Given: \(A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\) and \((3I + 4A)(3I - 4A) = x^2 I\)

First, expand the left side of the equation:

\((3I + 4A)(3I - 4A) = (3I)^2 - (4A)^2 = 9I^2 - 16A^2\)

Since \(I\) is the identity matrix, \(I^2 = I\). So, \(9I^2 = 9I\).

Now, calculate \(A^2\):

\(A^2 = A \cdot A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I\)

Substitute \(A^2 = -I\) into the equation:

\(9I - 16A^2 = 9I - 16(-I) = 9I + 16I = 25I\)

So, we have \(25I = x^2 I\).

Therefore, \(x^2 = 25\), which means \(x = \pm 5\).