Consider the matrices}
\[
A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}.
\]
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 _______.}
Correct Answer: 34
Solution and Explanation
Given:
\[ A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \]
Step 1: Find \(P\) such that \(PA = B\)
\[ P = BA^{-1} \] First, find \(A^{-1}\).
\[ \det(A) = (2)(-2) - (4)(-2) = -4 + 8 = 4 \] \[ A^{-1} = \frac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix} \] Now, \[ P = B A^{-1} \] \[ P = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix} \] Multiplying: \[ P = \begin{bmatrix} 3(-\frac{1}{2}) + 9(-1) & 3(\frac{1}{2}) + 9(\frac{1}{2}) \\ 1(-\frac{1}{2}) + 3(-1) & 1(\frac{1}{2}) + 3(\frac{1}{2}) \end{bmatrix} \] \[ P = \begin{bmatrix} -\frac{3}{2} - 9 & \frac{3}{2} + \frac{9}{2} \\ -\frac{1}{2} - 3 & \frac{1}{2} + \frac{3}{2} \end{bmatrix} = \begin{bmatrix} -\frac{21}{2} & 6 \\ -\frac{7}{2} & 2 \end{bmatrix} \]
Step 2: Find \(Q\) such that \(AQ = B\)
\[ Q = A^{-1}B \] \[ Q = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \] Multiplying: \[ Q = \begin{bmatrix} -\frac{1}{2}(3) + \frac{1}{2}(1) & -\frac{1}{2}(9) + \frac{1}{2}(3) \\ -1(3) + \frac{1}{2}(1) & -1(9) + \frac{1}{2}(3) \end{bmatrix} \] \[ Q = \begin{bmatrix} -1 & -3 \\ -\frac{5}{2} & -\frac{15}{2} \end{bmatrix} \]
Step 3: Compute \(P + Q\)
\[ P + Q = \begin{bmatrix} -\frac{21}{2} & 6 \\ -\frac{7}{2} & 2 \end{bmatrix} + \begin{bmatrix} -1 & -3 \\ -\frac{5}{2} & -\frac{15}{2} \end{bmatrix} \] \[ P + Q = \begin{bmatrix} -\frac{23}{2} & 3 \\ -6 & -\frac{11}{2} \end{bmatrix} \]
Step 4: Compute \(2(P+Q)\)
\[ 2(P+Q) = \begin{bmatrix} -23 & 6 \\ -12 & -11 \end{bmatrix} \]
Step 5: Sum of diagonal elements
\[ \text{Trace} = -23 + (-11) = -34 \]
Final Answer:
\[ \boxed{34} \]
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