To determine when the matrix $A = \begin{bmatrix} 5 & k \\ k & 5 \end{bmatrix}$ is invertible, we need to check when its determinant is non-zero.

The determinant of a 2x2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by:

$\text{det}(A) = ad - bc$

For the matrix $A = \begin{bmatrix} 5 & k \\ k & 5 \end{bmatrix}$, we can calculate the determinant as:

$\text{det}(A) = (5)(5) - (k)(k) = 25 - k^2$

To ensure that $A$ is invertible, we need:

$\text{det}(A) \neq 0$

This gives the inequality:

$25 - k^2 \neq 0$

$k^2 \neq 25$

$k \neq \pm 5$

Therefore, the matrix $A$ is invertible for all values of $k$ except $k = \pm 5$.

The correct answer is $a) \, k \neq \pm 5$.

Would you like further clarification or details on this process? Here are some related questions for deeper understanding:

1. What happens to the matrix if $k = \pm 5$?
2. Can you generalize the determinant formula for other symmetric matrices like $A$?
3. How would you interpret the determinant geometrically?
4. How would the matrix change if the diagonal elements were different?
5. How can we find the inverse of matrix $A$ if $k \neq \pm 5$?

Tip: To check if a matrix is invertible, always verify that its determinant is non-zero.