The products above are too big to fit in 32-bit integers. You can get the correct result by switching to 64-bit integers:
>> m64 = int64(A)
>> detm64 = m64(1)*m64(4)-m64(2)*m64(3)
detm64 = 1
Look at the L-U decomposition of A:
>> [L U P] =lu(A)
This shows that U(2,2) is numerically zero relative to U(1,1) and U(1,2). Thus U is numerically singular, and so is A.
Now look at the lower bound estimate of the 1-norm condition of A:
lbcond1A = condest(A) = 4.707074327130931e+016
This shows that A is very ill-conditioned. There is no hope of getting accurate results with double precision because log10(condest(A)) is about 17, the number of digits of precision lost when calculating the solution of Ax = b.
This example shows that there is no connection between det(A) and the condition of a matrix.