To determine the value of \( (a + b)^{ab} \), we are given the matrices \( P = \begin{bmatrix} 0 & -a \\ 2a & b \end{bmatrix} \), \( Q = \begin{bmatrix} b & a \\ -b & 0 \end{bmatrix} \), and their product \( PQ = \begin{bmatrix} 2 & 0 \\ 3 & 8 \end{bmatrix} \). First, we compute the product \( PQ \).
The computation of \( PQ \) is as follows:
\( PQ = \begin{bmatrix} 0 & -a \\ 2a & b \end{bmatrix} \begin{bmatrix} b & a \\ -b & 0 \end{bmatrix} = \begin{bmatrix} (0)(b) + (-a)(-b) & (0)(a) + (-a)(0) \\ (2a)(b) + (b)(-b) & (2a)(a) + (b)(0) \end{bmatrix} = \begin{bmatrix} ab & 0 \\ 2ab - b^2 & 2a^2 \end{bmatrix} \)
We equate this result with the given \( PQ \) matrix:
\(\begin{bmatrix} ab & 0 \\ 2ab - b^2 & 2a^2 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 3 & 8 \end{bmatrix}\)
By comparing the corresponding elements, we derive the following equations:
\( ab = 2 \)
\( 2ab - b^2 = 3 \)
\( 2a^2 = 8 \)
From \( 2a^2 = 8 \), we get \( a^2 = 4 \). Since \( a>0 \), \( a = 2 \).
Substituting \( a = 2 \) into \( ab = 2 \) yields:
\( 2b = 2 \), which means \( b = 1 \).
We verify these values using the second equation: \( 2(2)(1) - (1)^2 = 4 - 1 = 3 \). This equation holds true.
Finally, we calculate \( (a + b)^{ab} \):
\( a + b = 2 + 1 = 3 \), and \( ab = 2 \).
Therefore, \( (a + b)^{ab} = (3)^2 = 9 \).
The value of \( (a + b)^{ab} \) is 9.