How do I prove that the determinant map is a ring homomorphism

Question

I need to find the values of $n$ in the natural numbers, such that for $n×n$ matrices, the determinant map is a ring homomorphism.

I am stuck at proving this property:

$$\det A+\det B=\det (A+B).$$

Answer 1

Obviously this is true if $n=1$, since the determinant is then the same as the (unique) matrix entry.

Try different 2x2 matrices for A and B, see what happens. Diagonal matrices should be good enough.

Answer 2

I am stuck at proving this property: $\det A+ \det B= \det (A+B)$

You are stuck proving it because it is not true, for $n \times n$ matrices with $n \ge 2$. (Which is unfortunate, as it would make so many determinant calculations much easier.)

Instead, try to look for a counterexample in $2 \times 2$ matrices where this is not true. Just try out some simple matrices. Then for $n \times n$ matrices, you can just take your example of $2 \times 2$ matrices in the upper-left corner of the matrix, and fill the rest of the matrix with $0$s.

For $n = 1$, it is a homomorphism. Can you see why?