Sum of Prime Factors of |P⁻¹AP−2I| for Given 3×3 Matrices A, P

JEE Main 2024 — 29 January, Evening Shift. Previous Year Question.

Problem

Let $A=\begin{bmatrix}2&1&2\\6&2&11\\3&3&2\end{bmatrix}$ 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:

Key insight. Computing $P^{-1}$, then multiplying out $P^{-1}AP$, then subtracting $2I$, then finding that determinant, is an enormous amount of matrix arithmetic. But $2I = P^{-1}(2I)P$ trivially (since $P^{-1}P=I$), so $P^{-1}AP – 2I = P^{-1}(A-2I)P$ — and determinants of matrices related this way (“similar matrices”) are equal, so $P$ can be skipped entirely.

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Approach

Rewrite $P^{-1}AP-2I$ as $P^{-1}(A-2I)P$, then use the determinant property $|P^{-1}(A-2I)P| = |P^{-1}|\,|A-2I|\,|P| = |A-2I|$ (since $|P^{-1}||P|=1$). This means $P$ never needs to be inverted or multiplied at all — only $|A-2I|$ needs computing. Then factorise that number into primes and add them.

Solution

Step 1 — Rewrite the expression using P

$$P^{-1}AP – 2I = P^{-1}AP – P^{-1}(2I)P = P^{-1}(A-2I)P$$

Step 2 — Apply the determinant identity for similar matrices

$$\big|P^{-1}(A-2I)P\big| = |P^{-1}|\cdot|A-2I|\cdot|P| = \frac{1}{|P|}\cdot|A-2I|\cdot|P| = |A-2I|$$

So the entire problem reduces to computing $|A-2I|$ — $P$ is completely irrelevant to the final answer.

Step 3 — Compute A − 2I

$$A-2I = \begin{bmatrix}2-2&1&2\\6&2-2&11\\3&3&2-2\end{bmatrix} = \begin{bmatrix}0&1&2\\6&0&11\\3&3&0\end{bmatrix}$$

Step 4 — Compute the determinant

Expanding along the first row:

$$|A-2I| = 0\cdot(0\times0-11\times3) – 1\cdot(6\times0-11\times3) + 2\cdot(6\times3-0\times3)$$

$$= 0 – 1\times(-33) + 2\times18 = 33+36 = 69$$

Step 5 — Factorise and sum the prime factors

$$69 = 3 \times 23$$

$$\text{Sum of prime factors} = 3+23 = 26$$

Answer

Common mistakes

• Trying to compute $P^{-1}$ and multiply out the full matrix product. This works but is extremely long and error-prone for a $3\times3$ matrix — recognising the similarity-transformation shortcut avoids touching $P$ at all.
• Confusing “sum of prime factors” with “sum of all factors” or “number of prime factors.” $69=3\times23$ has exactly two distinct prime factors, and the question asks for their sum ($3+23=26$), not a count or the full divisor sum.

Practise next

• Let $A=\begin{bmatrix}3&1&1\\2&4&9\\1&1&3\end{bmatrix}$ and $Q$ be any invertible $3\times3$ matrix. Find the sum of the prime factors of $|Q^{-1}AQ-3I|$, using the same similarity-transformation shortcut.