Inverse Matrices.

Here's the result for a 3x3 matrix in case you ever want to use it.  (I didn't do this by hand!)

Thank you. Do the signs on c and d need to be changed when finding the determinant? What was the 1/|A| I was explained to do? Is there another way to find I? I was shown something multiply by first row, and add second row or something like that. Is that what I saw on Kahn Acadamy, reduced row echlon or something to solve for I? Thanks for the post.

Do the signs on c and d need to be changed when finding the determinant? 

No.  The determinant is the scalar value (i.e. a single number): ad-bc.

What was the 1/|A| I was explained to do?

This is just 1/(ad-bc).  Look at my expression for the inverse of A.  Every term has ad-bc on the bottom.  Because every term has a common factor this factor can be taken out of the matrix as a constant multiplier, so the inverse could be written as:

Look at the numerators of each term (including the signs).   

Is there another way to find I?

You don't really find I.  It's just defined as the unit matrix.  i.e. a square matrix with zeros everywhere except on the diagonal where there are 1s.

I was shown something multiply by first row, and add second row or something like that. Is that what I saw on Kahn Acadamy, reduced row echlon or something to solve for I? 

Yes there are several more complicated methods for finding the inverse of a matrix.  These are important, especially when large matrices are involved.  But for 2x2 matrices the first expression in the image above can be used (as long as ad-bc is not zero - if it is zero then A has no inverse).

Above you added a negative in the front of b and c in the last post ( I mistakingly said c and d) although in the originaly reply the negative do not exist. I see a difference but not why.

What is the maximum matrix dimension for the "method descibed" in this post. 

Is there a formula name please? I don't know that. 

The negative signs do exist in my original post, but they can be difficult to see!  They are in line with the horizontal "divide" lines.  Look closely.

What is the maximum matrix dimension for the "method descibed" in this post. 

Technically, there isn't a maximum, but the method becomes increasingly cumbersome extremely rapidly as the size increases.  I think I've seen it used on a 3x3 in the past, but I wouldn't use it on anything other than a 2x2.

Is there a formula name please? 

I don't think it has a special name (though I could be wrong - I'm not too concerned about what things are called. "A rose by any other name is still a rose"!). 

Thanks. This is enough for this week. lol.