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

Answer: (B)

9

Answer 2

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.