Absolute Value of Sum of Diagonal Elements of Matrix Equation

To find the absolute value of the sum of the diagonal elements (trace) of $2(P + Q)$, we start by finding the inverse of matrix $A$.

The given matrices are: $A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}$

Step 1: Find the inverse of matrix $A$

The determinant of $A$ is: $\det(A) = (2)(-2) - (-2)(4) = -4 + 8 = 4$

Since $\det(A) \neq 0$, the matrix $A$ is invertible. The inverse $A^{-1}$ is given by: $A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) = \frac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix}$

Step 2: Express $P$ and $Q$ in terms of $A^{-1}$ and $B$

From the given matrix equations:

1. $PA = B \implies P = B A^{-1}$
2. $AQ = B \implies Q = A^{-1} B$

Step 3: Calculate the trace of $P$ and $Q$

By the cyclic property of the trace function, $\text{Tr}(XY) = \text{Tr}(YX)$. Therefore: $\text{Tr}(P) = \text{Tr}(B A^{-1}) = \text{Tr}(A^{-1} B) = \text{Tr}(Q)$

Let's compute matrix $P$: $P = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \cdot \frac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix}$ $P = \frac{1}{4} \begin{bmatrix} 3(-2) + 9(-4) & 3(2) + 9(2) \\ 1(-2) + 3(-4) & 1(2) + 3(2) \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -42 & 24 \\ -14 & 8 \end{bmatrix}$

The trace of $P$ is: $\text{Tr}(P) = \frac{-42 + 8}{4} = \frac{-34}{4} = -\frac{17}{2}$

Thus, we also have: $\text{Tr}(Q) = -\frac{17}{2}$

Step 4: Find the sum of diagonal elements of $2(P + Q)$ and its absolute value

Using the linearity property of the trace operator: $\text{Tr}(2(P + Q)) = 2 \left[ \text{Tr}(P) + \text{Tr}(Q) \right]$ $\text{Tr}(2(P + Q)) = 2 \left( -\frac{17}{2} - \frac{17}{2} \right) = 2 (-17) = -34$

Taking the absolute value: $|\text{Tr}(2(P + Q))| = |-34| = 34$