Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to take the inverse of this matrix:

$\left[\begin{array}{rrrr} 2/7 &-6/7 &3/7 &1 \\ 6/7& 3/7 &2/7 &2 \\ -3/7 &2/7 &6/7 &3 \\ 0\ \ &0\ \ &0\ \ &\>\>\>1 \end{array}\right]$

I have been using this as a guide and I've seen people talk about row elimination (I don't know how to use it) to solve it but I am unsure which strategy would be the best way to attack this problem. I know its going to be a large amount of work either way.

share|improve this question
    
What do you mean by A^-1 = 1/detA? –  TMM Sep 18 '11 at 20:05
1  
@Thijs I understand that you are pointing out that the question is unclear. In any case, my guess is that the OP means this formula: en.wikipedia.org/wiki/Invertible_matrix#Analytic_solution. –  Srivatsan Sep 18 '11 at 20:10
1  
$A^{-1}$ is a matrix. $\det(A)$ is a scalar, so $1/\det(A)$ is a scalar. You cannot have $A^{-1}=1/\det(A)$. It's true that $A^{-1}$ equals $(1/\det(A))\mathrm{adj}(A)$, where $\mathrm{adj}(A)$ is the adjugate of $A$. Is this what you are talking about, and the process of finding $A^{-1}$ through row elimination instead of using the adjugate? –  Arturo Magidin Sep 18 '11 at 20:12
    
@Sri: No; I am pointing out that the statement $A^{-1} = 1/\det(A)$ is unclear, which does not seem to be his question. You should not have to guess what he means with that. (Also, I doubt he means that formula, as you don't learn that formula before learning about row elimination.) –  TMM Sep 18 '11 at 20:17
    
Yes, Sorry the link that Sri was pointing to was what I was referencing. I've also used the adjugate matrix to solve the 3x3 in a previous homework but thought there may be a better approach. –  Nick Sep 18 '11 at 20:30

4 Answers 4

up vote 5 down vote accepted

Here the best approach might be to observe that the upper left 3x3 block forms an orthogonal matrix. IOW, those three first columns form an orthonormal set of vectors.

Mind you, that was just a hint. It doesn't give you the inverse of the 4x4 matrix, but it is a good start!

[Edit] Extending the hint a little bit. The matrix is of the block form $$ A=\left(\begin{array}{cc}P&X\\0&1\end{array}\right), $$ where the 3x3 block $P$ is orthogonal, so $P^{-1}=P^T$. Using that observation it is easy to write down an inverse for the matrix $A'$ gotten from $A$ by replacing the the 3-vector $X$ with all zeros. The difference between $A$ and $A'$ amount adding multiples of the fourth row to the others. The way to invert that is...

Looks like the matrix represents a combination of a rotation and a translation of the 3D-space. Did this come from a 3D-graphics programming exercise by any chance?

[/Edit]

share|improve this answer
    
This was the exact method my professor showed me in order to solve the problem when I went in to him today. Thank you. And yes this is a for a robitics class where we use many translation and rotational matrices. –  Nick Sep 19 '11 at 19:01

Here's a method for inverting matrices using row reduction, which I think is what you are after:

Let $A$ be an $n\times n$ matrix. Let $B$ be the $n\times 2n$ matrix obtained by placing the $n\times n$ identity matrix to the right of $A$: $$B = \left( A\ |\ I_n\right).$$

Now, perform row reduction on this matrix until the left half is the identity matrix (or has a row of zeros). Then the matrix on the right half is $A^{-1}$ (or, if you got a row of zeros in the left half, then $A$ is not invertible).

The reason this works is that performing elementary row operations is equivalent to multiplying on the left by an elementary matrix. By doing the same operations to the identity matrix, you are computing the product of those elementary matrices. If $E_1,\ldots,E_n$ are elementary matrices such that $$E_nE_{n-1}\cdots E_1A = I_n,$$ then it follows that $E_nE_{n-1}\cdots E_1 = E_{n}E_{n-1}\cdots E_1I_n = A^{-1}$. (Because for $n\times n$ matrices over fields, if $CA=I_n$ then $AC=I_n$). Thus, the computations on the right half of the matrix $B$ give you the inverse of $A$.

For large matrices, this is certainly easier to do by hand than using the adjugate; it is probably faster in computers as well.

share|improve this answer

I think this is better explained in real time.

This video from YouTube seems popular. It demonstrates the row reduction method to find an inverse on a 3x3. Try to follow along with the example matrix first, and your matrix will not be much harder.

share|improve this answer
    
Would it be appropriate to choose the "community wiki" option when 1) this doesn't answer the question, and 2) the video isn't even yours? (I presume) –  The Chaz 2.0 Sep 18 '11 at 23:23
1  
This makes sense to me. Thanks for the suggestion. –  Hans Parshall Sep 19 '11 at 4:24

I always find row-reduction much useful and probably fastest for finding inverses.

You can use this dynamic linear algebra tool kit (finding inverse is the fifth feature after you click enter) to learn as well as visualize how your matrix is reduced to its inverse.


ADDED:The direct inverse hyperlink is working (as of now) and here is the self explained result for your matrix.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.