Let P be a square matrix such that P² = I - P. For α, β, γ, δ ∈ N, if P^α + P^β = γI - 29P and P^α -

Q: Wait, why can't we just solve for $P$ directly? Do we need to know the size of the matrix?

That is a fantastic question! In JEE Advanced problems, we often don't need the specific elements of the matrix. We are working with the 'algebraic identity' of the matrix. Since $P^2 = I - P$ holds true, any power of $P$ can be reduced to a linear form $aI + bP$. We don't need to know if $P$ is $2\times2$ or $100\times100$; the algebraic structure is universal.

Q: How did you know to stop at $P^8$? Could $\alpha$ or $\beta$ be much larger?

In an exam setting, we look for patterns. After calculating $P^4, P^6,$ and $P^8$, we noticed the coefficients of $P$ were $-3, -8,$ and $-21$. We needed a sum of $-29$ and a difference of $-13$. By testing the combinations of the powers we already derived, we found a perfect match. If we hadn't found it, we would have continued to $P^{10}$, but usually, these problems are designed to resolve within the first few iterations.

Q: Is there a faster way to do this without calculating every power manually?

Yes! If you look closely at the coefficients, you might notice they follow a Fibonacci-like sequence. However, for the scope of this problem, manual iteration is the safest and most reliable method to avoid calculation errors. Speed comes from practice, not just shortcuts!

Question

Let $P$ be a square matrix such that $P^2 = I - P$. For $\alpha, \beta, \gamma, \delta \in \mathbb{N}$, if $P^\alpha + P^\beta = \gamma I - 29P$ and $P^\alpha - P^\beta = \delta I - 13P$, then $\alpha + \beta + \gamma - \delta$ is equal to

Options:

(A) 18

(B) 40

(C) 22

(D) 24

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the fundamental relation $P^2 = I - P$. Think of this not as a static equation, but as a 'reduction machine'.

Every time you see a $P^2$, you can swap it for $I - P$. This is our key to unlocking higher powers. We are not looking for the matrix $P$ itself; we are looking for the structure of its powers.

### The Systematic Climb

We start with the simplest higher power, $P^4$. We know that $P^4 = (P^2)^2$. Substituting our reduction rule, we get $(I - P)^2$.

Expanding this, we get $I^2 - 2IP + P^2$. Since $I^2 = I$ and $IP = P$, this becomes $I - 2P + P^2$.

Now, we apply our machine again: replace $P^2$ with $I - P$. The result is:
$$P^4 = I - 2P + (I - P) = 2I - 3P$$

Now, let us push further to $P^6$. We can write this as $P^4 \cdot P^2$. Substituting our known values, we have $(2I - 3P)(I - P)$.

Expanding this carefully, we get $2I^2 - 2IP - 3PI + 3P^2$, which simplifies to $2I - 5P + 3P^2$. Again, we replace $P^2$ with $I - P$:
$$P^6 = 2I - 5P + 3(I - P) = 5I - 8P$$

Finally, let us reach $P^8$. This is $P^6 \cdot P^2 = (5I - 8P)(I - P)$.

Expanding this, we get $5I^2 - 5IP - 8PI + 8P^2 = 5I - 13P + 8P^2$. Substituting $P^2 = I - P$ one last time, we get:
$$P^8 = 5I - 13P + 8(I - P) = 13I - 21P$$

### The Detective Work

We are given the relations $P^\alpha + P^\beta = \gamma I - 29P$ and $P^\alpha - P^\beta = \delta I - 13P$. We need to find two powers whose coefficients of $P$ sum to $-29$ and differ by $-13$.

Looking at our results, $P^4$ has coefficient $-3$, $P^6$ has $-8$, and $P^8$ has $-21$. If we take $\alpha = 8$ and $\beta = 6$, let us check the sum:
$$P^8 + P^6 = (13I - 21P) + (5I - 8P) = 18I - 29P$$

This matches $\gamma I - 29P$ perfectly, giving us $\gamma = 18$. Now, let us check the difference:
$$P^8 - P^6 = (13I - 21P) - (5I - 8P) = 8I - 13P$$

This matches $\delta I - 13P$ perfectly, giving us $\delta = 8$.

### The Final Victory

We have successfully identified all our variables: $\alpha = 8$, $\beta = 6$, $\gamma = 18$, and $\delta = 8$. The question asks for the value of $\alpha + \beta + \gamma - \delta$.

Substituting our values, we get $8 + 6 + 18 - 8$. The $8$ and $-8$ cancel out beautifully, leaving us with $6 + 18 = 24$.

The final answer is **24**.

Animated Solution for Mathematics - Matrices and Determinants: Let $P$ be a square matrix such that $P^{2} = I - P$ . For $α, β, γ, δ \in N$ , if $P^{α} + P^{β} = γ I - 29 P$ and $P^{α} - P^{β} = δ I - 13 P$ , then $α + β + γ - δ$ is equal to

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Visualized Solution

Given Equation $P^{2} = I - P$

Calculating $P^{4}$

Calculating $P^{6}$

Calculating $P^{8}$

Matching Coefficients of $P$

Finding $α, β, γ$

Finding $δ$

Calculating $α + β + γ - δ$

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Systematic Climb

The Detective Work

The Final Victory

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Given Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2=I−P

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P4

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P6

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P8

Matching Coefficients of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α,β,γ

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } δ

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α+β+γ−δ