Question about finding the right inverse of a matirx

[Claire84]

Hey there, having a dizzy moment here at the mo about right and left inverses. We've got the matrix B given by first row (1 3 5) and second row (2 4 6). Normally I'd just have BC=I where C is 3x2 and we have the 2x2 identity matrix. However, I'm confused about solving it because the one we solved in class was B= first row (400) and second row (050), which was like a matrix that had been reduced by Gaussian elimination or something (or maybe it was justcoincidence!)... anyhoo, am I supposed to reduce my B matrix (the first one I mentioned), or do I just shove it stright into BC=I where C is the inverse of B? I've tried just shoving it in but I'm not sure how I'm supposed to solve it with all the unknowns. We're told to find the general form of the righ inverse of the matrix B, so does this mean we can have unknowns in it or something or something instead of them? I'm so confused. Any help would be really appreciated. Thanks!

Claire

 

[TenaliRaman]

Gee, i never came across anything like this before ...
For me most of the times B and C have been square matrices ...
Hmm reducing B does not seem an option at all .. as i don't see what to reduce it to...
My initial thoughts,
consider C as
[a b]
[c d]
[e f]
multiply out B with C
and then solve it as linear equations of 6 unknowns ..
use any linear equation solving method .. *simplest being Gaussian elimination method*
since there are 6 unknows and 4 equations,
there must be more than one solution to this ...

-- AI

 

[matt grime]

This is more part of a more general property of maps between sets:

if F is a surjection from X to Y, then there is a map g from Y to X such that fg is the identity from Y to Y. (nb, I may be assuming the axiom of choice here for those with a distaste for such things)
Here you can just work out the possible right inverse for a linear map (matrix). It will not in general be unique.

Anyway, you can greatly simplify your work here since you may assume e and f are zero and that

ab
cd

is the inverse of

13
24

Do you see why?

 

[TenaliRaman]

But ofcourse that's one solution of many

-- AI

 

[Claire84]

Thanks for your help. I rummaged around some past papers as well, and ended up doing it with the general solution is equal to the particular solution + homogenous solution. It wasn't exactly something I'd have thought of mind you (being a bit empty-headed most of the time!).

 

[Claire84]

Oh, and while I'm here (in a question asking mood here, as-per-usual), I was wondering if someone could help me with finding the deteminant of a matrix. I've done the other h/work questions okay, but this one has me fuzzled. Here's the link to the h/work sheet.

http://titus.phy.qub.ac.uk/group/Jorge/AMA203/assignments/ass03_4.pdf

It's question 2, matrix B.

If I do it the Gaussian elimination way, I end up with the determinant being 76, and if I do it the oher way (think it's called cofactor or something), I end up with zero! I don't understand how it could be zero though since I don't think the matrix is singular, but I don't know. We haven't actually covered determinants yet in class, but I've been working from a book (and online notes) and everything seems to be fine so far. I'll try and scan in what I've done so far...

 

[Claire84]

Attached is the first part of the Gaussian elimination, other bit just coming up.

 

[Claire84]

Here's the last bit of the elimination. If you can spot anywhere that I've made an error then I'd love to know. The cofactor method has bene started there but I'll not bother putting the whole thing up just yet because using a scanner is stressful for me!

 

[Claire84]

Btw, sorry if those are really difficult to see; it wouldn't let me have them any bigger!

 

[MiGUi]

Basically, you have to do lineal combinations of rows and columns to get the more zeroes as you can.

For example:

A = \left( \begin{array}{ccc} 1 &amp; -1 &amp; 3 \\ 2 &amp; 4 &amp; 5 \\ -3 &amp; 1 &amp; 2\end{array}\right) <br />

Now, I add the column 2 to the column 1 and I got a zero:

A = \left( \begin{array}{ccc} 0 &amp; -1 &amp; 3 \\ 6 &amp; 4 &amp; 5 \\ -2 &amp; 1 &amp; 2\end{array}\right) <br />

Then, I add the row 1 to the row 3 to get another zero:

A = \left( \begin{array}{ccc} 0 &amp; -1 &amp; 3 \\ 6 &amp; 4 &amp; 5 \\ -2 &amp; 0 &amp; 1\end{array}\right) <br />

Now, I multiply by 2 the third column and I add to it the first one, to get another zero:

A = \left( \begin{array}{ccc} 0 &amp; -1 &amp; 6 \\ 6 &amp; 4 &amp; 10 \\ -2 &amp; 0 &amp; 2\end{array}\right) \rightarrow^{1+3} <br /> <br /> \left( \begin{array}{ccc} 0 &amp; -1 &amp; 6 \\ 6 &amp; 4 &amp; 11 \\ -2 &amp; 0 &amp; 0\end{array}\right)<br />

And so, and so and so...

 

[Claire84]

What method is that one? I don't think we'e covering it. Just the cofactor one and the Gaussian elimination one. Thanks though.

 

[Claire84]

Eurgh, made a mistake during elimintion. Took me 6 hours to work out. Pfffft! Such a numpty!