The correct answer is $\boxed{\text{A}}$.
The eigenvalues of a matrix are the roots of its characteristic polynomial. The characteristic polynomial of a 2×2 matrix can be written as follows:
$$p(x) = |A – xI|$$
where $I$ is the identity matrix. In this case, we have:
$$p(x) = |A – xI| = \left| \begin{array}{cc} 1 & 4 \\ b & a \end{array} – x \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right| = \left| \begin{array}{cc} 1 – x & 4 \\ b & a – x \end{array} \right|$$
To find the eigenvalues, we need to solve the equation $p(x) = 0$. This gives us the following quadratic equation:
$$(1 – x)(a – x) – 4b = 0$$
$$x^2 – (a + 1)x + ab – 4b = 0$$
We can factor this equation as follows:
$$(x – a)(x – (1 – b)) = 0$$
Therefore, the eigenvalues of $A$ are $a$ and $1 – b$.
Since the eigenvalues are $-1$ and $7$, we have $a = -1$ and $1 – b = 7$. Solving for $b$, we get $b = 6$. Therefore, the values of $a$ and $b$ are $\boxed{a = 6, b = 4}$.
Here is a brief explanation of each option:
- Option A: $a = 6, b = 4$. This is the correct answer.
- Option B: $a = 4, b = 6$. This is not the correct answer, because $a$ cannot be greater than $1$.
- Option C: $a = 3, b = 5$. This is not the correct answer, because $a$ cannot be less than $-1$.
- Option D: $a = 5, b = 3$. This is not the correct answer, because $b$ cannot be less than $-1$.