You are told that a matrix $A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \mathbb{R}^{2 \times 2}$ has eigenvalue $\lambda_1 = 1$ and $\lambda_2 = 8$. Calculate the value of $a^2+7ad-5bc+d^2$.
What I have done so far is multiplied the identity ($2\times 2$) matrix by each eigenvalue and formed two equations by subtracting them from $a,b,c,d$ matrix and then performing $ad-bc = 0$ for each one, so that there are two equations that can be solved simultaneously. I hope that made some sense and I am not good at describing maths. Am I correct? Thank you!
Hint:
$$a^2+7ad-5bc+d^2=(a+d)^2+5(ad-bc)$$
Think of how determinant and trace relates to eigenvalues.
Remark:
To verify your answer, just work with $\begin{bmatrix} 1 & 0 \\ 0 & 8\end{bmatrix}$. Also, $a,b,c,d$ is not uniquely determined. For example, $\begin{bmatrix} 8 & 0 \\ 0 & 1\end{bmatrix}$ satisfy the condition too. You have to prove that the value of $a^2+7ad-5bc+d^2$ is unique even though there is some flexibility in choosing the value of $a,b,c,d$.