Question

Let $P$ be a square matrix such that $P^2=I-P$. For $\alpha$, $\beta$, $\gamma$, $\delta \in N$, if $P^\alpha+P^\beta+29P=\gamma I$ and $P^\alpha=P^\beta+\delta I -13P$, then $\frac{\alpha+\beta+\gamma-\delta}{5}$ is equal to

Answer 1

24/5

Answer 2

The given condition is $P^2 = I - P$. This can be rewritten as $P^2 + P - I = 0$. We need to express higher powers of $P$ in the form $aP + bI$.

$P^1 = P$

$P^2 = I - P$

$P^3 = P \cdot P^2 = P(I - P) = P - P^2 = P - (I - P) = 2P - I$

$P^4 = P \cdot P^3 = P(2P - I) = 2P^2 - P = 2(I - P) - P = 2I - 2P - P = 2I - 3P$

$P^5 = P \cdot P^4 = P(2I - 3P) = 2P - 3P^2 = 2P - 3(I - P) = 2P - 3I + 3P = 5P - 3I$

$P^6 = P \cdot P^5 = P(5P - 3I) = 5P^2 - 3P = 5(I - P) - 3P = 5I - 5P - 3P = 5I - 8P$

We observe a pattern related to Fibonacci numbers $F_n$, where $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, \dots$. The general form is $P^n = (-1)^{n+1} F_n P + (-1)^n F_{n-1}I$ for $n \ge 1$.

We are given two equations:

1. $P^\alpha + P^\beta + 29P = \gamma I \implies P^\alpha + P^\beta = \gamma I - 29P$
2. $P^\alpha = P^\beta + \delta I - 13P \implies P^\alpha - P^\beta = \delta I - 13P$

Add equation (1) and equation (2):

$(P^\alpha + P^\beta) + (P^\alpha - P^\beta) = (\gamma I - 29P) + (\delta I - 13P)$

$2P^\alpha = (\gamma + \delta)I - 42P$

Subtract equation (2) from equation (1):

$(P^\alpha + P^\beta) - (P^\alpha - P^\beta) = (\gamma I - 29P) - (\delta I - 13P)$

$2P^\beta = (\gamma - \delta)I - 16P$

Now, substitute the general form for $P^n$ into these two derived equations.

For $2P^\alpha = (\gamma + \delta)I - 42P$:

$2 [(-1)^{\alpha+1} F_\alpha P + (-1)^\alpha F_{\alpha-1}I] = -42P + (\gamma + \delta)I$

Comparing the coefficients of $P$ and $I$:

$2 (-1)^{\alpha+1} F_\alpha = -42 \implies (-1)^{\alpha+1} F_\alpha = -21$

Since $F_\alpha > 0$ for $\alpha \in N$, we must have $(-1)^{\alpha+1} = -1$, which implies $\alpha+1$ is odd, so $\alpha$ is even.

Thus, $F_\alpha = 21$. From the Fibonacci sequence, $F_8 = 21$, so $\alpha = 8$. This is consistent with $\alpha$ being even.

Comparing coefficients of $I$:

$2 (-1)^\alpha F_{\alpha-1} = \gamma + \delta$

Since $\alpha = 8$ (even), $(-1)^\alpha = 1$.

$2 F_{8-1} = \gamma + \delta \implies 2 F_7 = \gamma + \delta$

$2 \times 13 = \gamma + \delta \implies \gamma + \delta = 26$. (Eq. A)

For $2P^\beta = (\gamma - \delta)I - 16P$:

$2 [(-1)^{\beta+1} F_\beta P + (-1)^\beta F_{\beta-1}I] = -16P + (\gamma - \delta)I$

Comparing the coefficients of $P$ and $I$:

$2 (-1)^{\beta+1} F_\beta = -16 \implies (-1)^{\beta+1} F_\beta = -8$

Since $F_\beta > 0$ for $\beta \in N$, we must have $(-1)^{\beta+1} = -1$, which implies $\beta+1$ is odd, so $\beta$ is even.

Thus, $F_\beta = 8$. From the Fibonacci sequence, $F_6 = 8$, so $\beta = 6$. This is consistent with $\beta$ being even.

Comparing coefficients of $I$:

$2 (-1)^\beta F_{\beta-1} = \gamma - \delta$

Since $\beta = 6$ (even), $(-1)^\beta = 1$.

$2 F_{6-1} = \gamma - \delta \implies 2 F_5 = \gamma - \delta$

$2 \times 5 = \gamma - \delta \implies \gamma - \delta = 10$. (Eq. B)

Now we have a system of linear equations for $\gamma$ and $\delta$:

(A) $\gamma + \delta = 26$

(B) $\gamma - \delta = 10$

Adding (A) and (B):

$2\gamma = 36 \implies \gamma = 18$.

Subtracting (B) from (A):

$2\delta = 16 \implies \delta = 8$.

We have found the values:

$\alpha = 8$

$\beta = 6$

$\gamma = 18$

$\delta = 8$

All values are natural numbers, as required.

Finally, we need to calculate $\frac{\alpha+\beta+\gamma-\delta}{5}$:

$\frac{8+6+18-8}{5} = \frac{14+18-8}{5} = \frac{32-8}{5} = \frac{24}{5}$.

The final answer is $\boxed{\frac{24}{5}}$.

Explanation of the solution:

1. Derive the general form for $P^n$ using the given $P^2=I-P$ relation. It is found to be $P^n = (-1)^{n+1} F_n P + (-1)^n F_{n-1}I$, where $F_n$ are Fibonacci numbers ($F_0=0, F_1=1, \dots$).
2. Manipulate the given equations $P^\alpha+P^\beta+29P=\gamma I$ and $P^\alpha=P^\beta+\delta I -13P$ to get expressions for $2P^\alpha$ and $2P^\beta$ in the form $aP+bI$.
   • $2P^\alpha = -42P + (\gamma + \delta)I$
   • $2P^\beta = -16P + (\gamma - \delta)I$
3. Equate the coefficients of $P$ and $I$ from the general form of $P^n$ with the derived expressions for $2P^\alpha$ and $2P^\beta$.
4. Solve for $\alpha$ and $\beta$ using the coefficient of $P$. This involves identifying the correct Fibonacci number and checking the sign to determine parity.
   • $F_\alpha = 21 \implies \alpha = 8$ (even)
   • $F_\beta = 8 \implies \beta = 6$ (even)
5. Solve for $\gamma$ and $\delta$ using the coefficient of $I$.
   • $\gamma + \delta = 2F_7 = 2 \times 13 = 26$
   • $\gamma - \delta = 2F_5 = 2 \times 5 = 10$
   • Solving these gives $\gamma = 18, \delta = 8$.
6. Substitute the values of $\alpha, \beta, \gamma, \delta$ into the required expression $\frac{\alpha+\beta+\gamma-\delta}{5}$.
   • $\frac{8+6+18-8}{5} = \frac{24}{5}$.