| Line 5: | Line 5: | ||
Doesn't ad-bc also have to divide each of a, b, c, and d? Otherwise the inverse of the matrix would have fractional elements and therefore would not be in <math>M_2(Z)</math>.--[[User:Mkorb|Mkorb]] 23:17, 11 March 2009 (UTC) | Doesn't ad-bc also have to divide each of a, b, c, and d? Otherwise the inverse of the matrix would have fractional elements and therefore would not be in <math>M_2(Z)</math>.--[[User:Mkorb|Mkorb]] 23:17, 11 March 2009 (UTC) | ||
| + | |||
| + | Good point, I didn't even think of that. Thanks!<br> | ||
| + | --[[User:Jniederh|Jniederh]] 00:06, 12 March 2009 (UTC) | ||
Latest revision as of 19:06, 11 March 2009
Generally speaking, 2x2 matrices have the form {(a,b), (c,d)} where (a,b) is the first row and (c,d) is the second. The inverse of any 2x2 matrix, M, is just 1/det(M)*{(d,-b), (-c,a)} and the det(M) is just ad-bc. This means that every 2x2 matrix, M, has an inverse unless ad-bc=0.
--Jniederh 02:16, 11 March 2009 (UTC)
Doesn't ad-bc also have to divide each of a, b, c, and d? Otherwise the inverse of the matrix would have fractional elements and therefore would not be in $ M_2(Z) $.--Mkorb 23:17, 11 March 2009 (UTC)
Good point, I didn't even think of that. Thanks!
--Jniederh 00:06, 12 March 2009 (UTC)
