Question

数学疑难问题解答

Answer 1

问：数学疑难问题解答

问：三道题哦

答：好的，这边帮你看看哈

答：线性代数嘛对嘛亲

问：谢谢老师

问：是哒

答：那你等会，这个也要花时间

问：还有概率论的，这个等等再下一单

问：好的，不着急

答：先做完这个你再下单哈，这边做不赢

答：线代：1.(a) The augmented matrix is:[ M | xo ] = [ 2 1 -3 4 ][ 1 -2 1 -3 ][ 1 17 1 -1 ]We can find the reduced row echelon form of this matrix using elementary row operations:R2 = R2 - 2R1R3 = R3 - R1R1 = R1/2R3 = R3 - 17R2The matrix then becomes:[ 1 0 0 2 ][ 0 1 0 -3 ][ 0 0 1 1 ]Therefore, the reduced row echelon form of the augmented matrix is:[ M' | xo' ] = [ 1 0 0 2 ][ 0 1 0 -3 ][ 0 0 1 1 ](b) Using the result of part (a), the system of equations becomes:x1 + 2x4 = 4x2 - 3x4 = -3x3 + x4 = 1From the second equation, we can solve for x2 in terms of x4:x2 = -3x4 - 3Substituting this into the first and third equations, we get:x1 + 2x4 = 4x3 - 2x4 = 4Solving for x1 and x3 in terms of x4, we get:x1 = 4 - 2x4x3 = 1 + 2x4Therefore, the general solution is:x = [ 4 - 2t, -3t - 3, 1 + 2t, t ], where t is any real number.(c) (i) Let x and y be distinct solutions of equation (1). Then:Mx = xo (1)My = xo (2)Subtrac

答：Subtracting equation (2) from equation (1), we get:Mx - My = 0orM(x - y) = 0Since x and y are distinct solutions, we have x - y ≠ 0. Therefore, M(x - y) = 0 implies that (x - y) is an eigenvector of M with eigenvalue 0.(ii) From part (b), we know that the general solution to the system of equations is:x = [ 4 - 2t, -3t - 3, 1 + 2t, t ]Setting t = 1, we get:x = [ 2, -6, 3, 1 ]which is an eigenvector of M with eigenvalue 0, as shown in part (c)(i).

答：(a)i). the characteristic polynomial of t can be found by computing the determinant of the matrix t - xi, where i is the identity matrix of the same size as t.t - xi = |12-x loo 3||0 12-x 0||0 0 3-x|det(t - xi) = (12-x)(12-x)(3-x)= (12-x)^2(3-x)ii). eigenvalues of t are roots of the characteristic polynomial. thus,(12-x)^2(3-x) = 0so, the eigenvalues are x=12 (with multiplicity 2) and x=3.(b) to find an eigenvector for each eigenvalue:for x = 12:we need to solve the system of linear equations|0 -1 3| |a| |0||loo-1 0 | *|b| = |0||0 0| |c| |0|equivalently, we have-b + 3c = 0(loo-1)a = 00 = 0from the second equation, we know that a=kb for some scalar k; substituting this into the first equation gives:3c = kbsince we want a non-zero eigenvector, we can choose b=1 and obtain c=k/3. andfor a we get a = k(loo-1).therefore, any vector of the form (k(loo-1), 1, k/3) with k not equal to zero is an eigenvector associ

答：therefore, any vector of the form (k(loo-1), 1, k/3) with k not equal to zero is an eigenvector associated with the eigenvalue 12.for x = 3:we need to solve the system of linear equations|9 loo 3 ||0 9 0 ||0 0 0 |we can immediately see that the third row corresponds to the equation 0=0, which is redundant. the first two rows correspond to the system9a + loo b + 3c = 3a9b = 3bfrom the second equation we know b=0 or a non-zero multiple of b. choosing b=1, we obtain a=3/9 and c=0.therefore, any vector

答：3.(a) Consider any vector x in R^2 with norm equal to 1, i.e. ||x||=1. Then the image of x under A, Ax, is also a unit vector because det(A)=1 implies that A preserves area. Let y be the unit vector orthogonal to Ax, then {Ax,y} is an orthonormal basis for R^2. Since x belongs to the unit circle and y⊥x, we have y*=[0 1]T or y*=[0 -1]T. In either case, Ay=y*(Ay) * y=±[det(Ay),0]T and ||Ay||=1, which shows that Ay is another vector of the unit circle.(b) From part (a), we know that A maps the unit circle onto itself. Thus, for any vector on the unit circle represented by x, there exists some vector y on the unit circle such that Ax=(x y). Then, using the fact that det(A)=1, we obtain:(c) We define the cross product of two vectors in R^2 as follows:for any u, v ∈ R^2. Then we can show:i) xy = -yx. This follows directly from the definition of the cross product.

答：ii) (Ax)(Ay) = xy. Using the formula (b), we have:Expanding the dot product, we have:where θ is the angle between x and Ax. Since x and Ax span the same subspace, this angle is either 0 or π/2. If it's 0, then x=Ax and Ay is just the other vector on the unit circle orthogonal to x. In this case, Ay=y*[-det(A),0]T=±[0 det(A)]T =±y*(-det(A)) * Ay, and we get (Ax)(Ay)=xy. If θ=π/2, then x and Ax are orthogonal, so the term involving sin(θ) in the above equation vanishes. Since x and Ay both lie on the unit circle and are orthogonal to each other, it follows that Ay is a scalar multiple of y, i.e. Ay=ydet(Ay). Therefore, (Ax)(Ay) = det(Ay)x*y.In either case, we have shown that (Ax)(Ay)=xy.

答：概率论：1.a) Let F be the event that we pull out the fair die and L be the event that we pull out the loaded die. Then P(F)=P(L)=0.5, as we are equally likely to select either die. Let S be the event that a six is rolled. ThenP(S|F) = 1/6, since the fair die has an equal probability of rolling any of its faces P(S|L) = 1/2, since the loaded die only has three even faces and we know that a six was rolledUsing Bayes' rule, we can compute the updated probability of having chosen the loaded die given that a six was rolled:b) Let S' be the event that a five is rolled on the second die. We want to compute the updated probability that the first die is the loaded die given this new information. Using Bayes' rule again, we have:where D is the event that the first die is the loaded die. To calculate the denominator in this expression, we need to consider all possible ways that we could have obtained a 5 when rolling the second die. These are:

答：1.We selected the fair die and rolled a 5, which has probability (1/6) * (1/2) = 1/12.2.We selected the loaded die and rolled a 5, which has probability (1/3) * (1/2) = 1/6.Therefore,P(S') = (1/6)(1/2)+(1/3)(1/2) = 1/4Substituting in our values and simplifying, we get:P(D|S') = (1/3)/(1/3 + 1/24) = 8/9

答：2.a) Let D be the event that the finch is a Darwin finch and W be the event that it is a Wallace finch. We want to compute P(D|ID), the probability that the bird is a Darwin finch given that we identified it as such. Using Bayes' rule, we have:P(D|ID) = P(ID|D) * P(D) / [P(ID|D)*P(D) + P(ID|W)*P(W)]where P(ID|D)=0.95, P(ID|W)=0.05, P(D)=0.01, and P(W)=0.99. Substituting in these values and simplifying, we get:P(D|ID) = (0.95) * (0.01) / [(0.95)(0.01) + (0.05)(0.99)] ≈ 0.161Therefore, the probability that the bird is a Darwin finch given that we identified it as such is approximately 0.161.

答：2.（b) Using natural frequencies, we can reason as follows:Suppose there are 10,000 finches in the area. Then 1% or 100 of them are Darwin finches.Of these 100 Darwin finches, we expect to correctly identify 95% or 95 of them as Darwin finches. The remaining 5% or 5 of them would be misidentified as Wallace finches.Of the 9,900 Wallace finches, we expect to misidentify 5% or 495 of them as Darwin finches.Therefore, out of the total of 500 birds identified as Darwin finches, only 95 are true Darwin finches. Thus, the probability that the bird is a Darwin finch given that we identified it as such is approximately 95/500 = 0.19.

答：3.The heuristic that could lead to this behavior is the anchoring effect. The anchoring effect occurs when people rely too heavily on the first piece of information they receive when making decisions, even if that information is irrelevant or inaccurate. In this case, the salesperson provided an anchor of $100,000 when stating the price of the car, which may have influenced the buyer's perception of what a reasonable price for the car would be. When the buyer negotiated the price down to $20,000, they may have felt pleased with their negotiation skills because they were able to move the price significantly away from the initial anchor. However, when they later saw a similar car for sale for $15,000, they realized that they may have been overpaying for the car and that the original anchor had biased their perception of what was a reasonable price. This demonstrates how the anchoring effect can lead people to make suboptimal decisions by fixating too much on the initial anchor.

答：4.(a) The heuristic that could lead to this belief is the availability heuristic. The availability heuristic occurs when people make judgments about the likelihood of events based on how easily examples come to mind. In this case, the person's association between luxury cars, designer clothes, and wealth may be influenced by exposure to media, advertising, or cultural norms. These stimuli are frequently presented to us in a variety of contexts, making them readily available and easily accessible in our memory. This can lead people to overestimate the prevalence of wealth among those who display these symbols, without considering other factors that might contribute to the person's financial situation.

答：4.(b)The use of the availability heuristic differs from the use of Bayes' rule in that it relies less on statistical probability and more on subjective impressions. Bayes' rule involves the formal calculation of probabilities based on prior knowledge and conditional probabilities, whereas the availability heuristic draws on people's past experiences and salient examples to inform their beliefs. While both approaches may involve some degree of uncertainty or error, Bayes' rule provides a more systematic and objective method for updating beliefs based on new information, whereas the availability heuristic may be more prone to biases and stereotypes. Therefore, relying solely on the availability heuristic in this situation may not provide an accurate representation of the person's true financial status, as it may be influenced by superficial cues and cognitive shortcuts.