Question

\[\text{Let } A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix} \text{ and } P = \begin{bmatrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{bmatrix}.\]The sum of the prime factors of \( |P^{-1}AP - 2I| \) is equal to.

• A. 26
• B. 27
• C. 66
• D. 23

Answer 1

The Correct Option is A

Solution: We need to find \(|P^{-1}AP - 2I|\).  

Step 1. Calculating \(|P^{-1}AP - 2I|:\) 

 \(|P^{-1}AP - 2I| = |P^{-1}AP - 2P^{-1}P|\)
  \(= |P^{-1}(A - 2I)P|\) 
  \(= |P^{-1}| \cdot |A - 2I| \cdot |P|\) 
  \(= |A - 2I|\)
 

Step 2. Calculating \(|A - 2I|:\) 
 \(A - 2I = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}     = \begin{bmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{bmatrix}\)
\(|A - 2I| = 69\)

Step 3. The prime factors of 69 are 3 and 23, so the sum of the prime factors is:  
\(3 + 23 = 26\)
The Correct Answer is: 26