Question

If matrices \( A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 3 \\ 3 & 9 \end{bmatrix} \) are such that \( PA = B \) and \( AQ = B \), then \( \mathrm{tr}\bigl(2(P+Q)\bigr) \) is -

Hint:

Whenever equations involve matrices like \( PA=B \) or \( AQ=B \), first isolate the unknown matrix using the inverse of the given matrix. After that, use the trace as the sum of diagonal elements.

Answer 1

Correct Answer: 10

Step 1: Express \( P \) and \( Q \) in terms of \( A \) and \( B \).
Given \[ PA = B \qquad \text{and} \qquad AQ = B. \] From \( PA = B \), multiplying by \( A^{-1} \) on the right, we get \[ P = BA^{-1}. \] From \( AQ = B \), multiplying by \( A^{-1} \) on the left, we get \[ Q = A^{-1}B. \]
Step 2: Find \( A^{-1} \).
For \[ A=\begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}, \] its determinant is \[ |A| = (B)(-2)-(-2)(D) = -4+8 = 4. \] Therefore, \[ A^{-1}=\frac{1}{4}\begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix}. \]
Step 3: Calculate \( P=BA^{-1} \).
Now, \[ P= \begin{bmatrix} 1 & 3 \\ 3 & 9 \end{bmatrix} \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix}. \] Multiplying, \[ P= \begin{bmatrix} 1\left(-\frac{1}{2}\right)+3(-1) & 1\left(\frac{1}{2}\right)+3\left(\frac{1}{2}\right) \\ 3\left(-\frac{1}{2}\right)+9(-1) & 3\left(\frac{1}{2}\right)+9\left(\frac{1}{2}\right) \end{bmatrix} = \begin{bmatrix} -\frac{7}{2} & 2\\ -\frac{21}{2} & 6 \end{bmatrix}. \]
Step 4: Calculate \( Q=A^{-1}B \).
Now, \[ Q= \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 3 & 9 \end{bmatrix}. \] Multiplying, \[ Q= \begin{bmatrix} -\frac{1}{2}(A)+\frac{1}{2}(C) & -\frac{1}{2}(C)+\frac{1}{2}(9) \\ -1(A)+\frac{1}{2}(C) & -1(C)+\frac{1}{2}(9) \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ \frac{1}{2} & \frac{3}{2} \end{bmatrix}. \]
Step 5: Find \( \mathrm{tr}\bigl(2(P+Q)\bigr) \).
First, \[ P+Q= \begin{bmatrix} -\frac{7}{2} & 2 \\ -\frac{21}{2} & 6 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ \frac{1}{2} & \frac{3}{2} \end{bmatrix} = \begin{bmatrix} -\frac{5}{2} & 5
-10 & \frac{15}{2} \end{bmatrix}. \] So, \[ \mathrm{tr}(P+Q)= -\frac{5}{2}+\frac{15}{2} = \frac{10}{2}=5. \] Hence, \[ \mathrm{tr}\bigl(2(P+Q)\bigr)=2\,\mathrm{tr}(P+Q)=2\times 5=10. \]