💡Okay, so I have this system of equations to solve:[begin{cases}xy + yz + zx = 3n^2 - 1 x + y + z = 3nend{cases}]And I need to show that the only integral solution with (x geq y geq z) is (x = n+1), (y = n), and (z = n-1). Hmm, let's see how to approach this.First, I remember that sometimes when dealing with symmetric equations involving sums and products of variables, it's helpful to use identities or manipulate the equations to find more information. Maybe I can express (x^2 + y^2 + z^2) somehow because that often comes up in these kinds of problems.I recall the identity:[(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)]Given that (x + y + z = 3n), if I square both sides, I get:[(3n)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)]Which simplifies to:[9n^2 = x^2 + y^2 + z^2 + 2(3n^2 - 1)]Wait, because (xy + yz + zx = 3n^2 - 1), so substituting that in:[9n^2 = x^2 + y^2 + z^2 + 2(3n^2 - 1)]Let me compute the right-hand side:[x^2 + y^2 + z^2 + 6n^2 - 2]So, moving the (6n^2 - 2) to the left:[9n^2 - 6n^2 + 2 = x^2 + y^2 + z^2]Which simplifies to:[3n^2 + 2 = x^2 + y^2 + z^2]Alright, so now I have another equation: (x^2 + y^2 + z^2 = 3n^2 + 2). Hmm, how can I use this?I also remember another identity involving the squares of differences:[(x - y)^2 + (y - z)^2 + (x - z)^2 = 2(x^2 + y^2 + z^2) - 2(xy + yz + zx)]Let me plug in the values I have. I know (x^2 + y^2 + z^2 = 3n^2 + 2) and (xy + yz + zx = 3n^2 - 1). So substituting these in:[(x - y)^2 + (y - z)^2 + (x - z)^2 = 2(3n^2 + 2) - 2(3n^2 - 1)]Calculating the right-hand side:[6n^2 + 4 - 6n^2 + 2 = 6]So, the sum of the squares of the differences is 6. That is:[(x - y)^2 + (y - z)^2 + (x - z)^2 = 6]Now, since (x geq y geq z), all the differences (x - y) and (y - z) are non-negative. Also, (x - z = (x - y) + (y - z)), so it's the sum of the other two differences.Given that the sum of their squares is 6, and all terms are non-negative integers, I need to find integers (a = x - y) and (b = y - z) such that:[a^2 + b^2 + (a + b)^2 = 6]Let me compute that:[a^2 + b^2 + (a + b)^2 = a^2 + b^2 + a^2 + 2ab + b^2 = 2a^2 + 2b^2 + 2ab = 6]Divide both sides by 2:[a^2 + b^2 + ab = 3]So, I need to find non-negative integers (a) and (b) such that (a^2 + b^2 + ab = 3). Let's list possible small values for (a) and (b):- If (a = 0), then (0 + b^2 + 0 = b^2 = 3). But 3 is not a perfect square, so no solution here.- If (a = 1), then (1 + b^2 + b = 3). So, (b^2 + b - 2 = 0). Solving this quadratic: (b = [-1 pm sqrt{1 + 8}]/2 = [-1 pm 3]/2). So, (b = 1) or (b = -2). Since (b) is non-negative, (b = 1).- If (a = 2), then (4 + b^2 + 2b = 3). This simplifies to (b^2 + 2b + 1 = 0), which factors as ((b + 1)^2 = 0), so (b = -1). Not possible since (b) is non-negative.- If (a = 3), then (9 + b^2 + 3b = 3). This would require (b^2 + 3b + 6 = 0), which has no real solutions.So the only solution is (a = 1) and (b = 1). Therefore, (x - y = 1) and (y - z = 1). Thus, (x = y + 1) and (z = y - 1).Now, since (x + y + z = 3n), substituting (x = y + 1) and (z = y - 1):[(y + 1) + y + (y - 1) = 3n]Simplifying:[3y = 3n implies y = n]Therefore, (x = n + 1) and (z = n - 1). So, the solution is (x = n + 1), (y = n), (z = n - 1).I should check if this satisfies the original equation (xy + yz + zx = 3n^2 - 1):Compute (xy + yz + zx):[(n + 1)n + n(n - 1) + (n + 1)(n - 1)]Calculating each term:- ((n + 1)n = n^2 + n)- (n(n - 1) = n^2 - n)- ((n + 1)(n - 1) = n^2 - 1)Adding them up:[(n^2 + n) + (n^2 - n) + (n^2 - 1) = 3n^2 - 1]Which matches the given equation. So, yes, it works.Are there any other possible solutions? Well, from the earlier step, the only possible (a) and (b) are 1 and 1. Any other values would either not satisfy the equation or result in negative differences, which aren't allowed since (x geq y geq z). Therefore, this must be the only solution.I think that's it. So, the only integral solution with (x geq y geq z) is (x = n + 1), (y = n), and (z = n - 1).**Final Answer**The only integral solution is boxed{(x, y, z) = (n+1, n, n-1)}.

💡To determine the total percentage saved on the jacket, follow these steps:1. **Calculate the total savings:** - The store discount is 20. - The additional coupon discount is 16. - Total savings = 20 + 16 = 36.2. **Determine the percentage saved:** - The original price of the jacket is 160. - Percentage saved = (Total Savings / Original Price) × 100. - Percentage saved = (36 / 160) × 100 = 22.5%.Thus, the total percentage saved is 22.5%.

💡Alright, so I have this problem about real symmetric matrices and some trace conditions. It's divided into two parts, a) and b). Let me try to understand what's being asked here.Starting with part a): We're given that ( k ) is an even positive integer, and ( A ) is a real symmetric ( n times n ) matrix. The condition provided is that ( operatorname{tr}(A^k)^{k+1} = operatorname{tr}(A^{k+1})^k ). We need to prove that under these conditions, ( A^n = operatorname{tr}(A) A^{n-1} ).Hmm, okay. So, since ( A ) is a real symmetric matrix, I remember that it can be diagonalized. That means there exists an orthogonal matrix ( P ) such that ( P^T A P ) is a diagonal matrix. The diagonal entries are the eigenvalues of ( A ). Let me denote them as ( lambda_1, lambda_2, ldots, lambda_n ).Given that, the trace of ( A^k ) would just be the sum of the ( k )-th powers of the eigenvalues, right? So, ( operatorname{tr}(A^k) = lambda_1^k + lambda_2^k + ldots + lambda_n^k ). Similarly, ( operatorname{tr}(A^{k+1}) = lambda_1^{k+1} + lambda_2^{k+1} + ldots + lambda_n^{k+1} ).So, the condition ( operatorname{tr}(A^k)^{k+1} = operatorname{tr}(A^{k+1})^k ) translates to:[left( sum_{i=1}^n lambda_i^k right)^{k+1} = left( sum_{i=1}^n lambda_i^{k+1} right)^k]I need to analyze this equation. Since ( k ) is even, all the terms ( lambda_i^k ) and ( lambda_i^{k+1} ) are non-negative, regardless of the sign of ( lambda_i ). That might be helpful.Let me think about the inequality between the sums. Maybe I can use some inequality principles here. For example, if I have positive numbers, then certain inequalities hold. Since ( k ) is even, all the eigenvalues raised to the ( k )-th power are non-negative, so I can treat them as positive real numbers.I recall that for positive real numbers, there's an inequality related to power means. Specifically, for positive real numbers ( x_1, x_2, ldots, x_m ) and exponents ( p ) and ( q ), if ( p > q ), then the ( p )-th power mean is greater than or equal to the ( q )-th power mean.But in this case, the equation is an equality. So, maybe this suggests that all the terms are equal? Because in inequalities, equality often holds when all the elements are equal.Wait, but in our case, the equation is:[left( sum lambda_i^k right)^{k+1} = left( sum lambda_i^{k+1} right)^k]Let me denote ( S_k = sum lambda_i^k ) and ( S_{k+1} = sum lambda_i^{k+1} ). Then the equation becomes:[S_k^{k+1} = S_{k+1}^k]Taking both sides to the power of ( 1/k ), we get:[S_k^{(k+1)/k} = S_{k+1}]Which simplifies to:[S_{k+1} = S_k^{1 + 1/k}]Hmm, interesting. So, the sum of the ( (k+1) )-th powers is equal to the sum of the ( k )-th powers raised to the power ( 1 + 1/k ).I wonder if this implies that all the eigenvalues are equal. If all ( lambda_i ) are equal, say ( lambda ), then ( S_k = n lambda^k ) and ( S_{k+1} = n lambda^{k+1} ). Plugging into the equation:[(n lambda^k)^{k+1} = (n lambda^{k+1})^k]Simplify:[n^{k+1} lambda^{k(k+1)} = n^k lambda^{k(k+1)}]Divide both sides by ( n^k lambda^{k(k+1)} ):[n = 1]Wait, that suggests ( n = 1 ), which is trivial. But our matrix is ( n times n ), so ( n ) could be any positive integer. So, maybe not all eigenvalues are equal unless ( n = 1 ).Alternatively, perhaps only one eigenvalue is non-zero. Let me explore that.Suppose that only one eigenvalue is non-zero, say ( lambda_1 ), and the rest are zero. Then ( S_k = lambda_1^k ) and ( S_{k+1} = lambda_1^{k+1} ). Plugging into the equation:[(lambda_1^k)^{k+1} = (lambda_1^{k+1})^k]Which simplifies to:[lambda_1^{k(k+1)} = lambda_1^{k(k+1)}]Which is always true, regardless of ( lambda_1 ). So, if only one eigenvalue is non-zero, the condition holds.But in the case where more than one eigenvalue is non-zero, does the condition still hold? Let me test with two non-zero eigenvalues.Let’s say ( lambda_1 = a ) and ( lambda_2 = b ), and the rest are zero. Then:( S_k = a^k + b^k )( S_{k+1} = a^{k+1} + b^{k+1} )The equation becomes:[(a^k + b^k)^{k+1} = (a^{k+1} + b^{k+1})^k]Is this possible for some ( a ) and ( b )?Let me try specific numbers. Let’s take ( k = 2 ) (since it's even), ( a = 1 ), ( b = 1 ). Then:Left side: ( (1 + 1)^3 = 8 )Right side: ( (1 + 1)^2 = 4 )Not equal. So, that doesn't work.What if ( a = 2 ), ( b = 1 ):Left side: ( (4 + 1)^3 = 125 )Right side: ( (8 + 1)^2 = 81 )Still not equal.What if ( a = 1 ), ( b = -1 ). But since ( k ) is even, ( (-1)^k = 1 ), so:Left side: ( (1 + 1)^3 = 8 )Right side: ( (1 + (-1))^{2} = 0 )Not equal.Wait, but if ( a = 0 ), ( b = 0 ), then both sides are zero, but that's trivial.Alternatively, maybe ( a = 0 ), ( b neq 0 ). Then, as before, it reduces to the single eigenvalue case.So, it seems that unless only one eigenvalue is non-zero, the equation doesn't hold. Therefore, the condition ( operatorname{tr}(A^k)^{k+1} = operatorname{tr}(A^{k+1})^k ) implies that ( A ) has at most one non-zero eigenvalue.If ( A ) has at most one non-zero eigenvalue, then it's a rank-one matrix. For a real symmetric matrix, being rank-one implies that it can be written as ( A = lambda v v^T ), where ( v ) is a unit vector and ( lambda ) is the non-zero eigenvalue.Now, let's recall the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation. For a rank-one matrix ( A ), the characteristic polynomial is ( lambda^n - operatorname{tr}(A) lambda^{n-1} = 0 ). Therefore, by Cayley-Hamilton, ( A^n = operatorname{tr}(A) A^{n-1} ).So, that seems to be the conclusion.Moving on to part b): Does the assertion from a) also hold for odd positive integers ( k )?Hmm, so we need to check if the same conclusion ( A^n = operatorname{tr}(A) A^{n-1} ) holds when ( k ) is odd.Let me think. If ( k ) is odd, then the eigenvalues raised to the ( k )-th power retain their sign. So, negative eigenvalues would remain negative, and positive ones remain positive.In part a), the key was that the given condition forced all but one eigenvalue to be zero. But with ( k ) odd, maybe the behavior is different.Let me test with ( k = 1 ), which is odd. Let’s see if the condition ( operatorname{tr}(A)^2 = operatorname{tr}(A^2) ) implies that ( A^n = operatorname{tr}(A) A^{n-1} ).Wait, for ( k = 1 ), the condition is:[(operatorname{tr}(A))^2 = operatorname{tr}(A^2)]But for any matrix, ( operatorname{tr}(A^2) ) is equal to the sum of the squares of the eigenvalues, and ( (operatorname{tr}(A))^2 ) is the square of the sum of the eigenvalues.So, ( (sum lambda_i)^2 = sum lambda_i^2 ). Expanding the left side:[sum lambda_i^2 + 2 sum_{i < j} lambda_i lambda_j = sum lambda_i^2]Which simplifies to:[2 sum_{i < j} lambda_i lambda_j = 0]So, the sum of the products of the eigenvalues two at a time is zero.Does this imply that all but one eigenvalue are zero? Not necessarily. For example, consider a 3x3 matrix with eigenvalues ( a, b, c ) such that ( ab + ac + bc = 0 ). This doesn't force any of the eigenvalues to be zero unless specific conditions are met.For instance, take ( a = 1 ), ( b = 1 ), ( c = -2 ). Then ( ab + ac + bc = 1*1 + 1*(-2) + 1*(-2) = 1 - 2 - 2 = -3 neq 0 ). Hmm, not zero.Wait, maybe another example. Let’s take ( a = 1 ), ( b = -1 ), ( c = 0 ). Then ( ab + ac + bc = (1)(-1) + (1)(0) + (-1)(0) = -1 + 0 + 0 = -1 neq 0 ).Alternatively, take ( a = 2 ), ( b = -1 ), ( c = -1 ). Then ( ab + ac + bc = (2)(-1) + (2)(-1) + (-1)(-1) = -2 -2 + 1 = -3 neq 0 ).Wait, maybe it's tricky to find such eigenvalues. Let me think algebraically.Suppose ( a + b + c = S ), ( ab + ac + bc = 0 ), and ( abc = P ). Then, the characteristic equation is ( lambda^3 - S lambda^2 + 0 lambda - P = 0 ). So, ( lambda^3 - S lambda^2 - P = 0 ).But does this imply that one of the eigenvalues is zero? If ( c = 0 ), then ( ab = 0 ), so either ( a = 0 ) or ( b = 0 ). But then, if ( c = 0 ), and say ( a = 0 ), then ( b ) is arbitrary, but ( ab + ac + bc = 0 ) holds trivially. So, in this case, if one eigenvalue is zero, the condition holds.But in the case where none of the eigenvalues are zero, can ( ab + ac + bc = 0 ) hold? Let's see.Suppose ( a = 1 ), ( b = 1 ), ( c = -2 ). Then ( ab + ac + bc = 1 + (-2) + (-2) = -3 neq 0 ).Wait, maybe another set. Let’s take ( a = 3 ), ( b = -1 ), ( c = -2 ). Then ( ab + ac + bc = (-3) + (-6) + 2 = -7 neq 0 ).Alternatively, ( a = 1 ), ( b = -1 ), ( c = 0 ). Then ( ab + ac + bc = (-1) + 0 + 0 = -1 neq 0 ).Hmm, maybe it's not possible to have three non-zero eigenvalues with ( ab + ac + bc = 0 ). Wait, actually, let me think of a specific case where ( ab + ac + bc = 0 ) without any eigenvalue being zero.Let’s take ( a = 1 ), ( b = 1 ), ( c = -1 ). Then ( ab + ac + bc = 1 + (-1) + (-1) = -1 neq 0 ).Wait, maybe ( a = 2 ), ( b = -1 ), ( c = -1 ). Then ( ab + ac + bc = (-2) + (-2) + 1 = -3 neq 0 ).I'm starting to think that it might not be possible to have three non-zero eigenvalues with ( ab + ac + bc = 0 ). Maybe only two non-zero eigenvalues can satisfy this.Wait, let's try two non-zero eigenvalues. Suppose ( a = 1 ), ( b = -1 ), and ( c = 0 ). Then ( ab + ac + bc = (-1) + 0 + 0 = -1 neq 0 ).Alternatively, ( a = 2 ), ( b = -2 ), ( c = 0 ). Then ( ab + ac + bc = (-4) + 0 + 0 = -4 neq 0 ).Wait, maybe if ( a = 1 ), ( b = 1 ), ( c = -2 ), but as before, that doesn't work.Wait, perhaps if ( a = 1 ), ( b = -1 ), ( c = 0 ), but that still doesn't give zero.Wait, maybe I'm approaching this wrong. Let's consider the equation ( ab + ac + bc = 0 ). If ( a + b + c = S ), then ( ab + ac + bc = 0 ). Let me see if this can hold for non-zero ( a, b, c ).Suppose ( a = 1 ), ( b = t ), ( c = s ). Then ( ab + ac + bc = t + s + ts = 0 ). So, ( t + s + ts = 0 ). Let me solve for ( s ):( s(1 + t) = -t )If ( t neq -1 ), then ( s = -t / (1 + t) ).So, for example, let’s take ( t = 1 ), then ( s = -1/2 ). So, ( a = 1 ), ( b = 1 ), ( c = -1/2 ). Then ( ab + ac + bc = 1*1 + 1*(-1/2) + 1*(-1/2) = 1 - 1/2 - 1/2 = 0 ). Yes, that works.So, in this case, all eigenvalues are non-zero, and ( ab + ac + bc = 0 ). Therefore, the condition ( (operatorname{tr}(A))^2 = operatorname{tr}(A^2) ) can hold even when more than one eigenvalue is non-zero.Therefore, for ( k = 1 ), the condition does not necessarily imply that ( A ) has only one non-zero eigenvalue. Hence, the conclusion ( A^n = operatorname{tr}(A) A^{n-1} ) does not necessarily hold for odd ( k ).So, in part b), the assertion does not hold for odd positive integers ( k ).**Final Answer**a) boxed{A^n = operatorname{tr}(A) A^{n-1}}b) The assertion does not hold for odd positive integers ( k ).

💡First, Wendi starts with 12 chickens.Next, she brings home 3 times the number of her current chickens minus 8:3 * 12 - 8 = 36 - 8 = 28 chickens.Now, she has a total of 12 + 28 = 40 chickens.A neighbor's dog eats 2 chickens, leaving her with 40 - 2 = 38 chickens.Finally, Wendi finds twice the difference between three times her current number and ten chickens:3 * 38 - 10 = 114 - 10 = 104 chickens.She brings home twice this difference: 2 * 104 = 208 chickens.Adding the new chickens to her current number:38 + 208 = 246 chickens.

💡Okay, so I have this problem where a point M(x, y) lies on the line y = x + 2, and the y-coordinate is between 1 and 3. I need to calculate the value of A, which is the sum of two square roots: sqrt(y² - 8x) and sqrt(y² + 2x + 5). Hmm, let me try to figure this out step by step.First, since the point M lies on the line y = x + 2, I can express y in terms of x. That means y = x + 2. Also, it's given that 1 < y < 3. So, substituting y with x + 2, the inequality becomes 1 < x + 2 < 3. If I subtract 2 from all parts of this inequality, I get -1 < x < 1. So, x is between -1 and 1.Now, I need to find the value of A = sqrt(y² - 8x) + sqrt(y² + 2x + 5). Since y = x + 2, I can substitute y in the expression for A. Let me do that:A = sqrt((x + 2)² - 8x) + sqrt((x + 2)² + 2x + 5)Let me expand (x + 2)² first. That's x² + 4x + 4. So, substituting back in:A = sqrt(x² + 4x + 4 - 8x) + sqrt(x² + 4x + 4 + 2x + 5)Simplify the terms inside each square root:For the first square root: x² + 4x + 4 - 8x = x² - 4x + 4For the second square root: x² + 4x + 4 + 2x + 5 = x² + 6x + 9So now, A becomes:A = sqrt(x² - 4x + 4) + sqrt(x² + 6x + 9)Looking at these expressions inside the square roots, I notice that they might be perfect squares. Let me check:x² - 4x + 4 is (x - 2)² because (x - 2)² = x² - 4x + 4.Similarly, x² + 6x + 9 is (x + 3)² because (x + 3)² = x² + 6x + 9.So, A simplifies to:A = sqrt((x - 2)²) + sqrt((x + 3)²)The square root of a square is the absolute value, so:A = |x - 2| + |x + 3|Now, I need to evaluate this expression considering the range of x, which is -1 < x < 1.Let's analyze each absolute value term:1. |x - 2|: Since x is less than 1, x - 2 will always be negative (because 1 - 2 = -1, and x is less than 1). So, |x - 2| = -(x - 2) = 2 - x.2. |x + 3|: Since x is greater than -1, x + 3 will always be positive (because -1 + 3 = 2, and x is greater than -1). So, |x + 3| = x + 3.Putting these together:A = (2 - x) + (x + 3) = 2 - x + x + 3The x terms cancel out: -x + x = 0, so we're left with:A = 2 + 3 = 5So, regardless of the value of x within the given range, A is always 5.Wait, let me double-check to make sure I didn't make any mistakes. I substituted y correctly, expanded the squares properly, recognized the perfect squares, and evaluated the absolute values based on the range of x. Everything seems to add up. Yeah, I think that's correct.**Final Answer**The value of ( A ) is boxed{5}.

💡First, I need to simplify the given complex number ( z = i(-2 - i) ). To do this, I'll distribute the ( i ) across the terms inside the parentheses.Multiplying ( i ) by (-2 ) gives (-2i). Then, multiplying ( i ) by (-i ) results in (-i^2).Next, I recall that ( i^2 = -1 ), so (-i^2 = 1).Combining these results, the simplified form of the complex number is ( z = 1 - 2i ).In the complex plane, the real part of the complex number corresponds to the x-coordinate, and the imaginary part corresponds to the y-coordinate. Therefore, the point representing ( z ) is at coordinates ( (1, -2) ).Looking at the coordinates, the x-value is positive and the y-value is negative. This places the point in the fourth quadrant of the complex plane.Thus, the correct answer is D: The fourth quadrant.