Key Concepts and Formulas

1. Determinant of a Product: For any square matrices $X, Y, Z$ of the same order, the determinant of their product is the product of their determinants: $\det(XYZ) = \det(X) \det(Y) \det(Z)$
2. Determinant of an Inverse: If $P$ is an invertible matrix, then the determinant of its inverse is the reciprocal of its determinant: $\det(P^{-1}) = \frac{1}{\det(P)}$
3. Identity Matrix Property: For any invertible square matrix $P$ of order $n$, its inverse $P^{-1}$ satisfies $P^{-1}P = I_n$, where $I_n$ is the $n \times n$ identity matrix. This allows us to strategically replace $I$ with $P^{-1}P$ for factorization.
4. Determinant of Similar Matrices: A direct consequence of the above properties is that for any square matrix $B$ and an invertible matrix $P$ of the same order, $\det(P^{-1}BP) = \det(B)$. This is a powerful simplification that avoids calculating $P$ or $P^{-1}$.

Step-by-Step Solution

We need to find the sum of the prime factors of $\left|P^{-1} A P-2 I\right|$.

Step 1: Simplify the Determinant Expression $\left|P^{-1} A P-2 I\right|$

Our first and most crucial step is to simplify the expression inside the determinant using the properties outlined above. This avoids the tedious and error-prone direct computation of $P^{-1}$, $P^{-1}AP$, and subsequent matrix operations. We start with the given expression: $\left|P^{-1} A P-2 I\right|$ We use the identity matrix property $I = P^{-1}P$. By substituting $P^{-1}P$ for $I$, we prepare the expression for factoring: $\left|P^{-1} A P-2 (P^{-1}P)\right|$ Next, we factor out $P^{-1}$ from the left and $P$ from the right using the distributive property of matrix multiplication. This is similar to factoring $xa - xb = x(a-b)$ and $ay - by = (a-b)y$ in scalar algebra. $\left|P^{-1}(A-2I)P\right|$ Now, we apply the determinant of a product rule, $\det(XYZ) = \det(X)\det(Y)\det(Z)$: $\left|P^{-1}\right| \left|A-2I\right| \left|P\right|$ Using the property $\left|P^{-1}\right| = \frac{1}{|P|}$ (which is valid because $P$ is an invertible matrix, so its determinant $|P|$ is non-zero): $\frac{1}{|P|} \left|A-2I\right| |P|$ Since $|P|$ is a scalar value, we can cancel it from the numerator and denominator: $\left|A-2I\right|$ This crucial simplification means that the specific matrix $P$ is irrelevant to the final determinant value. We only need to calculate the determinant of $A-2I$.

Step 2: Calculate the Matrix $A-2I$}

Now we substitute the given matrix $A$ and the identity matrix $I$ (of order 3, since $A$ is $3 \times 3$). Given: $A=\left[\begin{array}{ccc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right]$ The $3 \times 3$ identity matrix $I$ is: $I=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ First, we calculate $2I$ by multiplying each element of $I$ by the scalar 2: $2I = 2 \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$ Next, we subtract $2I$ from $A$ by subtracting their corresponding elements: $A-2I = \left[\begin{array}{ccc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right] - \left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$ $A-2I = \left[\begin{array}{ccc}2-2 & 1-0 & 2-0 \\ 6-0 & 2-2 & 11-0 \\ 3-0 & 3-0 & 2-2\end{array}\right]$ $A-2I = \left[\begin{array}{ccc}0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0\end{array}\right]$

Step 3: Calculate the Determinant of $A-2I$

We now compute the determinant of the matrix obtained in Step 2: $\left|A-2I\right| = \left|\begin{array}{ccc}0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0\end{array}\right|$ We use the cofactor expansion method along the first row. The pattern of signs for cofactors along the first row is $+ - +$: $\left|A-2I\right| = 0 \cdot \left|\begin{array}{cc}0 & 11 \\ 3 & 0\end{array}\right| - 1 \cdot \left|\begin{array}{cc}6 & 11 \\ 3 & 0\end{array}\right| + 2 \cdot \left|\begin{array}{cc}6 & 0 \\ 3 & 3\end{array}\right|$ Now, we calculate each $2 \times 2$ determinant:

• $\left|\begin{array}{cc}0 & 11 \\ 3 & 0\end{array}\right| = (0 \times 0) - (11 \times 3) = 0 - 33 = -33$
• $\left|\begin{array}{cc}6 & 11 \\ 3 & 0\end{array}\right| = (6 \times 0) - (11 \times 3) = 0 - 33 = -33$
• $\left|\begin{array}{cc}6 & 0 \\ 3 & 3\end{array}\right| = (6 \times 3) - (0 \times 3) = 18 - 0 = 18$ Substitute these values back into the expansion: $\left|A-2I\right| = 0 \cdot (-33) - 1 \cdot (-33) + 2 \cdot (18)$ $\left|A-2I\right| = 0 + 33 + 36$ $\left|A-2I\right| = 69$

Step 4: Find the Sum of the Prime Factors of 69

The value of the determinant is 69. We need to find its unique prime factors and then sum them up. To find the prime factors of 69:

• 69 is an odd number, so it is not divisible by 2.
• The sum of its digits is $6+9=15$, which is divisible by 3. So, 69 is divisible by 3. $69 \div 3 = 23$.
• Now we have the number 23. We check if 23 is a prime number. 23 is not divisible by any prime numbers less than or equal to its square root ($\sqrt{23} \approx 4.8$, so we check 2 and 3). 23 is indeed a prime number. Thus, the unique prime factors of 69 are 3 and 23. The sum of these prime factors is $3 + 23 = 26$.

Common Mistakes & Tips

• Recognize the Structure: The expression $\left|P^{-1}BP\right|$ (where $B$ is $A-2I$ in this problem) is a strong indicator that the determinant simplifies to $\left|B\right|$. Always look for such structures in matrix problems, as they are common shortcuts in competitive exams like JEE.
• Careful with Signs in Determinants: When calculating $3 \times 3$ determinants using cofactor expansion, pay close attention to the alternating signs ($+ - +$ for the first row, for example). A single sign error can lead to an incorrect determinant value.
• Prime Factorization Definition: Ensure you find all unique prime factors and sum them correctly. For instance, if a number is $12 = 2^2 \cdot 3$, its prime factors are 2 and 3, and their sum is $2+3=5$, not $2+2+3=7$.

Summary This problem tested the understanding of determinant properties, particularly those related to similar matrices. The key insight was simplifying the expression $\left|P^{-1} A P-2 I\right|$ to $\left|A-2I\right|$ using properties of determinants and matrix inverses. After calculating $A-2I$, its determinant was found to be 69. Finally, the unique prime factors of 69 were identified as 3 and 23, and their sum was calculated as 26.

The final answer is $\boxed{26}$, which corresponds to option (D).