Consider the following matrix. \[{\text{A}} = \left[ {\begin{array}{*{20}{c}} 2&3 \\ {\text{x}}&{\text{y}} \end{array}} \right]\] If the eigen values of A are 4 and 8, then A. x = 4, y = 10 B. x = 5, y = 8 C. x = -3, y = 9 D. x = -4, y = 10

The correct answer is $\boxed{\text{C}}$.

The eigenvalues of a matrix are the roots of its characteristic polynomial. The characteristic polynomial of a matrix $A$ is given by $$p(x) = \det(xI – A)$$ where $I$ is the identity matrix.

In this case, we have $$p(x) = \det \left( \begin{array}{cc} x – 2 & -3 \\ x & x – y \end{array} \right) = x^2 – (2+y)x + 2y – 6$$

We are given that the eigenvalues of $A$ are 4 and 8. Therefore, we must have $$4^2 – (2+y)4 + 2y – 6 = 0$$ and $$8^2 – (2+y)8 + 2y – 6 = 0$$

Solving these equations, we find that $y = -3$ and $x = 9$.

Therefore, the correct answer is $\boxed{\text{C}}$.

Here is a brief explanation of each option:

• Option A: $x = 4$, $y = 10$. This is not possible, because the eigenvalues of a matrix must be distinct.
• Option B: $x = 5$, $y = 8$. This is also not possible, because the eigenvalues of a matrix must be distinct.
• Option C: $x = -3$, $y = 9$. This is possible, because the eigenvalues of a matrix can be any two numbers that are not equal to each other.
• Option D: $x = -4$, $y = 10$. This is not possible, because the eigenvalues of a matrix must be distinct.