Let A = [[2, 1, 2], [6, 2, 11], [3, 3, 2]] and P = [[1, 2, 0], [5, 0, 2], [7, 1, 5]]. The sum of the

Q: Wait, why can't we just calculate $P^{-1}$? It's just a $3 \times 3$ matrix, right?

You absolutely could, but in the high-stakes environment of JEE Advanced, time is your most precious resource. Calculating a $3 \times 3$ inverse is not only time-consuming but also a breeding ground for arithmetic errors. The examiner designed this problem to reward those who spot the algebraic structure rather than those who grind through the calculation.

Q: How did you know to write $2I$ as $P^{-1}(2I)P$?

That is a fantastic question! It comes from the goal of creating a common factor. We had $P^{-1}AP$ and we needed to subtract $2I$. By expressing $2I$ as $P^{-1}(2I)P$, we created a common $P^{-1}$ on the left and $P$ on the right, which allows us to use the distributive property of matrices: $P^{-1}AP - P^{-1}(2I)P = P^{-1}(A - 2I)P$.

Q: Is $|P^{-1}AP| = |A|$ always true?

Yes, it is! Because the determinant of a product is the product of the determinants, we have $|P^{-1}AP| = |P^{-1}| \cdot |A| \cdot |P|$. Since $|P^{-1}| = \frac{1}{|P|}$, they cancel out perfectly, leaving just $|A|$. This is a powerful property that simplifies many matrix problems.

Question

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

Options:

(A) 26

(B) 27

(C) 66

(D) 23

Sigma Percentile · Accepted Answer

### The Art of the Elegant Shortcut

Welcome, future engineers! Today, we are going to dismantle a problem that, at first glance, looks like a tedious exercise in matrix arithmetic. You see a $3 	imes 3$ matrix $A$ and another $3 	imes 3$ matrix $P$, and your brain immediately screams, "Calculate the inverse of $P$!"

But stop right there. In the world of JEE Advanced, the most elegant path is rarely the one that requires the most pencil lead. Let's look at the expression $|P^{-1}AP - 2I|$ and see if we can find the hidden beauty within.

### Phase 1: The Power of Algebraic Manipulation

The trap here is the brute-force approach. If you start calculating $P^{-1}$, you are already falling behind.

Instead, let's use the properties of the identity matrix $I$. We know that $I = P^{-1}P$. This means we can rewrite the term $2I$ as $2(P^{-1}P)$, which is equivalent to $P^{-1}(2I)P$.

Now, look at our expression again: $P^{-1}AP - P^{-1}(2I)P$. We have a common factor of $P^{-1}$ on the left and $P$ on the right. By the distributive property of matrices, we can factor this into $P^{-1}(A - 2I)P$.

This is the moment where the problem shifts from a calculation nightmare to a simple determinant property question.

### Phase 2: The Determinant Magic

Now we need the determinant of this new product: $|P^{-1}(A - 2I)P|$. The golden rule of determinants states that the determinant of a product of matrices is the product of their individual determinants: $|XYZ| = |X| \cdot |Y| \cdot |Z|$.

Applying this, we get:

$$|P^{-1}| \cdot |A - 2I| \cdot |P|$$

Since $|P^{-1}| = \frac{1}{|P|}$, the determinant of $P$ and its inverse cancel out completely! We are left with nothing but $|A - 2I|$. We have successfully bypassed the need to ever touch matrix $P$.

### Phase 3: Constructing and Solving

Now, let's construct $A - 2I$. We take matrix $A = \begin{bmatrix} 2 & 1 & 2 \ 6 & 2 & 11 \ 3 & 3 & 2 \end{bmatrix}$ and subtract $2$ from each diagonal element.

This gives us the matrix:

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

Now, we calculate the determinant by expanding along the first row:

$$|A - 2I| = 0(0 - 33) - 1(0 - 33) + 2(18 - 0)$$

This simplifies to $0 + 33 + 36 = 69$.

### Phase 4: The Final Stretch

The question asks for the sum of the prime factors of this determinant. Our determinant is $69$.

We know that $69 = 3 	imes 23$. Both $3$ and $23$ are prime numbers.

Therefore, the sum of the prime factors is $3 + 23 = \mathbf{26}$. By choosing the path of logic over the path of brute force, we solved a complex-looking matrix problem in just a few lines. Keep this mindset, and you will conquer any problem the exam throws at you!

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 2631232112$ and $P = 157201025$ . The sum of the prime factors of $∣ P^{- 1} A P - 2 I ∣$ is equal to

Select Answer:

Visualized Solution

Analyze the Expression $∣ P^{- 1} A P - 2 I ∣$

Rewrite the Identity Matrix

Factorize the Matrix Expression

Apply Determinant Properties

Construct Matrix $A - 2 I$

Calculate the Determinant

Find Prime Factors and Sum

The Sigma Insight: Properties of Determinants

The Art of the Elegant Shortcut

Phase 1

Phase 2

Phase 3

Phase 4

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