Question

For \( a, b > 0 \), if \( P = \begin{bmatrix} 0 & -a \\ 2a & b \end{bmatrix} \) and \( Q = \begin{bmatrix} b & a \\ -b & 0 \end{bmatrix} \) are two matrices such that \( PQ = \begin{bmatrix} 2 & 0 \\ 3 & 8 \end{bmatrix} \), then the value of \( (a + b)^{ab} \) is:

• A. 8
• B. 9
• C. \( \frac{1}{9} \)
• D. \( -\frac{1}{27} \)

Answer 1

The Correct Option is B

Compute the product \( PQ \):

\( PQ = \begin{bmatrix} 0 & -a \\ 2a & b \end{bmatrix} \cdot \begin{bmatrix} b & a \\ -b & 0 \end{bmatrix} \).

Perform the matrix multiplication:

\( PQ = \begin{bmatrix} (0)(b) + (-a)(-b) & (0)(a) + (-a)(0) \\ (2a)(b) + (b)(-b) & (2a)(a) + (b)(0) \end{bmatrix} \).

Simplify:

\( PQ = \begin{bmatrix} ab & 0 \\ 2ab - b^2 & 2a^2 \end{bmatrix} \).

Equating \( PQ \) to \(\begin{bmatrix} 2 & 0 \\ 3 & 8 \end{bmatrix}\), we get:

\( ab = 2 \), \( 2ab - b^2 = 3 \), \( 2a^2 = 8 \).

From \( 2a^2 = 8 \), solve for \( a \):

\( a^2 = 4 \implies a = 2 \quad \text{(as } a > 0 \text{)} \).

Substitute \( a = 2 \) into \( ab = 2 \):

\( 2b = 2 \implies b = 1 \).

Now calculate \( (a + b)^{ab} \): \( a + b = 2 + 1 = 3 \), \( ab = (2)(1) = 2 \). 
\( (a + b)^{ab} = 3^2 = 9 \).
Thus, the value of \( (a + b)^{ab} \) is 9.