We are given a matrix $A = \begin{pmatrix} 5 & a \\ -1 & -5 \end{pmatrix}$. We need to find the value of the parameter $a$ such that the trace of $e^A$ is equal to $2\cosh(1)$. That is, we need to solve for $a$ in the equation $tr(e^A) = 2\cosh(1)$.

1. Problem Description

We are given a matrix $A = \begin{pmatrix} 5 & a \\ -1 & -5 \end{pmatrix}$. We need to find the value of the parameter $a$ such that the trace of $e^A$ is equal to $2\cosh(1)$. That is, we need to solve for $a$ in the equation $tr(e^A) = 2\cosh(1)$.

2. Solution Steps

First, we need to find the eigenvalues of matrix $A$. To do this, we need to solve for $\lambda$ in the characteristic equation $|A - \lambda I| = 0$, where $I$ is the identity matrix.

So we have:

$\begin{vmatrix} 5 - \lambda & a \\ -1 & -5 - \lambda \end{vmatrix} = (5-\lambda)(-5-\lambda) - (a)(-1) = 0$

This simplifies to:

$-25 - 5\lambda + 5\lambda + \lambda^2 + a = 0$

$\lambda^2 + a - 25 = 0$

So, $\lambda^2 = 25 - a$

Therefore, the eigenvalues are $\lambda_1 = \sqrt{25-a}$ and $\lambda_2 = -\sqrt{25-a}$.

Now we know that the eigenvalues of $e^A$ are $e^{\lambda_1}$ and $e^{\lambda_2}$. Thus,

$tr(e^A) = e^{\lambda_1} + e^{\lambda_2} = e^{\sqrt{25-a}} + e^{-\sqrt{25-a}}$

We are given that $tr(e^A) = 2\cosh(1)$.

Also, we know that $\cosh(x) = \frac{e^x + e^{-x}}{2}$, so $2\cosh(x) = e^x + e^{-x}$.

Therefore, $2\cosh(1) = e^1 + e^{-1}$.

So, we have $e^{\sqrt{25-a}} + e^{-\sqrt{25-a}} = e^1 + e^{-1}$.

This implies that $\sqrt{25-a} = 1$.

Squaring both sides, we get $25-a = 1$.

Solving for $a$, we get $a = 25 - 1 = 24$.

3. Final Answer

The value of the parameter $a$ is 24.