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 $tr(e^A) = 2\cosh(1)$.

Linear AlgebraMatrix ExponentialsEigenvaluesTraceCharacteristic Equation
2025/5/14

1. Problem Description

We are given a matrix A=(5a15)A = \begin{pmatrix} 5 & a \\ -1 & -5 \end{pmatrix}. We need to find the value of the parameter aa such that the trace of eAe^A is equal to 2cosh(1)2\cosh(1). That is, we need to solve for aa in the equation tr(eA)=2cosh(1)tr(e^A) = 2\cosh(1).

2. Solution Steps

First, we need to find the eigenvalues of matrix AA. To do this, we need to solve for λ\lambda in the characteristic equation AλI=0|A - \lambda I| = 0, where II is the identity matrix.
So we have:
5λa15λ=(5λ)(5λ)(a)(1)=0\begin{vmatrix} 5 - \lambda & a \\ -1 & -5 - \lambda \end{vmatrix} = (5-\lambda)(-5-\lambda) - (a)(-1) = 0
This simplifies to:
255λ+5λ+λ2+a=0-25 - 5\lambda + 5\lambda + \lambda^2 + a = 0
λ2+a25=0\lambda^2 + a - 25 = 0
So, λ2=25a\lambda^2 = 25 - a
Therefore, the eigenvalues are λ1=25a\lambda_1 = \sqrt{25-a} and λ2=25a\lambda_2 = -\sqrt{25-a}.
Now we know that the eigenvalues of eAe^A are eλ1e^{\lambda_1} and eλ2e^{\lambda_2}. Thus,
tr(eA)=eλ1+eλ2=e25a+e25atr(e^A) = e^{\lambda_1} + e^{\lambda_2} = e^{\sqrt{25-a}} + e^{-\sqrt{25-a}}
We are given that tr(eA)=2cosh(1)tr(e^A) = 2\cosh(1).
Also, we know that cosh(x)=ex+ex2\cosh(x) = \frac{e^x + e^{-x}}{2}, so 2cosh(x)=ex+ex2\cosh(x) = e^x + e^{-x}.
Therefore, 2cosh(1)=e1+e12\cosh(1) = e^1 + e^{-1}.
So, we have e25a+e25a=e1+e1e^{\sqrt{25-a}} + e^{-\sqrt{25-a}} = e^1 + e^{-1}.
This implies that 25a=1\sqrt{25-a} = 1.
Squaring both sides, we get 25a=125-a = 1.
Solving for aa, we get a=251=24a = 25 - 1 = 24.

3. Final Answer

The value of the parameter aa is 24.

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