🧠:<think>Okay, so I'm trying to solve this problem about a professional bug fixer analyzing a software system represented by a directed graph. The graph has modules as vertices and dependencies as edges, each with a probability that a bug will propagate from one module to another. First, part 1 asks me to formulate the problem of finding the expected number of affected modules when a bug is identified in module M. Hmm. I remember that expectation can be calculated by summing the probabilities of each module being affected. So, if module M has a bug, the expected number of affected modules would be 1 (for M itself) plus the sum of the probabilities that each dependent module gets the bug, considering the propagation probabilities.Let me think more formally. Let’s denote E[M] as the expected number of modules affected starting from M. Then, E[M] = 1 + sum over all v in V, v ≠ M of P(v is affected | M has a bug). But how do I calculate P(v is affected)? Since the graph is directed, the propagation can happen through multiple paths. So, it's not just direct edges but all possible paths from M to v. This seems like a problem where each edge contributes to the probability, but since the propagation is probabilistic, the overall probability isn't just the product of probabilities along a path but something more complex.Wait, maybe I can model this using linearity of expectation. The expected number of affected modules is the sum over all modules v of the probability that v is affected. So, E[M] = sum_{v ∈ V} P(v is affected). For each module v, P(v is affected) is the probability that there exists at least one path from M to v where all edges along the path propagate the bug. But calculating this directly seems complicated because of the dependencies between paths.Alternatively, maybe we can model this as a system of equations. For each module v, the probability that it's affected is the probability that it's directly affected by M or by any of its predecessors. So, for each v, P(v) = 1 if v = M, otherwise P(v) = sum_{u ∈ predecessors(v)} p_{uv} * P(u). Wait, no, that might not be correct because it's not just the direct predecessors but all possible paths.Hmm, actually, I think this is similar to the problem of finding the expected number of reachable nodes in a probabilistic graph. There's a concept called the "expected number of reachable nodes" which can be calculated using a system of equations where each node's probability is the sum of the probabilities of its incoming edges multiplied by the probability of the source node.So, maybe for each node v, P(v) = 1 if v = M, otherwise P(v) = sum_{u ∈ predecessors(v)} p_{uv} * P(u). But wait, that seems recursive. It might not account for multiple paths correctly because if a node can be reached through multiple paths, the probabilities could add up incorrectly.Alternatively, perhaps it's better to model this using the inclusion-exclusion principle, but that can get complicated for a general graph.Wait, maybe the correct approach is to model this as a Markov chain where each node's probability is the sum of the probabilities of being reached through each edge. So, starting from M, the probability that a bug propagates to v is the sum over all paths from M to v of the product of the probabilities along the edges of the path. But this is essentially the expected number of times the bug reaches v, but since we just care about whether it's affected or not, it's the probability that at least one path propagates the bug.But calculating this exactly is difficult because of overlapping paths. However, for small probabilities, the probability that at least one path propagates the bug can be approximated by the sum of the probabilities of each path, assuming that the events are independent, which they are not. So, this might not be accurate.Alternatively, maybe we can use the fact that the expected number of affected modules is the sum over all modules of the probability that the module is reachable from M in the graph where edges are included with probability p_{uv}. This is similar to the expected size of the reachable set in a random graph.I think in this case, the expected number of affected modules can be formulated as the sum over all modules v of the probability that there exists a path from M to v where each edge in the path is present with probability p_{uv}. But how do we compute this? It seems like it's related to the reliability polynomial, which gives the probability that a graph remains connected, but here it's about reachability.Wait, maybe we can use dynamic programming. Starting from M, we can compute the probability that each node is reachable by considering all possible paths. Let me try to formalize this.Let’s denote f(v) as the probability that module v is affected. Then, f(M) = 1, since it's definitely affected. For other modules v, f(v) is the probability that at least one of its predecessors u has a bug and the edge (u, v) propagates the bug. But since the propagation is probabilistic, f(v) = 1 - product_{u ∈ predecessors(v)} (1 - p_{uv} * f(u)). Wait, that seems right because the probability that v is not affected is the product of the probabilities that none of its predecessors propagate the bug to it. So, f(v) = 1 - product_{u ∈ predecessors(v)} (1 - p_{uv} * f(u)).But this is a system of equations that needs to be solved for all f(v). However, solving this system can be complex because it's non-linear.Alternatively, if the probabilities are small, we can approximate f(v) using the inclusion of only the first-order terms, ignoring higher-order terms. So, f(v) ≈ sum_{u ∈ predecessors(v)} p_{uv} * f(u). This is a linear system which can be solved using methods like Gaussian elimination.But the problem doesn't specify any approximations, so I think the exact expectation would require solving the non-linear system. Therefore, the expected number of affected modules is the sum over all v of f(v), where f(v) satisfies f(M) = 1 and for v ≠ M, f(v) = 1 - product_{u ∈ predecessors(v)} (1 - p_{uv} * f(u)).Alternatively, another way to express this is using the concept of reachability in probabilistic graphs. The expected number of reachable nodes from M is the sum over all v of the probability that v is reachable from M, considering the edge probabilities.So, putting it all together, the expected number of affected modules is the sum over all modules v of the probability that there exists a path from M to v such that all edges on the path propagate the bug. This can be expressed as E = sum_{v ∈ V} P(v is reachable from M).But to compute P(v is reachable from M), we can use the system of equations I mentioned earlier, which might be the most precise way.Now, moving on to part 2. The bug fixer wants to implement fixes on some edges to reduce the propagation probabilities, with each fix having a cost C(p_{uv}, q_{uv}) = k*(p_{uv} - q_{uv})^2. The goal is to minimize the total expected number of affected modules subject to a budget B.So, we need to formulate an optimization problem where we choose which edges to fix (i.e., choose q_{uv} for each edge (u, v)) such that the total cost sum_{(u,v)} C(p_{uv}, q_{uv}) ≤ B, and the total expected number of affected modules is minimized.But how do we model the expected number of affected modules after fixing some edges? It would be similar to part 1, but with the updated probabilities q_{uv} for the fixed edges.So, the optimization problem would involve variables q_{uv} for each edge (u, v), which are constrained to satisfy 0 ≤ q_{uv} ≤ p_{uv}. The objective is to minimize the expected number of affected modules, which depends on the q_{uv} values, subject to the total cost sum_{(u,v)} k*(p_{uv} - q_{uv})^2 ≤ B.But the expected number of affected modules is a function of the q_{uv} values, which in turn affects the probabilities f(v) as defined earlier. So, the optimization problem would need to minimize sum_{v} f(v) subject to the cost constraint and the constraints on q_{uv}.However, since f(v) is defined recursively, this becomes a non-linear optimization problem with potentially many variables and constraints. It might be challenging to solve exactly for large graphs, but for the purposes of formulation, we can describe it as such.So, in summary, the optimization problem is:Minimize E = sum_{v ∈ V} f(v)Subject to:1. For each edge (u, v), 0 ≤ q_{uv} ≤ p_{uv}2. sum_{(u,v) ∈ E} k*(p_{uv} - q_{uv})^2 ≤ B3. For each module v, f(v) = 1 - product_{u ∈ predecessors(v)} (1 - q_{uv} * f(u)) if v ≠ M, and f(M) = 1.This is a non-linear optimization problem because the f(v) are non-linear functions of the q_{uv}.Alternatively, if we use the linear approximation for small probabilities, where f(v) ≈ sum_{u ∈ predecessors(v)} q_{uv} * f(u), then the problem becomes linear, which might be easier to handle. But since the problem doesn't specify approximations, I think we should stick with the exact formulation.So, to recap:1. The expected number of affected modules is the sum over all modules v of the probability that v is reachable from M, considering the propagation probabilities. This can be expressed using a system of equations where each module's probability is a function of its predecessors.2. The optimization problem involves choosing new propagation probabilities q_{uv} for some edges to minimize the expected number of affected modules, subject to a budget constraint on the total cost of implementing the fixes.I think that's the gist of it. Now, let me try to write this more formally.</think>

🧠:<think>Okay, so I have two problems to solve here related to a retired tennis player's training regimen. The first one is about calculating the total distance covered during a physical training session, and the second is about finding the probability of a certain number of successful serves using a Poisson distribution. Let me tackle them one by one.Starting with the first problem: Physical Training Analysis. The player runs back and forth along the baseline, covering 10 meters each time. Their speed is modeled by the function ( v(t) = 2t - 0.1t^2 ) meters per second. We need to find the total distance covered from ( t = 0 ) to ( t = 20 ) seconds.Hmm, okay. So, I remember that distance is the integral of speed over time. Since speed is given as a function of time, I can compute the total distance by integrating ( v(t) ) from 0 to 20 seconds. That makes sense because integration gives the area under the curve, which in this case is distance.So, let's write that down. The total distance ( D ) is:[D = int_{0}^{20} v(t) , dt = int_{0}^{20} (2t - 0.1t^2) , dt]Alright, now I need to compute this integral. Let me recall how to integrate term by term.The integral of ( 2t ) with respect to ( t ) is ( t^2 ), and the integral of ( -0.1t^2 ) is ( -0.033333t^3 ) because ( int t^2 dt = frac{t^3}{3} ), so multiplying by -0.1 gives ( -0.033333t^3 ). Let me write that out:[int (2t - 0.1t^2) dt = t^2 - frac{0.1}{3} t^3 + C]Simplifying ( frac{0.1}{3} ), which is approximately 0.033333. So, the antiderivative is:[F(t) = t^2 - frac{1}{30} t^3]Now, I need to evaluate this from 0 to 20 seconds. So, compute ( F(20) - F(0) ).First, ( F(20) ):[F(20) = (20)^2 - frac{1}{30}(20)^3 = 400 - frac{1}{30}(8000)]Calculating ( frac{8000}{30} ). Let me do that division: 8000 divided by 30 is 266.666..., right? So, 400 minus 266.666... is 133.333... meters.Now, ( F(0) ) is just 0 because plugging in 0 gives 0 - 0 = 0.Therefore, the total distance ( D ) is 133.333... meters. Since the question mentions the player runs back and forth along the baseline, covering 10 meters each time. Wait, does that affect the total distance? Hmm, maybe not, because the speed function already accounts for the movement. So, perhaps the 10 meters each time is just context, but the integral gives the total distance regardless of direction. So, I think 133.333 meters is the answer.But let me double-check my calculations. Integrating ( 2t ) gives ( t^2 ), correct. Integrating ( -0.1t^2 ) gives ( -0.033333t^3 ), correct. Plugging in 20: 20 squared is 400, 20 cubed is 8000, times 1/30 is 266.666..., subtracting gives 133.333... So, 133.333 meters is correct.Wait, but the player is running back and forth, so does the integral account for the direction? Because speed is scalar, so integrating speed over time gives total distance, regardless of direction. So, if the player is moving back and forth, the integral still gives the total distance covered, not the displacement. So, that should be fine. So, 133.333 meters is the total distance.But let me think again: the function ( v(t) = 2t - 0.1t^2 ). Is this a speed function that could potentially become negative? Because if the speed becomes negative, that would imply the player is moving in the opposite direction, but since we're integrating speed, which is a scalar, the negative speed would contribute negatively to the integral, which would actually represent displacement, not distance. Wait, no, hold on.Wait, actually, if the function ( v(t) ) is speed, it should be non-negative. But let me check: ( v(t) = 2t - 0.1t^2 ). Let's see when this becomes zero or negative.Setting ( v(t) = 0 ):( 2t - 0.1t^2 = 0 )( t(2 - 0.1t) = 0 )So, ( t = 0 ) or ( 2 - 0.1t = 0 ) => ( t = 20 ) seconds.So, the speed is zero at t=0 and t=20, and positive in between. So, from t=0 to t=20, the speed is non-negative, meaning the player is moving in one direction the entire time. But wait, the problem says the player runs back and forth along the baseline, covering 10 meters each time. So, that suggests that the player is moving back and forth, but the speed function is given as a function of time, which might not account for direction.Wait, maybe I need to clarify: if the player is running back and forth, the speed function might be modeling the magnitude of velocity, so the total distance would still be the integral of speed. But in this case, since the speed function is given as ( v(t) = 2t - 0.1t^2 ), which is a quadratic function opening downward, peaking at t = 10 seconds (since the vertex is at t = -b/(2a) = -2/(2*(-0.1)) = 10 seconds). So, the speed increases until t=10, then decreases.But since the speed is non-negative throughout the interval [0,20], integrating it gives the total distance covered, regardless of direction. So, even though the player is moving back and forth, the integral of speed over time is the total distance. So, 133.333 meters is the total distance.Wait, but the problem says the player runs back and forth along the baseline, covering a distance of 10 meters each time. So, does that mean each run is 10 meters? So, maybe the player goes 10 meters forward, then 10 meters back, so each round trip is 20 meters? But the speed function is given as 2t - 0.1t^2, which is a function that starts at 0, increases to a maximum at t=10, then decreases back to 0 at t=20. So, perhaps the player is moving in one direction for the first 10 seconds, then turns around and moves back for the next 10 seconds? But the speed function is given as 2t - 0.1t^2, which is positive throughout, so maybe it's just the speed regardless of direction.Wait, maybe I'm overcomplicating. The problem says the player runs back and forth along the baseline, covering a distance of 10 meters each time. So, each time they go back and forth, it's 10 meters? Or each run is 10 meters? Hmm, the wording is a bit unclear. Maybe it's just context, and the main thing is the speed function given. So, perhaps the 10 meters each time is just to set the scene, but the actual distance is computed via the integral.So, I think my initial approach is correct: integrate the speed function from 0 to 20 seconds to get the total distance, which is 133.333 meters. So, approximately 133.33 meters.So, that's the first problem. Now, moving on to the second problem: Statistical Performance Analysis.The player keeps track of successful serves during training sessions. The probability of a successful serve is modeled by a Poisson distribution with an average rate of 4 successful serves per session. If the player has 10 sessions, what is the probability that they will have exactly 40 successful serves in total over these sessions?Alright, so Poisson distribution is typically used for counting the number of events happening in a fixed interval of time or space. The Poisson distribution is given by:[P(k; lambda) = frac{lambda^k e^{-lambda}}{k!}]where ( lambda ) is the average rate (the expected number of occurrences), and ( k ) is the number of occurrences.But in this case, we have 10 sessions, each with an average of 4 successful serves. So, the total number of successful serves over 10 sessions would have a Poisson distribution with ( lambda = 4 times 10 = 40 ).Wait, is that correct? Because if each session is independent and has a Poisson distribution with ( lambda = 4 ), then the sum of 10 independent Poisson variables is also Poisson with ( lambda = 40 ). So, yes, the total number of successful serves over 10 sessions is Poisson with ( lambda = 40 ).Therefore, the probability of exactly 40 successful serves is:[P(40; 40) = frac{40^{40} e^{-40}}{40!}]So, that's the formula we need to compute.But calculating this directly might be challenging because factorials of large numbers can be computationally intensive, and 40 is a pretty large number. However, we can use approximations or properties of the Poisson distribution to estimate this probability.Alternatively, since ( lambda ) is large (40), the Poisson distribution can be approximated by a normal distribution with mean ( mu = lambda = 40 ) and variance ( sigma^2 = lambda = 40 ), so standard deviation ( sigma = sqrt{40} approx 6.3246 ).But since we're looking for the probability of exactly 40, which is a discrete value, the normal approximation might not be the best approach because the normal distribution is continuous. However, we can use the continuity correction. So, the probability ( P(X = 40) ) can be approximated by ( P(39.5 < X < 40.5) ) in the normal distribution.But before I go into that, let me recall that for large ( lambda ), the Poisson distribution is approximately normal, but the exact probability can be quite small because the distribution is spread out. However, the mode of the Poisson distribution is at ( lfloor lambda rfloor ) or ( lfloor lambda rfloor - 1 ). Since ( lambda = 40 ), the mode is at 40. So, the probability at 40 is the highest single probability, but it's still a small number.Alternatively, maybe using Stirling's approximation for the factorial to compute the exact probability.Stirling's approximation is:[n! approx sqrt{2pi n} left( frac{n}{e} right)^n]So, applying this to 40!:[40! approx sqrt{2pi times 40} left( frac{40}{e} right)^{40}]Similarly, ( 40^{40} ) is just ( 40^{40} ), and ( e^{-40} ) is a constant.So, plugging into the Poisson formula:[P(40; 40) = frac{40^{40} e^{-40}}{40!} approx frac{40^{40} e^{-40}}{sqrt{2pi times 40} left( frac{40}{e} right)^{40}} = frac{40^{40} e^{-40}}{sqrt{80pi} times frac{40^{40}}{e^{40}}}} = frac{e^{-40} times e^{40}}{sqrt{80pi}} = frac{1}{sqrt{80pi}}]Calculating ( sqrt{80pi} ):First, ( 80pi approx 80 times 3.1416 approx 251.3274 )Then, ( sqrt{251.3274} approx 15.85 )So, ( frac{1}{15.85} approx 0.063 )Therefore, the approximate probability is about 6.3%.But wait, let me check this because I might have made a mistake in the approximation.Wait, using Stirling's formula:[P(k; lambda) = frac{lambda^k e^{-lambda}}{k!} approx frac{lambda^k e^{-lambda}}{sqrt{2pi k} left( frac{k}{e} right)^k} = frac{e^{-lambda}}{sqrt{2pi k}} left( frac{lambda}{k} right)^k]But when ( lambda = k = 40 ), this simplifies to:[frac{e^{-40}}{sqrt{80pi}} times 1 = frac{1}{sqrt{80pi}} approx frac{1}{15.85} approx 0.063]So, that seems correct.Alternatively, perhaps using the normal approximation with continuity correction.The exact probability ( P(X = 40) ) can be approximated by ( P(39.5 < X < 40.5) ) where ( X ) is normal with ( mu = 40 ) and ( sigma = sqrt{40} approx 6.3246 ).So, we can compute the Z-scores for 39.5 and 40.5.Z1 = (39.5 - 40)/6.3246 ≈ (-0.5)/6.3246 ≈ -0.079Z2 = (40.5 - 40)/6.3246 ≈ 0.5/6.3246 ≈ 0.079Now, looking up these Z-scores in the standard normal distribution table.The probability between Z = -0.079 and Z = 0.079 is approximately the difference between the cumulative probabilities at 0.079 and -0.079.Looking up Z = 0.079: approximately 0.5319Looking up Z = -0.079: approximately 0.4681So, the probability between them is 0.5319 - 0.4681 = 0.0638, or about 6.38%.That's very close to the approximation using Stirling's formula, which gave us about 6.3%. So, that seems consistent.Therefore, the probability is approximately 6.3%.But let me check if there's a more precise way to calculate this without approximation. Maybe using the exact Poisson formula with a calculator or computational tool.But since I don't have a calculator here, I can use the fact that for Poisson distribution with ( lambda = 40 ), the probability mass function at 40 is:[P(40; 40) = frac{40^{40} e^{-40}}{40!}]This is a very small number, but using the approximations, we saw it's about 6.3%. Wait, actually, 6.3% is not that small. Wait, but in reality, for Poisson distribution with large ( lambda ), the probability at the mean is roughly ( frac{1}{sqrt{2pi lambda}} ), which in this case is ( frac{1}{sqrt{80pi}} approx 0.063 ), which is 6.3%. So, that seems correct.Alternatively, if I use the exact formula, I can write:[P(40; 40) = frac{40^{40} e^{-40}}{40!}]But calculating this exactly would require computing 40^40, which is a huge number, and 40!, which is also a huge number. However, using logarithms, we can compute the natural log of the probability and then exponentiate.Let me try that.Compute ( ln(P) = 40 ln(40) - 40 - ln(40!) )Using Stirling's approximation for ( ln(40!) ):[ln(40!) approx 40 ln(40) - 40 + frac{1}{2} ln(2pi times 40)]So,[ln(P) = 40 ln(40) - 40 - left(40 ln(40) - 40 + frac{1}{2} ln(80pi)right) = - frac{1}{2} ln(80pi)]Therefore,[P = e^{- frac{1}{2} ln(80pi)} = frac{1}{sqrt{80pi}} approx 0.063]So, that confirms the earlier result.Therefore, the probability is approximately 6.3%.But let me think again: is this the exact value? Because using Stirling's approximation, we get an approximate value, but the exact value might be slightly different.However, for the purposes of this problem, I think using the approximation is acceptable, especially since calculating the exact value would require more precise computation.So, to summarize:1. The total distance covered by the player is approximately 133.33 meters.2. The probability of exactly 40 successful serves in 10 sessions is approximately 6.3%.But let me write the exact forms for both answers.For the first problem, the exact distance is ( frac{400}{3} ) meters, which is approximately 133.333... meters.For the second problem, the exact probability is ( frac{40^{40} e^{-40}}{40!} ), which is approximately 0.063 or 6.3%.So, I think that's the solution.</think>

🧠:<think>Okay, so I have this problem about nutrient diffusion in a printed organ, and I need to find the steady-state concentration and then the steady-state growth rate. Let me try to break this down step by step.Starting with part 1: The scientist uses the partial differential equation (PDE) for nutrient diffusion:[ frac{partial C}{partial t} = D nabla^2 C - kC ]And we need to find the steady-state concentration ( C(x, y, z) ) in a spherical organ of radius ( R ) with the boundary condition ( C(R) = C_0 ).Hmm, steady-state means that the concentration doesn't change with time anymore. So, in the steady state, the time derivative ( frac{partial C}{partial t} ) should be zero. That simplifies the equation to:[ 0 = D nabla^2 C - kC ][ Rightarrow D nabla^2 C = kC ][ Rightarrow nabla^2 C = frac{k}{D} C ]So, this is an elliptic PDE, specifically a Helmholtz equation. Since the problem is spherically symmetric (the organ is a sphere), I can assume that the concentration ( C ) depends only on the radial distance ( r ) from the center. That means ( C = C(r) ), and the Laplacian in spherical coordinates simplifies.In spherical coordinates, the Laplacian of a radially symmetric function is:[ nabla^2 C = frac{1}{r^2} frac{d}{dr} left( r^2 frac{dC}{dr} right) ]So substituting that into our equation:[ frac{1}{r^2} frac{d}{dr} left( r^2 frac{dC}{dr} right) = frac{k}{D} C ]Let me denote ( lambda = frac{k}{D} ) for simplicity. Then the equation becomes:[ frac{1}{r^2} frac{d}{dr} left( r^2 frac{dC}{dr} right) = lambda C ]This is a second-order linear ordinary differential equation (ODE). Let's write it out:[ frac{d}{dr} left( r^2 frac{dC}{dr} right) = lambda r^2 C ]Expanding the left side:[ 2r frac{dC}{dr} + r^2 frac{d^2 C}{dr^2} = lambda r^2 C ]Divide both sides by ( r^2 ) (assuming ( r neq 0 )):[ frac{2}{r} frac{dC}{dr} + frac{d^2 C}{dr^2} = lambda C ]This is a Bessel equation. The standard form of Bessel's equation is:[ r^2 frac{d^2 C}{dr^2} + r frac{dC}{dr} + (r^2 lambda - n^2) C = 0 ]But our equation is:[ frac{d^2 C}{dr^2} + frac{2}{r} frac{dC}{dr} - lambda C = 0 ]Multiplying through by ( r^2 ):[ r^2 frac{d^2 C}{dr^2} + 2r frac{dC}{dr} - lambda r^2 C = 0 ]Comparing with the standard Bessel equation, which has a term ( r frac{dC}{dr} ) instead of ( 2r frac{dC}{dr} ), it seems like we have a modified Bessel equation. Specifically, it's similar to the modified Bessel equation of order zero.The standard modified Bessel equation is:[ r^2 frac{d^2 C}{dr^2} + r frac{dC}{dr} - (r^2 + n^2) C = 0 ]But our equation is:[ r^2 frac{d^2 C}{dr^2} + 2r frac{dC}{dr} - lambda r^2 C = 0 ]Hmm, so it's a bit different. Let me check if I can manipulate it into a known form.Let me make a substitution. Let ( u = r C ). Then:( C = frac{u}{r} )Compute the derivatives:First derivative:[ frac{dC}{dr} = frac{du}{dr} cdot frac{1}{r} - frac{u}{r^2} ]Second derivative:[ frac{d^2 C}{dr^2} = frac{d}{dr} left( frac{du}{dr} cdot frac{1}{r} - frac{u}{r^2} right) ][ = frac{d^2 u}{dr^2} cdot frac{1}{r} - frac{du}{dr} cdot frac{1}{r^2} - frac{du}{dr} cdot frac{1}{r^2} + frac{2u}{r^3} ][ = frac{1}{r} frac{d^2 u}{dr^2} - frac{2}{r^2} frac{du}{dr} + frac{2u}{r^3} ]Substitute back into the original equation:[ r^2 left( frac{1}{r} frac{d^2 u}{dr^2} - frac{2}{r^2} frac{du}{dr} + frac{2u}{r^3} right) + 2r left( frac{du}{dr} cdot frac{1}{r} - frac{u}{r^2} right) - lambda r^2 cdot frac{u}{r} = 0 ]Simplify term by term:First term:[ r^2 cdot frac{1}{r} frac{d^2 u}{dr^2} = r frac{d^2 u}{dr^2} ][ r^2 cdot left( - frac{2}{r^2} frac{du}{dr} right) = -2 frac{du}{dr} ][ r^2 cdot frac{2u}{r^3} = frac{2u}{r} ]Second term:[ 2r cdot frac{du}{dr} cdot frac{1}{r} = 2 frac{du}{dr} ][ 2r cdot left( - frac{u}{r^2} right) = - frac{2u}{r} ]Third term:[ - lambda r^2 cdot frac{u}{r} = - lambda r u ]Putting all together:[ r frac{d^2 u}{dr^2} - 2 frac{du}{dr} + frac{2u}{r} + 2 frac{du}{dr} - frac{2u}{r} - lambda r u = 0 ]Simplify:- The ( -2 frac{du}{dr} ) and ( +2 frac{du}{dr} ) cancel out.- The ( frac{2u}{r} ) and ( - frac{2u}{r} ) also cancel out.So we're left with:[ r frac{d^2 u}{dr^2} - lambda r u = 0 ][ Rightarrow frac{d^2 u}{dr^2} - lambda u = 0 ]Oh, that's a much simpler equation! It's a second-order linear ODE with constant coefficients. The characteristic equation is:[ m^2 - lambda = 0 ][ Rightarrow m = pm sqrt{lambda} ]So the general solution is:[ u(r) = A e^{sqrt{lambda} r} + B e^{-sqrt{lambda} r} ]But since ( u = r C ), we have:[ C(r) = frac{u(r)}{r} = frac{A e^{sqrt{lambda} r} + B e^{-sqrt{lambda} r}}{r} ]Now, let's recall that ( lambda = frac{k}{D} ). So,[ C(r) = frac{A e^{sqrt{frac{k}{D}} r} + B e^{-sqrt{frac{k}{D}} r}}{r} ]But we need to consider the behavior of ( C(r) ) as ( r to 0 ) and ( r to R ).First, as ( r to 0 ), the concentration should remain finite. Let's see what happens to the solution as ( r to 0 ):The term ( e^{sqrt{frac{k}{D}} r} ) approaches 1, and ( e^{-sqrt{frac{k}{D}} r} ) also approaches 1. So,[ C(r) approx frac{A + B}{r} ]As ( r to 0 ), this would go to infinity unless ( A + B = 0 ). So, to avoid a singularity at the origin, we must have:[ A + B = 0 ][ Rightarrow B = -A ]So, the solution simplifies to:[ C(r) = frac{A (e^{sqrt{frac{k}{D}} r} - e^{-sqrt{frac{k}{D}} r})}{r} ]Notice that ( e^{x} - e^{-x} = 2 sinh(x) ), so we can write:[ C(r) = frac{2A sinhleft( sqrt{frac{k}{D}} r right)}{r} ]Now, applying the boundary condition at ( r = R ): ( C(R) = C_0 ).So,[ C_0 = frac{2A sinhleft( sqrt{frac{k}{D}} R right)}{R} ][ Rightarrow A = frac{C_0 R}{2 sinhleft( sqrt{frac{k}{D}} R right)} ]Therefore, substituting back into ( C(r) ):[ C(r) = frac{2 cdot frac{C_0 R}{2 sinhleft( sqrt{frac{k}{D}} R right)} cdot sinhleft( sqrt{frac{k}{D}} r right)}{r} ][ = frac{C_0 R sinhleft( sqrt{frac{k}{D}} r right)}{r sinhleft( sqrt{frac{k}{D}} R right)} ]So, the steady-state concentration is:[ C(r) = C_0 frac{R}{r} frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)} ]Alternatively, we can write this as:[ C(r) = C_0 frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)} cdot frac{R}{r} ]Wait, but is this correct? Let me double-check.We started with the PDE, transformed it into an ODE, solved it, applied the boundary condition, and got this solution. It seems consistent.But let me think about the behavior. As ( r ) increases from 0 to ( R ), the concentration should decrease because nutrients are being consumed. Let's see:At ( r = 0 ), ( sinh(0) = 0 ), so ( C(0) ) would be zero? Wait, no, because we have ( sinh(sqrt{frac{k}{D}} r) ) over ( r ). As ( r to 0 ), ( sinh(x) approx x + x^3/6 ), so ( sinh(x)/x approx 1 + x^2/6 ). So,[ C(r) approx C_0 frac{R}{r} cdot frac{sqrt{frac{k}{D}} r}{sqrt{frac{k}{D}} R} ][ = C_0 frac{R}{r} cdot frac{r}{R} ][ = C_0 ]Wait, that can't be right. If ( C(r) ) approaches ( C_0 ) as ( r to 0 ), but we have a boundary condition at ( r = R ) as ( C(R) = C_0 ). So, actually, the concentration is highest at the boundary and decreases towards the center? That seems counterintuitive because usually, in diffusion, the concentration is higher at the source and decreases away from it.Wait, in this case, the boundary condition is given at ( r = R ) as ( C(R) = C_0 ). So, if the organ is being supplied with nutrients at the boundary, the concentration would be highest at the boundary and decrease towards the center. That makes sense.So, at ( r = 0 ), the concentration would be:[ C(0) = C_0 frac{R}{0} cdot frac{sinh(0)}{sinh(sqrt{frac{k}{D}} R)} ]But this is undefined because of division by zero. However, as ( r to 0 ), ( C(r) ) approaches ( C_0 ) as we saw earlier because the ( sinh ) terms cancel out the ( r ) in the denominator.Wait, actually, no. Let me recast the expression:[ C(r) = C_0 frac{R}{r} cdot frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)} ]As ( r to 0 ), ( sinh(x) approx x + x^3/6 ), so:[ sinhleft( sqrt{frac{k}{D}} r right) approx sqrt{frac{k}{D}} r + left( sqrt{frac{k}{D}} r right)^3 / 6 ]So,[ C(r) approx C_0 frac{R}{r} cdot frac{sqrt{frac{k}{D}} r}{sinhleft( sqrt{frac{k}{D}} R right)} ][ = C_0 frac{R sqrt{frac{k}{D}}}{sinhleft( sqrt{frac{k}{D}} R right)} ]So, as ( r to 0 ), ( C(r) ) approaches a constant value, which is ( C_0 frac{R sqrt{frac{k}{D}}}{sinhleft( sqrt{frac{k}{D}} R right)} ). That seems okay because it doesn't go to infinity.But wait, if ( k/D ) is small, ( sqrt{frac{k}{D}} R ) might be small, so ( sinh(x) approx x ), so ( C(r) approx C_0 frac{R sqrt{frac{k}{D}}}{sqrt{frac{k}{D}} R} = C_0 ). So, in that case, the concentration is approximately uniform, which makes sense if the consumption rate is low.On the other hand, if ( k/D ) is large, ( sqrt{frac{k}{D}} R ) is large, so ( sinh(x) approx frac{1}{2} e^x ), so:[ C(r) approx C_0 frac{R}{r} cdot frac{frac{1}{2} e^{sqrt{frac{k}{D}} r}}{frac{1}{2} e^{sqrt{frac{k}{D}} R}} ][ = C_0 frac{R}{r} e^{sqrt{frac{k}{D}} (r - R)} ][ = C_0 frac{R}{r} e^{-sqrt{frac{k}{D}} (R - r)} ]So, as ( k/D ) increases, the concentration decays exponentially from the boundary towards the center, which also makes sense because the consumption is high, so nutrients are quickly used up as you move inward.Okay, so the solution seems to behave correctly in the limits. Therefore, I think the steady-state concentration is:[ C(r) = C_0 frac{R}{r} frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)} ]Now, moving on to part 2: We need to find the steady-state growth rate ( G(x, y, z) ) given the steady-state nutrient concentration ( C(r) ) from part 1.The equation given is:[ frac{partial G}{partial t} = alpha C - beta G ]Again, steady-state implies that ( frac{partial G}{partial t} = 0 ), so:[ 0 = alpha C - beta G ][ Rightarrow G = frac{alpha}{beta} C ]So, the steady-state growth rate is proportional to the nutrient concentration. Therefore, substituting the ( C(r) ) we found:[ G(r) = frac{alpha}{beta} C_0 frac{R}{r} frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)} ]So, the steady-state growth rate ( G(r) ) is:[ G(r) = frac{alpha C_0 R}{beta r} frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)} ]Alternatively, we can write this as:[ G(r) = frac{alpha}{beta} C(r) ]Which is a direct proportionality.Let me just verify if this makes sense. If the nutrient concentration is highest at the boundary, then the growth rate should also be highest there, which aligns with our solution. As you move towards the center, both ( C(r) ) and ( G(r) ) decrease, which is logical because there's less nutrients available for the cells to consume, so their growth rate diminishes.So, summarizing:1. The steady-state nutrient concentration is:[ C(r) = C_0 frac{R}{r} frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)} ]2. The steady-state growth rate is:[ G(r) = frac{alpha}{beta} C(r) = frac{alpha C_0 R}{beta r} frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)} ]I think that's it. I don't see any mistakes in the reasoning, and the behavior of the solutions makes sense physically.Final Answer1. The steady-state concentration is (boxed{C(r) = C_0 frac{R}{r} frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)}}).2. The steady-state growth rate is (boxed{G(r) = frac{alpha C_0 R}{beta r} frac{sinhleft( sqrt{frac{k}{D}} r right)}{sinhleft( sqrt{frac{k}{D}} R right)}}).</think>

🧠:<think>Alright, so I've got these two problems to solve. Let me start with the first one about the encryption algorithm. It says that the encryption uses a matrix ( A ) such that ( E = A^k times D ), and the pattern repeats every 4 cycles, meaning ( A^4 = I ). I need to find possible integer matrices ( A ) that satisfy this condition.Hmm, okay. So ( A ) is a 3x3 integer matrix, and when raised to the 4th power, it becomes the identity matrix. That means ( A ) is a matrix of finite order, specifically order dividing 4. So, ( A^4 = I ), but maybe ( A^2 = I ) or ( A = I ) as well? Although, if ( A^2 = I ), then ( A^4 = I ) as well, so that's possible. But the problem says the period is 4, so maybe the minimal polynomial of ( A ) divides ( x^4 - 1 ) but doesn't divide ( x^k - 1 ) for any ( k < 4 ). So, the minimal polynomial could be factors like ( x^4 - 1 ), ( x^2 + 1 ), etc.But since we're dealing with integer matrices, perhaps we can think of permutation matrices or rotation matrices. Wait, permutation matrices have finite order, right? For example, a permutation matrix corresponding to a 4-cycle would have order 4. But permutation matrices are orthogonal, so their inverses are their transposes. But since we're dealing with integer matrices, maybe that's a way to go.Alternatively, think of diagonal matrices where each diagonal entry is a root of unity. But since we need integer entries, the only roots of unity possible are 1 and -1. So, if we have a diagonal matrix with entries 1 and -1, then ( A^2 = I ), which would have period 2, not 4. So that might not work.Wait, but maybe if we have a combination of different roots. But since we're restricted to integers, it's hard to get higher roots of unity. So perhaps permutation matrices are the way to go.Let me think. A permutation matrix is a matrix where each row and column has exactly one 1 and the rest are 0s. These correspond to permutations of the standard basis vectors. So, if I can find a permutation matrix that has order 4, that would satisfy ( A^4 = I ).In symmetric group ( S_3 ), the possible orders of elements are 1, 2, 3. There are no elements of order 4 in ( S_3 ). So, permutation matrices corresponding to ( S_3 ) can't have order 4. Hmm, that complicates things.So, maybe permutation matrices aren't the way to go. Alternatively, maybe consider rotation matrices, but in 3D. A rotation matrix in 3D that's a rotation by 90 degrees around some axis would have order 4. But rotation matrices have determinant 1, and they are orthogonal, so their inverses are their transposes. But would such a matrix have integer entries?Wait, a rotation matrix in 3D with integer entries? That seems unlikely because rotation matrices typically involve cosines and sines, which are rarely integers unless the angle is 0 or 180 degrees, which would give determinant 1 or -1, but not necessarily integer entries.Alternatively, maybe consider block matrices. For example, a block diagonal matrix where each block is a 2x2 rotation matrix of order 4, and the third diagonal entry is 1. But again, the 2x2 rotation matrix for 90 degrees is:[begin{pmatrix}0 & -1 1 & 0end{pmatrix}]Which has integer entries. So, if I construct a 3x3 matrix as a block diagonal matrix with this 2x2 matrix and a 1 on the bottom right, that would be:[A = begin{pmatrix}0 & -1 & 0 1 & 0 & 0 0 & 0 & 1end{pmatrix}]Then, ( A^4 ) would be the identity matrix because the 2x2 block has order 4. Let me check:( A^2 ) would be:[begin{pmatrix}-1 & 0 & 0 0 & -1 & 0 0 & 0 & 1end{pmatrix}]And ( A^4 ) would be:[begin{pmatrix}1 & 0 & 0 0 & 1 & 0 0 & 0 & 1end{pmatrix}]Yes, that works. So this is one possible matrix. But the problem says "determine the possible integer matrices ( A )", so there might be more than one.Alternatively, maybe the 2x2 block could be on the top right instead of top left, but that would require the matrix to have a different structure. Wait, no, because the block needs to be diagonal. Alternatively, maybe the 1 could be in different positions.Wait, actually, the 1 could be in any of the diagonal positions, but since the 2x2 block is already taking two positions, the third one is fixed. So, another possibility is having the 2x2 block in the middle or at the bottom, but that would require shifting the 1 accordingly.For example:[A = begin{pmatrix}1 & 0 & 0 0 & 0 & -1 0 & 1 & 0end{pmatrix}]Or:[A = begin{pmatrix}0 & 0 & 1 0 & 0 & 0 1 & 0 & 0end{pmatrix}]Wait, no, that last one isn't a rotation matrix. Let me think again.Alternatively, maybe consider matrices that are direct sums of smaller matrices of order 4. Since we're in 3x3, we can have a 2x2 block of order 4 and a 1x1 block which is 1 or -1. But since 1 has order 1, the overall order would still be 4 because the LCM of 4 and 1 is 4. Similarly, if the 1x1 block is -1, which has order 2, the LCM of 4 and 2 is 4, so that still works.So, the 2x2 block can be any matrix of order 4 with integer entries. The standard one is the rotation matrix I mentioned earlier. Are there others?Well, another 2x2 integer matrix of order 4 is:[begin{pmatrix}0 & 1 -1 & 0end{pmatrix}]Which is similar to the previous one but transposed. So, embedding this into 3x3 would give another matrix.Alternatively, maybe consider the negative of the identity matrix. But ( (-I)^4 = I ), so that's another possibility. But that's just the scalar matrix -1, but in 3x3, it's:[A = begin{pmatrix}-1 & 0 & 0 0 & -1 & 0 0 & 0 & -1end{pmatrix}]But this has order 2, since ( A^2 = I ). So, it doesn't have order 4, so it's not suitable.Alternatively, maybe consider a permutation matrix combined with sign changes. For example, a matrix that permutes basis vectors and also flips signs. But in 3x3, it's tricky because the permutation part might interfere with the sign flips.Wait, let me think of another approach. The characteristic polynomial of ( A ) must divide ( x^4 - 1 ). So, the minimal polynomial divides ( x^4 - 1 ). Since ( A ) is a 3x3 matrix, its minimal polynomial can't have degree higher than 3. So, possible minimal polynomials are factors of ( x^4 - 1 ) of degree at most 3.Factoring ( x^4 - 1 ), we get ( (x^2 - 1)(x^2 + 1) ), which further factors as ( (x - 1)(x + 1)(x^2 + 1) ). So, possible minimal polynomials could be ( x - 1 ), ( x + 1 ), ( x^2 + 1 ), or products of these, but without repeating factors.But since we want ( A ) to have order 4, the minimal polynomial should be such that the order is 4. So, if the minimal polynomial is ( x^2 + 1 ), then ( A^2 = -I ), so ( A^4 = I ). That would work. Similarly, if the minimal polynomial is ( (x - 1)(x^2 + 1) ), then ( A ) would have eigenvalues 1 and ( i, -i ), but since we're dealing with integer matrices, the characteristic polynomial must have integer coefficients, so complex eigenvalues would come in conjugate pairs.But wait, if the minimal polynomial is ( x^2 + 1 ), then the characteristic polynomial would be ( (x^2 + 1)(x - c) ) for some integer ( c ). But since the characteristic polynomial must have integer coefficients, ( c ) must be an integer. However, for the matrix to have order 4, the eigenvalues must be roots of unity, so ( c ) must be either 1 or -1.Wait, but if the minimal polynomial is ( x^2 + 1 ), then the characteristic polynomial must be a multiple of ( x^2 + 1 ). Since it's a 3x3 matrix, the characteristic polynomial would be ( (x^2 + 1)(x - d) ), where ( d ) is an integer. But for the matrix to have finite order, ( d ) must be a root of unity, so ( d = 1 ) or ( d = -1 ).So, possible characteristic polynomials are ( (x^2 + 1)(x - 1) ) or ( (x^2 + 1)(x + 1) ). Let's see if such matrices exist.For example, take the companion matrix of ( x^3 - x^2 + x - 1 ), which is ( (x - 1)(x^2 + 1) ). The companion matrix would be:[begin{pmatrix}0 & 0 & 1 1 & 0 & -1 0 & 1 & 1end{pmatrix}]Wait, no, the companion matrix for polynomial ( x^3 + a x^2 + b x + c ) is:[begin{pmatrix}0 & 0 & -c 1 & 0 & -b 0 & 1 & -aend{pmatrix}]So, for ( x^3 - x^2 + x - 1 ), the companion matrix is:[begin{pmatrix}0 & 0 & 1 1 & 0 & -1 0 & 1 & 1end{pmatrix}]Let me check if this matrix satisfies ( A^4 = I ). It might be a bit tedious, but let's try computing ( A^2 ):First, compute ( A^2 ):[A^2 = A times A = begin{pmatrix}0 & 0 & 1 1 & 0 & -1 0 & 1 & 1end{pmatrix}timesbegin{pmatrix}0 & 0 & 1 1 & 0 & -1 0 & 1 & 1end{pmatrix}]Compute each element:First row:- (1,1): 0*0 + 0*1 + 1*0 = 0- (1,2): 0*0 + 0*0 + 1*1 = 1- (1,3): 0*1 + 0*(-1) + 1*1 = 1Second row:- (2,1): 1*0 + 0*1 + (-1)*0 = 0- (2,2): 1*0 + 0*0 + (-1)*1 = -1- (2,3): 1*1 + 0*(-1) + (-1)*1 = 0Third row:- (3,1): 0*0 + 1*1 + 1*0 = 1- (3,2): 0*0 + 1*0 + 1*1 = 1- (3,3): 0*1 + 1*(-1) + 1*1 = 0So, ( A^2 ) is:[begin{pmatrix}0 & 1 & 1 0 & -1 & 0 1 & 1 & 0end{pmatrix}]Now, compute ( A^3 = A^2 times A ):[begin{pmatrix}0 & 1 & 1 0 & -1 & 0 1 & 1 & 0end{pmatrix}timesbegin{pmatrix}0 & 0 & 1 1 & 0 & -1 0 & 1 & 1end{pmatrix}]First row:- (1,1): 0*0 + 1*1 + 1*0 = 1- (1,2): 0*0 + 1*0 + 1*1 = 1- (1,3): 0*1 + 1*(-1) + 1*1 = 0Second row:- (2,1): 0*0 + (-1)*1 + 0*0 = -1- (2,2): 0*0 + (-1)*0 + 0*1 = 0- (2,3): 0*1 + (-1)*(-1) + 0*1 = 1Third row:- (3,1): 1*0 + 1*1 + 0*0 = 1- (3,2): 1*0 + 1*0 + 0*1 = 0- (3,3): 1*1 + 1*(-1) + 0*1 = 0So, ( A^3 ) is:[begin{pmatrix}1 & 1 & 0 -1 & 0 & 1 1 & 0 & 0end{pmatrix}]Now, compute ( A^4 = A^3 times A ):[begin{pmatrix}1 & 1 & 0 -1 & 0 & 1 1 & 0 & 0end{pmatrix}timesbegin{pmatrix}0 & 0 & 1 1 & 0 & -1 0 & 1 & 1end{pmatrix}]First row:- (1,1): 1*0 + 1*1 + 0*0 = 1- (1,2): 1*0 + 1*0 + 0*1 = 0- (1,3): 1*1 + 1*(-1) + 0*1 = 0Second row:- (2,1): (-1)*0 + 0*1 + 1*0 = 0- (2,2): (-1)*0 + 0*0 + 1*1 = 1- (2,3): (-1)*1 + 0*(-1) + 1*1 = 0Third row:- (3,1): 1*0 + 0*1 + 0*0 = 0- (3,2): 1*0 + 0*0 + 0*1 = 0- (3,3): 1*1 + 0*(-1) + 0*1 = 1So, ( A^4 ) is:[begin{pmatrix}1 & 0 & 0 0 & 1 & 0 0 & 0 & 1end{pmatrix}]Yes! So this matrix ( A ) satisfies ( A^4 = I ). Therefore, this is another possible matrix.But wait, is this matrix similar to the block diagonal matrix I mentioned earlier? Let me check. The block diagonal matrix had a 2x2 rotation block and a 1. This companion matrix seems to be a different structure. So, there are at least two different types of matrices that satisfy ( A^4 = I ).Another thought: maybe consider matrices that are direct sums of smaller matrices. For example, a 1x1 block and a 2x2 block. The 1x1 block could be 1 or -1, and the 2x2 block could be the rotation matrix. So, as I thought earlier, the block diagonal matrix is one possibility.Additionally, maybe consider matrices that are not block diagonal but still satisfy ( A^4 = I ). For example, the companion matrix we just saw is one such example.So, to summarize, possible integer matrices ( A ) that satisfy ( A^4 = I ) include:1. Block diagonal matrices with a 2x2 rotation matrix (order 4) and a 1x1 block of 1 or -1.2. Companion matrices of polynomials like ( x^3 - x^2 + x - 1 ), which have minimal polynomial ( x^2 + 1 ) and thus satisfy ( A^4 = I ).Are there more? Maybe, but these are two distinct types. Since the problem asks for possible matrices, I can present these as examples.Now, moving on to the second problem. The hacker used a hash function ( H(x) = (a x + b) mod p ), where ( p = 23 ). We're given two data blocks ( B_1 = 5 ) and ( B_2 = 9 ), with hash values ( h_1 = 15 ) and ( h_2 = 7 ). We need to find ( a ) and ( b ).So, we have the system of congruences:1. ( 5a + b equiv 15 mod 23 )2. ( 9a + b equiv 7 mod 23 )We can solve this system for ( a ) and ( b ).Subtracting the first equation from the second:( (9a + b) - (5a + b) equiv 7 - 15 mod 23 )Simplify:( 4a equiv -8 mod 23 )Since ( -8 mod 23 ) is ( 15 ), we have:( 4a equiv 15 mod 23 )Now, we need to solve for ( a ). To do this, we can find the modular inverse of 4 modulo 23.Find ( 4^{-1} mod 23 ). We need an integer ( x ) such that ( 4x equiv 1 mod 23 ).Using the extended Euclidean algorithm:23 = 5*4 + 34 = 1*3 + 13 = 3*1 + 0So, backtracking:1 = 4 - 1*3But 3 = 23 - 5*4, so:1 = 4 - 1*(23 - 5*4) = 6*4 - 1*23Therefore, ( 4^{-1} equiv 6 mod 23 ).So, multiply both sides of ( 4a equiv 15 mod 23 ) by 6:( a equiv 15*6 mod 23 )Calculate 15*6 = 9090 mod 23: 23*3 = 69, 90 - 69 = 21So, ( a equiv 21 mod 23 )Now, substitute ( a = 21 ) into the first equation to find ( b ):( 5*21 + b equiv 15 mod 23 )Calculate 5*21 = 105105 mod 23: 23*4 = 92, 105 - 92 = 13So, 13 + b ≡ 15 mod 23Thus, b ≡ 15 - 13 = 2 mod 23So, ( b = 2 )Let me verify:For ( B_1 = 5 ):( H(5) = 21*5 + 2 = 105 + 2 = 107 )107 mod 23: 23*4 = 92, 107 - 92 = 15. Correct.For ( B_2 = 9 ):( H(9) = 21*9 + 2 = 189 + 2 = 191 )191 mod 23: 23*8 = 184, 191 - 184 = 7. Correct.So, ( a = 21 ) and ( b = 2 ).Final Answer1. One possible integer matrix ( A ) is (boxed{begin{pmatrix} 0 & -1 & 0 1 & 0 & 0 0 & 0 & 1 end{pmatrix}}).2. The coefficients are ( a = boxed{21} ) and ( b = boxed{2} ).</think>