# Proving determinant equality \begin{equation}\det{((A+B)^2)} = [\det(A+B)]^2\end{equation}

This is what we have to prove or disprove:

\begin{equation}\det{((A+B)^2)} = [\det(A+B)]^2\end{equation}

However, I really have no idea where to start - I tried plugging in two random sets of matrices A, B, and both seemed to check out for this, so I assume this is true.

In terms of proving it in a general way, how should I go about approaching this? I'm thinking about proving the equivalence of the numeric operations involved (plus and multiply), but if that works, it would only do so with very small matrices, and I'd find it hard to generalise it to matrices of an n x n scale.

Also have a similar question to prove:

\begin{equation}\det((A+B)^2) = \det(A^2+2AB+B^2)\end{equation}

Would appreciate any tips.

• What do you know about $\det(XY)$ and $\det(X+Y)$ compared to $\det(X)$ and $\det(Y)$? – Arthur Dec 11 '14 at 8:39
• Hmm, we've never explored that in class. I've learnt recently though that det(XY) = det(X)det(Y) if my memory serves me right. Is det(X+Y) = det(X) + det(Y) as well? And I wouldn't know how to go about proving det(XY) = det(X)det(Y) either – user1746848 Dec 11 '14 at 8:42

Let $A$ and $B$ be two square matrices of equal size then$$\det(AB)=\det(A)\cdot\det(B)$$ so $$\det(A+B)^2=\det((A+B)\cdot(A+B))=\det(A+B)\cdot\det(A+B)=(\det(A+B))^2$$