💡Okay, so I need to figure out how many two-digit numbers are "wonderful." A "wonderful" number is defined as having exactly 3 different odd natural divisors, and it can have any number of even divisors. Hmm, interesting. Let me break this down step by step.First, I know that divisors come in pairs, and for a number to have exactly 3 odd divisors, there must be something special about its prime factorization. Since 3 is a prime number, it suggests that the number of odd prime factors is limited in a specific way. Maybe it's related to the exponents in the prime factorization?Let me recall that the total number of divisors of a number is found by adding one to each of the exponents in its prime factorization and then multiplying those together. For example, if a number ( n = p^a times q^b ), then the number of divisors is ( (a+1)(b+1) ). So, if we're only considering odd divisors, we need to look at the odd part of the number, which means ignoring the factor of 2.So, if a number has exactly 3 odd divisors, the number of divisors contributed by the odd part must be 3. Since 3 is a prime number, the only way to get exactly 3 divisors is if the odd part is a square of a prime number. Because ( (2+1) = 3 ). So, the odd part of the number must be ( p^2 ), where ( p ) is an odd prime.Therefore, the number can be expressed as ( n = 2^k times p^2 ), where ( p ) is an odd prime and ( k ) is a non-negative integer. This makes sense because the number of odd divisors is determined by ( p^2 ), which has exactly 3 divisors: 1, ( p ), and ( p^2 ). The factor ( 2^k ) accounts for the even divisors, which can be any number.Now, I need to find all two-digit numbers that fit this form. So, let's consider different values of ( p ) and ( k ) such that ( n ) is between 10 and 99.Let's start with the smallest odd prime, which is 3.1. **Case 1: ( p = 3 )** - ( p^2 = 9 ) - So, ( n = 2^k times 9 ) - Let's find values of ( k ) such that ( n ) is a two-digit number. - ( k = 0 ): ( n = 1 times 9 = 9 ) (Not a two-digit number) - ( k = 1 ): ( n = 2 times 9 = 18 ) (Valid) - ( k = 2 ): ( n = 4 times 9 = 36 ) (Valid) - ( k = 3 ): ( n = 8 times 9 = 72 ) (Valid) - ( k = 4 ): ( n = 16 times 9 = 144 ) (Too big, exceeds two digits) - So, for ( p = 3 ), we have 18, 36, and 72.2. **Case 2: ( p = 5 )** - ( p^2 = 25 ) - So, ( n = 2^k times 25 ) - Let's find values of ( k ): - ( k = 0 ): ( n = 1 times 25 = 25 ) (Valid) - ( k = 1 ): ( n = 2 times 25 = 50 ) (Valid) - ( k = 2 ): ( n = 4 times 25 = 100 ) (Too big) - So, for ( p = 5 ), we have 25 and 50.3. **Case 3: ( p = 7 )** - ( p^2 = 49 ) - So, ( n = 2^k times 49 ) - Let's find values of ( k ): - ( k = 0 ): ( n = 1 times 49 = 49 ) (Valid) - ( k = 1 ): ( n = 2 times 49 = 98 ) (Valid) - ( k = 2 ): ( n = 4 times 49 = 196 ) (Too big) - So, for ( p = 7 ), we have 49 and 98.4. **Case 4: ( p = 11 )** - ( p^2 = 121 ) - So, ( n = 2^k times 121 ) - Even ( k = 0 ): ( n = 121 ) (Too big) - So, no valid two-digit numbers here.5. **Case 5: ( p = 13 )** - ( p^2 = 169 ) - Similarly, ( n = 2^k times 169 ) is already too big for two digits. - So, no valid numbers here either.I think primes larger than 7 will result in ( p^2 ) being greater than 100, which is beyond two digits, so we can stop here.Now, compiling all the valid numbers we found:- From ( p = 3 ): 18, 36, 72- From ( p = 5 ): 25, 50- From ( p = 7 ): 49, 98So, that's a total of 7 numbers: 18, 25, 36, 49, 50, 72, 98.Let me double-check to ensure none of these numbers have more than 3 odd divisors.- **18**: Divisors are 1, 2, 3, 6, 9, 18. Odd divisors: 1, 3, 9. Exactly 3.- **25**: Divisors are 1, 5, 25. Odd divisors: 1, 5, 25. Exactly 3.- **36**: Divisors are 1, 2, 3, 4, 6, 9, 12, 18, 36. Odd divisors: 1, 3, 9. Exactly 3.- **49**: Divisors are 1, 7, 49. Odd divisors: 1, 7, 49. Exactly 3.- **50**: Divisors are 1, 2, 5, 10, 25, 50. Odd divisors: 1, 5, 25. Exactly 3.- **72**: Divisors are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Odd divisors: 1, 3, 9. Exactly 3.- **98**: Divisors are 1, 2, 7, 14, 49, 98. Odd divisors: 1, 7, 49. Exactly 3.All of them satisfy the condition of having exactly 3 odd divisors. Therefore, there are 7 such numbers.**Final Answer**The number of "wonderful" two-digit numbers is boxed{7}.

💡Alright, so I have this problem here about two functions, f(x) and g(x), both defined for all real numbers. It says that g(x) is an odd function, and we have two equations: f(x) + g(x) = 2 and f(x) + g(x - 2) = 2. Then, I need to figure out which of the options A, B, C, or D are correct.First, let me recall what an odd function is. An odd function satisfies the property g(-x) = -g(x) for all x in its domain. This means that if I plug in -x into the function, I get the negative of the original function value. A classic example is g(x) = x, because g(-x) = -x = -g(x).Now, looking at the given equations:1. f(x) + g(x) = 22. f(x) + g(x - 2) = 2Hmm, both of these equal 2. So, if I subtract the second equation from the first, I get:[f(x) + g(x)] - [f(x) + g(x - 2)] = 2 - 2Simplifying this:f(x) + g(x) - f(x) - g(x - 2) = 0So, the f(x) terms cancel out, leaving:g(x) - g(x - 2) = 0Which means:g(x) = g(x - 2)Interesting. So, g(x) is equal to g(x - 2). That suggests that g is periodic with period 2. In other words, shifting x by 2 doesn't change the value of g(x). So, g is a periodic function with period 2.But we also know that g is an odd function. So, combining these two properties: g is odd and periodic with period 2.Let me think about what that implies. If g is odd, then g(-x) = -g(x). And since it's periodic with period 2, g(x + 2) = g(x). So, combining these, g(-x) = -g(x) and g(x + 2) = g(x).Let me see if I can find specific values or properties of g(x) from this.First, let's consider x = 0. Plugging into the first equation: f(0) + g(0) = 2.But since g is odd, g(0) = 0. Because for odd functions, g(-0) = -g(0), but g(-0) is just g(0), so g(0) = -g(0), which implies g(0) = 0.Therefore, f(0) + 0 = 2, so f(0) = 2.Looking at option A: f(0) = 0. But we just found f(0) = 2, so A is incorrect.Next, let's look at option B: g(1) = 0.Hmm, let's see. Since g is periodic with period 2, g(1) = g(1 - 2) = g(-1). But g is odd, so g(-1) = -g(1). Therefore, g(1) = -g(1), which implies that 2g(1) = 0, so g(1) = 0.So, option B is correct.Moving on to option C: the sum from i=1 to n of f(i) = 0.Hmm, let's see. We know that f(x) + g(x) = 2, so f(x) = 2 - g(x). Therefore, f(i) = 2 - g(i). So, the sum from i=1 to n of f(i) is equal to the sum from i=1 to n of (2 - g(i)).Which is equal to 2n - sum from i=1 to n of g(i).Now, if we can find the sum of g(i) from i=1 to n, we can see if this equals 2n.But wait, from option B, we know that g(1) = 0. What about other integer points?Given that g is periodic with period 2, g(1) = g(1 - 2) = g(-1) = -g(1) = 0, as we saw earlier.Similarly, g(2) = g(0) = 0, since g is periodic with period 2, and g(0) = 0.g(3) = g(1) = 0, and so on. So, for any integer i, g(i) = 0.Therefore, the sum from i=1 to n of g(i) is 0.Therefore, the sum from i=1 to n of f(i) = 2n - 0 = 2n.So, unless n = 0, this sum is not zero. Since n is a positive integer (as it's the upper limit of the summation starting from 1), the sum is 2n, which is not zero unless n = 0, which it isn't.Therefore, option C is incorrect.Finally, option D: the sum from i=1 to n of g(i) = 0.From the above reasoning, since g(i) = 0 for all integers i, the sum is indeed 0.Therefore, option D is correct.So, summarizing:A: IncorrectB: CorrectC: IncorrectD: CorrectThus, the correct options are B and D.**Final Answer**The correct choices are boxed{BD}.

💡Okay, so I've got this system of equations to solve:[begin{aligned}& (x + 2y)(x + 3y) = x + y & (2x + y)(3x + y) = -99(x + y)end{aligned}]Hmm, let's see. I need to find all pairs (x, y) that satisfy both equations. First, I notice that both equations are quadratic in nature because when I expand them, they'll have terms with (x^2) and (y^2). That might make things a bit tricky, but maybe there's a way to simplify or factor them.Let me start by expanding the first equation:[(x + 2y)(x + 3y) = x + y]Expanding the left side:[x cdot x + x cdot 3y + 2y cdot x + 2y cdot 3y = x + y]Simplifying:[x^2 + 3xy + 2xy + 6y^2 = x + y]Combine like terms:[x^2 + 5xy + 6y^2 = x + y]Okay, so that's the first equation simplified. Now, let's do the same for the second equation:[(2x + y)(3x + y) = -99(x + y)]Expanding the left side:[2x cdot 3x + 2x cdot y + y cdot 3x + y cdot y = -99(x + y)]Simplifying:[6x^2 + 2xy + 3xy + y^2 = -99x - 99y]Combine like terms:[6x^2 + 5xy + y^2 = -99x - 99y]Now, let's write both equations together:1. (x^2 + 5xy + 6y^2 = x + y)2. (6x^2 + 5xy + y^2 = -99x - 99y)Hmm, both equations have similar terms, especially the (5xy) term. Maybe I can subtract one equation from the other to eliminate some variables or simplify.Let me subtract equation 1 from equation 2:[(6x^2 + 5xy + y^2) - (x^2 + 5xy + 6y^2) = (-99x - 99y) - (x + y)]Simplifying the left side:[6x^2 - x^2 + 5xy - 5xy + y^2 - 6y^2 = -99x - x - 99y - y]Which simplifies to:[5x^2 - 5y^2 = -100x - 100y]I can factor out a 5 from the left side and a -100 from the right side:[5(x^2 - y^2) = -100(x + y)]Divide both sides by 5:[x^2 - y^2 = -20(x + y)]Hmm, (x^2 - y^2) is a difference of squares, so I can factor that:[(x - y)(x + y) = -20(x + y)]Now, if I move all terms to one side:[(x - y)(x + y) + 20(x + y) = 0]Factor out (x + y):[(x + y)(x - y + 20) = 0]So, this gives me two possibilities:1. (x + y = 0)2. (x - y + 20 = 0)Let me consider each case separately.**Case 1: (x + y = 0)**If (x + y = 0), then (y = -x). Let's substitute this into the first equation:[x^2 + 5x(-x) + 6(-x)^2 = x + (-x)]Simplify:[x^2 - 5x^2 + 6x^2 = 0]Combine like terms:[(1 - 5 + 6)x^2 = 0 implies 2x^2 = 0 implies x^2 = 0 implies x = 0]So, if (x = 0), then (y = -x = 0). Therefore, one solution is ((0, 0)).Let me check this solution in the second equation to make sure:[(2(0) + 0)(3(0) + 0) = -99(0 + 0) implies 0 = 0]Yes, it works.**Case 2: (x - y + 20 = 0)**This implies (x = y - 20). Let's substitute (x = y - 20) into the first equation:[(y - 20)^2 + 5(y - 20)y + 6y^2 = (y - 20) + y]First, expand each term:1. ((y - 20)^2 = y^2 - 40y + 400)2. (5(y - 20)y = 5y^2 - 100y)3. (6y^2) remains as is.So, substituting back:[(y^2 - 40y + 400) + (5y^2 - 100y) + 6y^2 = (y - 20) + y]Combine like terms on the left side:[y^2 + 5y^2 + 6y^2 - 40y - 100y + 400 = 2y - 20]Simplify:[12y^2 - 140y + 400 = 2y - 20]Bring all terms to one side:[12y^2 - 140y + 400 - 2y + 20 = 0]Combine like terms:[12y^2 - 142y + 420 = 0]Hmm, this is a quadratic equation in terms of y. Let me try to simplify it. Maybe divide all terms by 2 to make it easier:[6y^2 - 71y + 210 = 0]Now, let's try to factor this quadratic. I'm looking for two numbers that multiply to (6 times 210 = 1260) and add up to -71.Hmm, 1260 is a big number. Let me see:- 35 and 36: 35*36=1260, but 35+36=71. Since the middle term is -71y, these would work if both are negative.So, I can write:[6y^2 - 35y - 36y + 210 = 0]Group the terms:[(6y^2 - 35y) - (36y - 210) = 0]Factor out common terms:[y(6y - 35) - 6(6y - 35) = 0]Now, factor out (6y - 35):[(y - 6)(6y - 35) = 0]So, the solutions are:1. (y - 6 = 0 implies y = 6)2. (6y - 35 = 0 implies y = frac{35}{6})Now, let's find the corresponding x values using (x = y - 20):1. If (y = 6), then (x = 6 - 20 = -14)2. If (y = frac{35}{6}), then (x = frac{35}{6} - 20 = frac{35}{6} - frac{120}{6} = -frac{85}{6})So, the solutions from this case are ((-14, 6)) and (left(-frac{85}{6}, frac{35}{6}right)).Let me check these solutions in the second equation to make sure.**Checking ((-14, 6)):**Second equation:[(2(-14) + 6)(3(-14) + 6) = -99((-14) + 6)]Simplify:[(-28 + 6)(-42 + 6) = -99(-8)][(-22)(-36) = 792][792 = 792]Yes, it works.**Checking (left(-frac{85}{6}, frac{35}{6}right)):**Second equation:[left(2left(-frac{85}{6}right) + frac{35}{6}right)left(3left(-frac{85}{6}right) + frac{35}{6}right) = -99left(-frac{85}{6} + frac{35}{6}right)]Simplify each part:First part:[2left(-frac{85}{6}right) = -frac{170}{6} = -frac{85}{3}][-frac{85}{3} + frac{35}{6} = -frac{170}{6} + frac{35}{6} = -frac{135}{6} = -frac{45}{2}]Second part:[3left(-frac{85}{6}right) = -frac{255}{6} = -frac{85}{2}][-frac{85}{2} + frac{35}{6} = -frac{255}{6} + frac{35}{6} = -frac{220}{6} = -frac{110}{3}]Left side:[left(-frac{45}{2}right)left(-frac{110}{3}right) = frac{45 times 110}{6} = frac{4950}{6} = 825]Right side:[-99left(-frac{85}{6} + frac{35}{6}right) = -99left(-frac{50}{6}right) = -99left(-frac{25}{3}right) = 99 times frac{25}{3} = 33 times 25 = 825]Yes, it works.So, all solutions are:1. ((0, 0))2. ((-14, 6))3. (left(-frac{85}{6}, frac{35}{6}right))I think that's all the solutions. I don't think there are any other cases because when I factored (x + y) and (x - y + 20), I covered all possibilities. So, these should be the only solutions.**Final Answer**The solutions to the system are (boxed{(0, 0)}), (boxed{(-14, 6)}), and (boxed{left(-frac{85}{6}, frac{35}{6}right)}).

💡Okay, so I need to write down a quadratic function that opens upward and has its vertex at the point (2,1). Hmm, quadratic functions... I remember they have the general form of y = ax² + bx + c, but there's also something called vertex form, right? Vertex form is probably easier to use here because I know the vertex coordinates.Let me recall, the vertex form of a quadratic function is y = a(x - h)² + k, where (h,k) is the vertex. So, in this case, h is 2 and k is 1. That means I can plug those values into the vertex form equation. So it would look like y = a(x - 2)² + 1. Now, I need to determine the value of 'a'. The problem says the parabola opens upward, which means the coefficient 'a' has to be positive. If 'a' were negative, the parabola would open downward. So, I can choose any positive number for 'a'. The simplest choice is probably 1 because it makes the equation as straightforward as possible. If I set a = 1, then the equation becomes y = (x - 2)² + 1. Let me check if this makes sense. The vertex is at (2,1), which matches what the problem asked for. Also, since 'a' is positive, it opens upward, which is exactly what we need. Wait, could there be other possible values for 'a'? I mean, technically, any positive number would work, right? For example, if I chose a = 2, the equation would be y = 2(x - 2)² + 1, and it would still open upward with the same vertex. But since the problem doesn't specify any other conditions, like passing through a certain point, I think the simplest form with a = 1 is acceptable. Just to make sure, let me expand this equation to standard form and see if it still looks like a quadratic function. Starting with y = (x - 2)² + 1, I can expand (x - 2)² to x² - 4x + 4. Adding 1 gives y = x² - 4x + 5. Yep, that's a quadratic function in standard form, and the coefficient of x² is positive, so it definitely opens upward. I think I'm confident with this answer. It meets all the requirements: it's a quadratic function, it opens upward, and its vertex is at (2,1). Plus, it's in the simplest form possible, which is usually what is expected unless specified otherwise.

💡Okay, so I have this problem where I'm given the inverse of a matrix A, and I need to find the eigenvalues of A itself. Hmm, let me think about how to approach this. First, I remember that if I have a matrix A and its inverse A⁻¹, then A * A⁻¹ = I, where I is the identity matrix. So, if I can find matrix A from its inverse, I can then find its eigenvalues. But how do I get A from A⁻¹? Well, I think it's just taking the inverse of A⁻¹, right? So, A should be (A⁻¹)⁻¹. That makes sense because inverting the inverse should give me back the original matrix.Alright, so let me write down the given inverse matrix:A⁻¹ = [ -1/4 3/4 ] [ 1/2 -1/2 ]I need to find A, which is (A⁻¹)⁻¹. To find the inverse of a 2x2 matrix, I recall the formula:If B = [ a b ] [ c d ]then B⁻¹ = (1/det(B)) * [ d -b ] [ -c a ]where det(B) is the determinant of B, calculated as ad - bc.So, let me compute the determinant of A⁻¹ first. The determinant is:det(A⁻¹) = (-1/4)(-1/2) - (3/4)(1/2) = (1/8) - (3/8) = (-2/8) = -1/4.Okay, so the determinant of A⁻¹ is -1/4. Therefore, the determinant of A, which is (A⁻¹)⁻¹, should be the reciprocal of det(A⁻¹). Because det(A⁻¹) = 1/det(A), so det(A) = 1 / det(A⁻¹) = 1 / (-1/4) = -4.Wait, hold on, is that correct? Let me double-check. If det(A⁻¹) = 1/det(A), then det(A) = 1 / det(A⁻¹). Since det(A⁻¹) is -1/4, then det(A) is indeed -4. Okay, that seems right.Now, using the inverse formula, let me compute A = (A⁻¹)⁻¹.Given A⁻¹ is:[ -1/4 3/4 ][ 1/2 -1/2 ]So, applying the inverse formula:A = (1 / det(A⁻¹)) * [ -1/2 -3/4 ] [ -1/2 1/4 ]Wait, let me make sure. The formula is:B⁻¹ = (1/det(B)) * [ d -b ] [ -c a ]So, for A⁻¹, which is B, we have:a = -1/4, b = 3/4, c = 1/2, d = -1/2.So, B⁻¹ = (1/det(B)) * [ d -b ] [ -c a ]Plugging in the values:B⁻¹ = (1 / (-1/4)) * [ -1/2 -3/4 ] [ -1/2 1/4 ]Simplify the scalar multiplication:1 / (-1/4) is -4, so:B⁻¹ = -4 * [ -1/2 -3/4 ] [ -1/2 1/4 ]Now, multiply each element by -4:First row: -4 * (-1/2) = 2, and -4 * (-3/4) = 3.Second row: -4 * (-1/2) = 2, and -4 * (1/4) = -1.So, putting it all together:A = [ 2 3 ] [ 2 -1 ]Okay, so now I have matrix A:A = [ 2 3 ] [ 2 -1 ]Now, to find the eigenvalues of A, I need to solve the characteristic equation, which is det(A - λI) = 0.Let me write down A - λI:A - λI = [ 2 - λ 3 ] [ 2 -1 - λ ]The determinant of this matrix is:(2 - λ)(-1 - λ) - (3)(2) = ?Let me compute that step by step.First, expand (2 - λ)(-1 - λ):= 2*(-1) + 2*(-λ) - λ*(-1) - λ*λ= -2 - 2λ + λ + λ²= λ² - λ - 2Then, subtract 3*2 = 6:So, determinant = (λ² - λ - 2) - 6 = λ² - λ - 8.Wait, hold on, that doesn't seem right. Let me check my multiplication again.Wait, no, actually, the determinant is (2 - λ)(-1 - λ) - (3)(2). So, I computed (2 - λ)(-1 - λ) as λ² - λ - 2, which is correct. Then subtract 6, so λ² - λ - 2 - 6 = λ² - λ - 8.Wait, but earlier, when I thought about the determinant, I thought det(A) was -4, but if the characteristic polynomial is λ² - λ - 8, then the determinant is the constant term, which is -8, but earlier I thought det(A) was -4. Hmm, that's a discrepancy. Did I make a mistake somewhere?Let me go back. The determinant of A⁻¹ was -1/4, so det(A) should be 1 / (-1/4) = -4. But according to the characteristic polynomial, det(A) is -8. That's a problem.Wait, maybe I made a mistake in computing the determinant of A - λI.Let me recalculate:A - λI = [ 2 - λ 3 ] [ 2 -1 - λ ]Determinant is (2 - λ)(-1 - λ) - (3)(2).Compute (2 - λ)(-1 - λ):= 2*(-1) + 2*(-λ) - λ*(-1) - λ*λ= -2 - 2λ + λ + λ²= λ² - λ - 2Then, subtract 6:= λ² - λ - 2 - 6= λ² - λ - 8So, determinant is λ² - λ - 8. Therefore, det(A) is -8, but earlier I thought det(A) was -4. That's conflicting.Wait, so where did I go wrong? Let me check the determinant of A⁻¹ again.A⁻¹ = [ -1/4 3/4 ] [ 1/2 -1/2 ]det(A⁻¹) = (-1/4)(-1/2) - (3/4)(1/2) = (1/8) - (3/8) = (-2/8) = -1/4. That's correct.So, det(A) = 1 / det(A⁻¹) = 1 / (-1/4) = -4. But according to the characteristic polynomial, det(A) is -8. That's a problem.Wait, maybe I made a mistake in computing A. Let me double-check the inverse computation.Given A⁻¹ = [ -1/4 3/4 ] [ 1/2 -1/2 ]Compute A = (A⁻¹)⁻¹.Using the inverse formula:det(A⁻¹) = -1/4So, A = (1 / det(A⁻¹)) * [ d -b ] [ -c a ]Where a = -1/4, b = 3/4, c = 1/2, d = -1/2.So,A = (1 / (-1/4)) * [ -1/2 -3/4 ] [ -1/2 1/4 ]Which is:-4 * [ -1/2 -3/4 ] [ -1/2 1/4 ]Compute each element:First row:-4 * (-1/2) = 2-4 * (-3/4) = 3Second row:-4 * (-1/2) = 2-4 * (1/4) = -1So, A = [ 2 3 ] [ 2 -1 ]That seems correct. So, determinant of A is (2)(-1) - (3)(2) = -2 -6 = -8. Which matches the characteristic polynomial. But earlier, I thought det(A) was -4 because det(A⁻¹) is -1/4, and det(A) = 1 / det(A⁻¹) = -4. But that's conflicting.Wait, hold on, actually, det(A⁻¹) is 1 / det(A), so det(A) = 1 / det(A⁻¹). So, if det(A⁻¹) is -1/4, then det(A) is 1 / (-1/4) = -4. But according to the matrix A I computed, det(A) is -8. That's a contradiction. So, I must have made a mistake somewhere.Wait, maybe I made a mistake in computing A. Let me check again.A⁻¹ = [ -1/4 3/4 ] [ 1/2 -1/2 ]Compute A = (A⁻¹)⁻¹.Using the inverse formula:det(A⁻¹) = (-1/4)(-1/2) - (3/4)(1/2) = (1/8) - (3/8) = (-2/8) = -1/4.So, A = (1 / (-1/4)) * [ -1/2 -3/4 ] [ -1/2 1/4 ]Which is:-4 * [ -1/2 -3/4 ] [ -1/2 1/4 ]Compute each element:First row:-4 * (-1/2) = 2-4 * (-3/4) = 3Second row:-4 * (-1/2) = 2-4 * (1/4) = -1So, A = [ 2 3 ] [ 2 -1 ]Which gives det(A) = (2)(-1) - (3)(2) = -2 -6 = -8.But according to the determinant relationship, det(A) should be -4. So, there's a mistake here.Wait, maybe I confused the formula for the inverse. Let me double-check the inverse formula.For a 2x2 matrix B = [ a b ] [ c d ]The inverse is (1/det(B)) * [ d -b ] [ -c a ]So, for A⁻¹, which is B, we have a = -1/4, b = 3/4, c = 1/2, d = -1/2.So, B⁻¹ = (1/det(B)) * [ d -b ] [ -c a ]Which is (1 / (-1/4)) * [ -1/2 -3/4 ] [ -1/2 1/4 ]Which is -4 * [ -1/2 -3/4 ] [ -1/2 1/4 ]Which gives:First row: (-4)*(-1/2) = 2, (-4)*(-3/4) = 3Second row: (-4)*(-1/2) = 2, (-4)*(1/4) = -1So, A = [ 2 3 ] [ 2 -1 ]But then det(A) is -8, which contradicts det(A) = -4.Wait, maybe the determinant relationship is different? Let me recall: det(A⁻¹) = 1 / det(A). So, if det(A⁻¹) is -1/4, then det(A) is -4. But according to the matrix A I computed, det(A) is -8. That's a problem.Wait, perhaps I made a mistake in computing the inverse. Let me try another method to compute A from A⁻¹.Alternatively, since A * A⁻¹ = I, maybe I can set up equations to solve for A.Let me denote A as:A = [ a b ] [ c d ]Then, A * A⁻¹ = I.So,[ a b ] [ -1/4 3/4 ] = [ 1 0 ][ c d ] [ 1/2 -1/2 ] [ 0 1 ]Multiplying the matrices:First row of A times first column of A⁻¹: a*(-1/4) + b*(1/2) = 1First row of A times second column of A⁻¹: a*(3/4) + b*(-1/2) = 0Second row of A times first column of A⁻¹: c*(-1/4) + d*(1/2) = 0Second row of A times second column of A⁻¹: c*(3/4) + d*(-1/2) = 1So, we have four equations:1. (-1/4)a + (1/2)b = 12. (3/4)a - (1/2)b = 03. (-1/4)c + (1/2)d = 04. (3/4)c - (1/2)d = 1Let me solve equations 1 and 2 first to find a and b.From equation 2: (3/4)a - (1/2)b = 0Let me multiply both sides by 4 to eliminate denominators:3a - 2b = 0 => 3a = 2b => b = (3/2)aNow, plug b = (3/2)a into equation 1:(-1/4)a + (1/2)*(3/2)a = 1Simplify:(-1/4)a + (3/4)a = 1Combine like terms:( (-1 + 3)/4 )a = 1 => (2/4)a = 1 => (1/2)a = 1 => a = 2Then, b = (3/2)*2 = 3So, a = 2, b = 3Now, let's solve equations 3 and 4 for c and d.From equation 3: (-1/4)c + (1/2)d = 0Multiply both sides by 4:- c + 2d = 0 => -c + 2d = 0 => c = 2dFrom equation 4: (3/4)c - (1/2)d = 1Substitute c = 2d into equation 4:(3/4)*(2d) - (1/2)d = 1Simplify:(6/4)d - (2/4)d = 1 => (4/4)d = 1 => d = 1Then, c = 2d = 2*1 = 2So, c = 2, d = 1Therefore, matrix A is:A = [ 2 3 ] [ 2 1 ]Wait, but earlier I thought A was [2 3; 2 -1]. That was a mistake. So, actually, A is [2 3; 2 1].So, let me correct that. A is:A = [ 2 3 ] [ 2 1 ]Now, let's compute det(A):det(A) = (2)(1) - (3)(2) = 2 - 6 = -4Which matches the determinant relationship det(A) = -4. Good, that's consistent.Now, let's compute the characteristic polynomial of A.A - λI = [ 2 - λ 3 ] [ 2 1 - λ ]Determinant is:(2 - λ)(1 - λ) - (3)(2) = ?Compute (2 - λ)(1 - λ):= 2*1 + 2*(-λ) - λ*1 + λ²= 2 - 2λ - λ + λ²= λ² - 3λ + 2Subtract 6:= λ² - 3λ + 2 - 6= λ² - 3λ - 4So, the characteristic equation is λ² - 3λ - 4 = 0Now, solve for λ:λ = [3 ± sqrt(9 + 16)] / 2 = [3 ± sqrt(25)] / 2 = [3 ± 5]/2So, λ₁ = (3 + 5)/2 = 8/2 = 4λ₂ = (3 - 5)/2 = (-2)/2 = -1Therefore, the eigenvalues of A are 4 and -1.Wait, but earlier when I computed A incorrectly as [2 3; 2 -1], I got eigenvalues 4 and -1 as well. But that was a mistake because the determinant was wrong. However, after correctly computing A as [2 3; 2 1], the determinant is -4, and the characteristic polynomial is λ² - 3λ -4, leading to eigenvalues 4 and -1. So, the eigenvalues are the same despite the initial mistake in computing A.But wait, that's interesting. Even though I initially made a mistake in computing A, the eigenvalues turned out the same. Is that a coincidence? Or is there a deeper reason?Well, eigenvalues are properties of the matrix, so if I correctly compute A, the eigenvalues should be consistent. But in this case, even with an incorrect A, I got the same eigenvalues. That might be a coincidence or perhaps because the determinant and trace are preserved in some way.But to be thorough, let me verify the eigenvalues using another method. Since I have A⁻¹, maybe I can find the eigenvalues of A⁻¹ and then take their reciprocals to get the eigenvalues of A.I remember that if λ is an eigenvalue of A, then 1/λ is an eigenvalue of A⁻¹, provided that λ ≠ 0.So, let me find the eigenvalues of A⁻¹ first.Given A⁻¹ = [ -1/4 3/4 ] [ 1/2 -1/2 ]Compute its characteristic polynomial.A⁻¹ - μI = [ -1/4 - μ 3/4 ] [ 1/2 -1/2 - μ ]Determinant:(-1/4 - μ)(-1/2 - μ) - (3/4)(1/2) = ?Compute (-1/4 - μ)(-1/2 - μ):= (-1/4)(-1/2) + (-1/4)(-μ) + (-μ)(-1/2) + (-μ)(-μ)= 1/8 + (1/4)μ + (1/2)μ + μ²= μ² + (3/4)μ + 1/8Subtract (3/4)(1/2) = 3/8:Determinant = μ² + (3/4)μ + 1/8 - 3/8 = μ² + (3/4)μ - 2/8 = μ² + (3/4)μ - 1/4So, the characteristic equation is μ² + (3/4)μ - 1/4 = 0Multiply both sides by 4 to eliminate fractions:4μ² + 3μ - 1 = 0Now, solve for μ:μ = [-3 ± sqrt(9 + 16)] / 8 = [-3 ± sqrt(25)] / 8 = [-3 ± 5]/8So, μ₁ = (-3 + 5)/8 = 2/8 = 1/4μ₂ = (-3 - 5)/8 = (-8)/8 = -1Therefore, the eigenvalues of A⁻¹ are 1/4 and -1.Since eigenvalues of A are reciprocals of eigenvalues of A⁻¹, then eigenvalues of A are 4 and -1.Which matches the earlier result. So, despite the initial confusion, the eigenvalues are indeed 4 and -1.So, to summarize, the eigenvalues of matrix A are 4 and -1.

💡Alright, so I'm trying to figure out this tennis tournament problem. Let me read it again carefully.We have a tournament with more than 4 athletes. Each day, every player plays exactly one game. By the end, each player has played against every other player exactly once. So, it's a round-robin tournament where everyone faces everyone else once.Now, they define a "persistent" player as someone who has won at least one match and, after their first victory, never lost again. The others are called "non-persistent." The question is asking whether there were more game days when non-persistent players played against each other than half of the total game days.Hmm, okay. So, first, I need to understand what a persistent player is. It's someone who has at least one win, and after their first win, they don't lose any more matches. So, they might have some losses before their first win, but after that, they keep winning.Non-persistent players are everyone else, meaning either they never won a match or they lost at least one match after their first win.The tournament has more than 4 players, so let's say n players where n > 4.Each day, every player plays exactly one game, so each day there are n/2 matches (since each match involves two players). The total number of matches in the tournament is C(n,2) = n(n-1)/2. Therefore, the total number of days is C(n,2) divided by n/2, which simplifies to (n-1) days.So, total game days are n-1.Now, the question is whether the number of days when non-persistent players played against each other is more than half of the total game days. Half of the total game days would be (n-1)/2.So, we need to determine if the number of days with non-persistent vs. non-persistent matches is greater than (n-1)/2.Let me think about how to approach this. Maybe I can model the tournament as a directed graph where each edge represents a match, and the direction indicates who won.Persistent players have a certain structure in this graph. After their first win, they never lose again. So, their wins form a consecutive sequence from their first win onwards.Non-persistent players either never won or lost after their first win.I need to find the number of days where both players in a match are non-persistent.Let me consider the structure of the tournament.First, let's denote the number of persistent players as k. Then, the number of non-persistent players is n - k.Each persistent player has at least one win, and after their first win, they never lose. So, their wins are in a consecutive sequence.Now, in the tournament, each player plays every other player exactly once. So, for persistent players, their matches against other persistent players must all be wins after their first win.Wait, but if two persistent players play each other, one of them must win. So, if both are persistent, their match must have been won by one of them after their first win.But this might complicate things. Maybe I should think about the order of the matches.Since each day, every player plays exactly one game, the tournament can be thought of as a sequence of rounds, each consisting of n/2 matches.Now, the key is to figure out how the persistent and non-persistent players are distributed across these days.Persistent players, after their first win, never lose. So, once they start winning, they continue to win all their subsequent matches.Non-persistent players either never win or lose at least once after their first win.I think the crux is to realize that the persistent players must have their wins concentrated in the latter part of the tournament.But how does that affect the number of days when non-persistent players play against each other?Maybe I can think about the matches between non-persistent players. These are matches where both players are non-persistent, meaning either they never won or lost after their first win.If I can show that these matches happen more than half the time, then the answer is true.Alternatively, maybe I can use an averaging argument or some combinatorial counting.Let me try to count the number of matches between non-persistent players.The total number of matches is C(n,2). The number of matches between persistent players is C(k,2). The number of matches between persistent and non-persistent players is k(n - k).So, the number of matches between non-persistent players is C(n - k, 2).Now, each match occurs on a specific day. So, the number of days when non-persistent players play against each other is equal to the number of such matches divided by the number of matches per day, which is n/2.Wait, no. Actually, each day has n/2 matches, but the number of days when a particular match occurs is just 1. So, the number of days when non-persistent players play against each other is equal to the number of such matches, which is C(n - k, 2).But that doesn't make sense because the total number of days is n - 1, and C(n - k, 2) could be larger than n - 1.Wait, no. Actually, each match occurs on exactly one day, so the number of days when non-persistent players play against each other is equal to the number of matches between non-persistent players, which is C(n - k, 2).But the total number of days is n - 1, so we need to compare C(n - k, 2) with (n - 1)/2.But this seems off because C(n - k, 2) is the number of matches, not the number of days. Each day has multiple matches.Wait, I think I'm confusing matches and days. Each day has n/2 matches, so the total number of days is C(n,2) / (n/2) = (n - 1).So, the number of days when non-persistent players play against each other is equal to the number of matches between non-persistent players divided by the number of matches per day, but that's not quite right.Actually, each match is on a specific day, so the number of days when non-persistent players play against each other is equal to the number of matches between non-persistent players, which is C(n - k, 2). But since each day has n/2 matches, the number of days is C(n - k, 2) / (n/2).Wait, no. That's not correct either. Because multiple matches can happen on the same day, but the number of days when at least one non-persistent vs. non-persistent match occurs is what we're interested in.But the question is about the number of days when non-persistent players played against each other, meaning days where at least one such match occurred. But actually, the question is more precise: it's about the number of days when non-persistent players played against each other, which I think means the number of days where a match between two non-persistent players occurred.But each such match occurs on exactly one day, so the number of such days is equal to the number of matches between non-persistent players, which is C(n - k, 2).But the total number of days is n - 1, so we need to compare C(n - k, 2) with (n - 1)/2.But C(n - k, 2) = (n - k)(n - k - 1)/2.We need to check if (n - k)(n - k - 1)/2 > (n - 1)/2.Simplifying, we get (n - k)(n - k - 1) > n - 1.But this depends on the value of k.Wait, but we don't know k. So, maybe we need to find the minimum value of C(n - k, 2) over possible k.Alternatively, perhaps we can find a lower bound for the number of non-persistent vs. non-persistent matches.Wait, another approach: consider that each persistent player must have at least one win, and after their first win, they never lose. So, their wins are in a consecutive sequence.This implies that the persistent players must have their wins concentrated in the latter part of the tournament.Therefore, the matches between persistent players must have occurred after their first wins, meaning that these matches are in the latter part of the tournament.But how does that affect the number of days when non-persistent players play against each other?Alternatively, maybe we can think about the fact that non-persistent players must have lost at least one match after their first win, which means that their losses are spread out.But I'm not sure.Wait, maybe I can use the fact that the number of persistent players is limited.Since each persistent player must have at least one win, and after their first win, they never lose, the number of persistent players cannot exceed n/2.Because if more than n/2 players were persistent, then on the last day, each persistent player would need to play against a non-persistent player, but there aren't enough non-persistent players.Wait, let's think about that.On the last day, each persistent player must have won their match, because after their first win, they never lose. So, if there are k persistent players, they must have k wins on the last day. But each match has one winner and one loser, so the number of wins on the last day is n/2 (since there are n/2 matches).Therefore, k <= n/2.So, the maximum number of persistent players is n/2.Therefore, the number of non-persistent players is at least n - n/2 = n/2.So, there are at least n/2 non-persistent players.Now, the number of matches between non-persistent players is C(n - k, 2) >= C(n/2, 2).So, C(n/2, 2) = (n/2)(n/2 - 1)/2 = n(n - 2)/8.We need to compare this with (n - 1)/2.So, is n(n - 2)/8 > (n - 1)/2?Let's solve the inequality:n(n - 2)/8 > (n - 1)/2Multiply both sides by 8:n(n - 2) > 4(n - 1)Expand:n^2 - 2n > 4n - 4Bring all terms to left:n^2 - 6n + 4 > 0Solve the quadratic inequality:n^2 - 6n + 4 > 0The roots are n = [6 ± sqrt(36 - 16)]/2 = [6 ± sqrt(20)]/2 = [6 ± 2sqrt(5)]/2 = 3 ± sqrt(5).Approximately, sqrt(5) is about 2.236, so the roots are approximately 3 + 2.236 = 5.236 and 3 - 2.236 = 0.764.Since n > 4, we are considering n >= 5.For n >= 6, n^2 - 6n + 4 is positive because n=6: 36 - 36 + 4 = 4 > 0.For n=5: 25 - 30 + 4 = -1 < 0.So, for n >=6, n(n - 2)/8 > (n - 1)/2.But for n=5, it's not.But the problem states more than 4 athletes, so n >=5.But for n=5, the inequality doesn't hold. So, does the statement hold for n=5?Wait, let's check for n=5.n=5, total days = 4.Number of non-persistent players is at least 5 - 2 = 3 (since k <=2).Number of matches between non-persistent players is C(3,2)=3.But total days=4.So, 3 matches, but each day has 2 matches.Wait, no, each day has 5/2=2.5 matches, which is not possible. Wait, n must be even?Wait, the problem says more than 4 athletes, but doesn't specify even or odd.Wait, in a tournament with n players, each day, every player plays exactly one game, so n must be even because you can't have an odd number of players each playing one game.Wait, actually, no. If n is odd, one player would have to sit out each day, but the problem says "each day, every tennis player participated in exactly one game." So, n must be even.Therefore, n is even and greater than 4, so n >=6.So, for n=6, total days=5.Number of non-persistent players is at least 3.Number of matches between non-persistent players is C(3,2)=3.But total days=5.So, 3 matches, but each day has 3 matches (since n=6, n/2=3).Wait, no, n=6, each day has 3 matches.So, the number of days when non-persistent players play against each other is equal to the number of such matches divided by the number of matches per day.Wait, no, each match is on a specific day, so the number of days when non-persistent players play against each other is equal to the number of such matches, which is 3.But total days=5, so 3 days vs. 5 days.Is 3 > 5/2=2.5? Yes, 3>2.5.So, for n=6, it's true.Similarly, for n=8, total days=7.Number of non-persistent players is at least 4.Number of matches between non-persistent players is C(4,2)=6.Number of days when non-persistent players play against each other is 6.But total days=7.Is 6 > 7/2=3.5? Yes.So, it seems to hold for n=6 and n=8.But wait, earlier when I tried to derive it, I saw that for n=5, it doesn't hold, but n must be even, so n=5 is invalid.Therefore, for n>=6, the number of days when non-persistent players play against each other is at least C(n/2,2), which is greater than (n-1)/2.Wait, but actually, the number of days is equal to the number of matches between non-persistent players, which is C(n - k,2), and since k <=n/2, n -k >=n/2.Therefore, C(n -k,2) >=C(n/2,2).And we saw that for n>=6, C(n/2,2) > (n-1)/2.Therefore, the number of days when non-persistent players play against each other is greater than half of the total game days.So, the answer is True.