Consider the matrices A = bmatrix 2 & -2 4 & -2 bmatrix and B =...

Given $A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}$, we find its determinant:

$|A| = (2)(-2) - (-2)(4) = 4$

Since $|A| \neq 0$, $A$ is invertible. The inverse of $A$ is:

$A^{-1} = \dfrac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$

We are given $PA = B \Rightarrow P = BA^{-1}$ and $AQ = B \Rightarrow Q = A^{-1}B$.

Using the cyclic property of trace $\text{Tr}(XY) = \text{Tr}(YX)$, we have:

$\text{Tr}(P) = \text{Tr}(BA^{-1}) = \text{Tr}(A^{-1}B) = \text{Tr}(Q)$

Let us calculate $\text{Tr}(P)$ by finding the diagonal elements of $P = BA^{-1}$:

$P = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -1/2 & 1/2 \\ -1 & 1/2 \end{bmatrix}$

The diagonal elements of $P$ are:

$P_{11} = 3\left(-\dfrac{1}{2}\right) + 9(-1) = -\dfrac{21}{2}$

$P_{22} = 1\left(\dfrac{1}{2}\right) + 3\left(\dfrac{1}{2}\right) = 2$

$\text{Tr}(P) = P_{11} + P_{22} = -\dfrac{21}{2} + 2 = -\dfrac{17}{2}$

Since $\text{Tr}(P) = \text{Tr}(Q) = -\dfrac{17}{2}$, the sum of the diagonal elements of $2(P + Q)$ is:

$\text{Tr}(2(P + Q)) = 2(\text{Tr}(P) + \text{Tr}(Q)) = 2\left(-\dfrac{17}{2} - \dfrac{17}{2}\right) = 2(-17) = -34$

The absolute value is $|-34| = 34$.

Answer: $34$