Invariant determinant for a specific class of matrices Let $A = \begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}$     be a such real $2\times{2}$ matrix  that the sum of elements on diagonal and anti-diagonal are equal i.e. $a+d=b+c$. 
$\\$
Question: 


*

*How to prove (if it is always true) that for this kind of matrices:
$\det(A) = \det(A+k\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix})$
for any real $k$ ?

 A: You may simply expand 
$$
\det\left(A+k\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}\right)
$$
and see what you get.
You will get something like
$$
\det\left(\begin{bmatrix}
a+k & b+k \\
c+k & d+k \\
\end{bmatrix}\right)
=
ad - bc + k (\dots).
$$
A: We have that determinant of $A = ad-bc...(1)$. 
Now the matrix $A +k\begin{bmatrix}
  1 & 1\\
  1 & 1
\end{bmatrix}$. It can be simplified to $\begin{bmatrix} a+k & b+k\\
c+k & d+k \end{bmatrix}$. It's determinant is $((a+k)(d+k)-(b+k)(c+k)) = ad-bc +k((a+d)-(b+c))...(2)$.  

If $(1)=(2)$, we will have $$a+d=b+c$$ which is true. Hope it helps.
A: You can use linearity of the determinant obtaining
\begin{eqnarray}
\det\left[\begin{array}{cc}a +k & b+k\\ c +k & d+k\end{array}\right]&=&\det\left[\begin{array}{cc}a  & b\\ c  & d\end{array}\right]+k\det\left[\begin{array}{cc}a  & b\\ 1  & 1\end{array}\right]+k\det\left[\begin{array}{cc}1  & 1\\ c  & d\end{array}\right]+k^2\det\left[\begin{array}{cc}1  & 1\\ 1  & 1\end{array}\right]\\
&=& \det\left[\begin{array}{cc}a  & b\\ c  & d\end{array}\right]+k(a-b+d-c)\\
&=& \det\left[\begin{array}{cc}a  & b\\ c  & d\end{array}\right].
\end{eqnarray}
This can be probably generalized in dimensions larger than 2.
