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

To solve the problem, we need to find matrices \( P \) and \( Q \) such that \( PA = B \) and \( AQ = B \). Then, we calculate \(\mathrm{tr}\bigl(2(P+Q)\bigr)\).
Let \( A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 3 \\ 3 & 9 \end{bmatrix} \).
The problem gives \( PA = B \) and \( AQ = B \). Hence, \( P = BA^{-1} \) and \( Q = A^{-1}B \).
First, find \( A^{-1} \):
\( \det(A) = 2(-2) - (-2)(4) = -4 + 8 = 4 \).
The inverse is \( A^{-1} = \frac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix} \).
Now, calculate \( P = BA^{-1} \):
\( BA^{-1} = \begin{bmatrix} 1 & 3 \\ 3 & 9 \end{bmatrix} \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} - 3 & \frac{1}{2} + \frac{3}{2} \\ -\frac{3}{2} - 9 & \frac{3}{2} + \frac{9}{2} \end{bmatrix} = \begin{bmatrix} -\frac{7}{2} & 2 \\ -\frac{21}{2} & 6 \end{bmatrix} \).
Similarly, calculate \( Q = A^{-1}B \):
\( A^{-1}B = \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ -1 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 3 & 9 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2}(1) + \frac{1}{2}(3) & -\frac{1}{2}(3) + \frac{1}{2}(9) \\ -1(1) + \frac{1}{2}(3) & -1(3) + \frac{1}{2}(9) \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ \frac{1}{2} & \frac{3}{2} \end{bmatrix} \).
Next, compute \( P+Q \):
\( P+Q = \begin{bmatrix} -\frac{7}{2} + 1 & 2 + 3 \\ -\frac{21}{2} + \frac{1}{2} & 6 + \frac{3}{2} \end{bmatrix} = \begin{bmatrix} -\frac{5}{2} & 5 \\ -\frac{20}{2} & \frac{15}{2} \end{bmatrix} = \begin{bmatrix} -\frac{5}{2} & 5 \\ -10 & \frac{15}{2} \end{bmatrix} \).
Calculate the trace: \(\mathrm{tr}(P+Q) = -\frac{5}{2} + \frac{15}{2} = 5\).
Finally, compute \(\mathrm{tr}\bigl(2(P+Q)\bigr)\):
\(\mathrm{tr}\bigl(2(P+Q)\bigr) = 2 \times 5 = 10\).
Verify the result is within the range [10,10]. It fits the range perfectly.
Thus, the answer is indeed \( 10 \).