I can read it. Anyway we know the rank is at most 2 because it is a 2x3 matrix. One way to show that it is exactly 2 is to show that a 2x2 subdeterminant is not zero which they did. It is obvious by inspection that the rank is two anyway. In fact as you point out all three of the 2x2 subdeterminants are nonzero so any could be used.

 

[Simon Bridge]

It is obvious by inspection...

huh - when I was a student struggling with a topic, I wanted to smack profs who said stuff like that ;)

But yeh - to show the rank, they could have used any of three pairs of rows ... but they have to use at least one pair - why not the top one?

 

[HallsofIvy]

$x_1=2u-v \\x_2= u+v \\ x_3=v-u\\$

So the book has this arranged in a 2-by-3 matrix below
$\left[ \begin{array} { c c } 2 & -1 \\ 1 & 1 \\ -1 & 1\\ \end{array} \right]$
Then end up taking the determinant of this matrix,

No, they don't only "square" matrices have a determinant.
You are leaving out an important point- what question are they trying to answer? Why do they take this "determinant"?

This linear transformation, from $R^2$ to $R^3$ can only map $R^2$ into an at most two dimensional subspace of $R^2$. I suspect the question was whether it really does map it into a two dimensional subspace or only a one dimensional subspace or even a single point.

but the two by two part at the top which looks like the below
$det\left| \begin{array} { c c } 2 & -1 \\ 1 & 1 \\ \end{array} \right|$
The question is how did they do this (I see why it is to find the rank of this transformation) but why would they only use the top part, the bottom part has a determinant of 2 which is less than three... this is sooo confusing

 

[Tenshou]

Well, I guess it was asking to find the rank of the matrix(and demonstrate a proof of a theorem), but they took the determinant of the submatrix, which threw me off, because what I learned in Linear Algebra was you know, Row Reduce echelon form and then see the left over free variables and you got your rank. So, my question is, to be more succincty and to clarify any confusion, Why in hell did they take the determinant of this matrix which ended up to being 3 and then deduce that the rank was only 2( Well at the end they ended up looking at elimination to deduce this), I guess I was just really thrown off by the determinant of the submatrix

The book is a schaums outline on differential geoemtry, I want to get caught up because I am falling behind on my studies. The rank signifies the number of "fixed" solutions so, it will tell you if the transformaiton is either a bijective one, injective one or surjective one.

I think lurfy answered my question.

 

[WannabeNewton]

huh - when I was a student struggling with a topic, I wanted to smack profs who said stuff like that ;)

Well in this case it is obvious by inspection because there are only two column vectors in the matrix and they form a linearly independent set so the rank must be 2.

 

[Simon Bridge]

they took the determinant of the submatrix, which threw me off, because what I learned in Linear Algebra was you know, Row Reduce echelon form and then see the left over free variables and you got your rank.

... and if you knew a shorter way to find the same information, you'd use it right? That's what they did.

The rank is the number of linearly independent rows - which is also the order of the largest square sub-matrix that has a non-zero determinant.

http://www.math4all.in/public_html/linear algebra/chapter4.5.html
... this one has a proof.

http://www.vitutor.com/alg/determinants/matrix_rank.html
... this one is easier to read.

 

[Tenshou]

Thanks for the resource Simon, that was very helpful :D

 

[Simon Bridge]

No worries.
The matrix approach gets faster with a bit of practice - experienced mathematicians get so they can look at a matrix and get a good idea which square sub-matrix to pick that won't be singular. Beats tedious row-reducing every time. You saw how WannabeNewton thought the result was "obvious by inspection" (by another means) - it's experience.

Of course, get it wrong and you just replace the tedium of row reducing with the tedium of computing determinants :)