Consider the matrices $$A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}$$ and $$B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}$$. If matrices $$P$$ and $$Q$$ are such that $$PA = B$$ and $$AQ = B$$, then the absolute value of the sum of the diagonal elements of $$2(P + Q)$$ is _________.

We have to find two matrices $$P$$ and $$Q$$ satisfying
$$PA = B \qquad\text{and}\qquad AQ = B$$
for $$A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix},\; B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}$$.

Since $$A$$ is a square matrix, first compute its inverse.

Determinant of $$A$$:
$$\det A = 2(-2) - (-2)(4) = -4 + 8 = 4$$

Adjugate of $$A$$:
$$\operatorname{adj}(A)=\begin{bmatrix}-2 & 2 \\ -4 & 2\end{bmatrix}$$

Hence
$$A^{-1}= \frac{1}{\det A}\,\operatorname{adj}(A) =\frac{1}{4}\begin{bmatrix}-2 & 2 \\ -4 & 2\end{bmatrix} =\begin{bmatrix}-\tfrac12 & \tfrac12 \\ -1 & \tfrac12\end{bmatrix}$$

Matrix $$P$$ is obtained from $$PA = B$$:
$$P = BA^{-1} =\begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \begin{bmatrix}-\tfrac12 & \tfrac12 \\ -1 & \tfrac12\end{bmatrix}$$

Multiplying row by column:

First row:
$$\bigl(3,\;9\bigr)\cdot\begin{bmatrix} -\tfrac12 \\ -1 \end{bmatrix} =-\,\tfrac{3}{2}-9=-\tfrac{21}{2},\qquad \bigl(3,\;9\bigr)\cdot\begin{bmatrix} \tfrac12 \\ \tfrac12 \end{bmatrix} =\tfrac{3}{2}+ \tfrac{9}{2}=6$$

Second row:
$$\bigl(1,\;3\bigr)\cdot\begin{bmatrix} -\tfrac12 \\ -1 \end{bmatrix} =-\tfrac12-3=-\tfrac72,\qquad \bigl(1,\;3\bigr)\cdot\begin{bmatrix} \tfrac12 \\ \tfrac12 \end{bmatrix} =\tfrac12+\tfrac32=2$$

Thus
$$P=\begin{bmatrix}-\tfrac{21}{2} & 6 \\[2pt] -\tfrac72 & 2\end{bmatrix}$$

Matrix $$Q$$ is obtained from $$AQ = B$$:
$$Q = A^{-1}B =\begin{bmatrix}-\tfrac12 & \tfrac12 \\ -1 & \tfrac12\end{bmatrix} \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}$$

Multiplying:

First row:
$$\bigl(-\tfrac12,\;\tfrac12\bigr)\cdot\begin{bmatrix}3\\1\end{bmatrix} =-\tfrac32+\tfrac12=-1,\qquad \bigl(-\tfrac12,\;\tfrac12\bigr)\cdot\begin{bmatrix}9\\3\end{bmatrix} =-\tfrac92+\tfrac32=-3$$

Second row:
$$\bigl(-1,\;\tfrac12\bigr)\cdot\begin{bmatrix}3\\1\end{bmatrix} =-3+\tfrac12=-\tfrac52,\qquad \bigl(-1,\;\tfrac12\bigr)\cdot\begin{bmatrix}9\\3\end{bmatrix} =-9+\tfrac32=-\tfrac{15}{2}$$

So
$$Q=\begin{bmatrix}-1 & -3 \\[2pt] -\tfrac52 & -\tfrac{15}{2}\end{bmatrix}$$

Add $$P$$ and $$Q$$:
$$P+Q=\begin{bmatrix} -\tfrac{23}{2} & 3 \\ -6 & -\tfrac{11}{2} \end{bmatrix}$$

Compute $$2(P+Q)$$:
$$2(P+Q)=\begin{bmatrix} -23 & 6 \\ -12 & -11 \end{bmatrix}$$

The sum of its diagonal elements (its trace) is
$$\operatorname{tr}\bigl(2(P+Q)\bigr)= -23 + (-11) = -34$$

Taking the absolute value:
$$|\,\operatorname{tr}(2(P+Q))\,| = 34$$

Therefore, the required value is 34.