Invariant determinant for a specific class of matrices

Question

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. $a+d=b+c$.

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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 $k$ ?

Answer 1

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 $((a+k)(d+k)-(b+k)(c+k)) = ad-bc +k((a+d)-(b+c))...(2)$.

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If $(1)=(2)$, we will have $$a+d=b+c$$ which is true. Hope it helps.

Answer 2

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 $$ \det\left(\begin{bmatrix} a+k & b+k \\ c+k & d+k \\ \end{bmatrix}\right) = ad - bc + k (\dots). $$

Answer 3

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.