answer：Okay, so I need to find the angle between two planes. Hmm, I remember that the angle between two planes is related to their normal vectors. Let me think. Each plane has a normal vector, right? For the first plane, the equation is 3x - y + 2z + 15 = 0. So the coefficients of x, y, and z give the normal vector. That would be <3, -1, 2>. Similarly, for the second plane, 5x + 9y - 3z - 1 = 0, the normal vector is <5, 9, -3>. Alright, so I have two normal vectors: n1 = <3, -1, 2> and n2 = <5, 9, -3>. I think the angle between the planes is the same as the angle between their normal vectors. Or wait, is it the supplementary angle? I might need to double-check that. I recall that the angle θ between two planes can be found using the dot product formula. The formula is cosθ = (n1 · n2) / (|n1| |n2|). So, first, I need to compute the dot product of n1 and n2. Let me calculate that. n1 · n2 = (3)(5) + (-1)(9) + (2)(-3) = 15 - 9 - 6. Let me add that up: 15 - 9 is 6, and 6 - 6 is 0. Wait, the dot product is zero? That means the normal vectors are perpendicular to each other. So, does that mean the planes are also perpendicular? But hold on, if the normal vectors are perpendicular, then the planes themselves must be perpendicular as well because the angle between the planes is equal to the angle between their normals. Or is it the supplementary angle? Hmm, I think if the normals are perpendicular, the planes are also perpendicular. So, the angle between the planes should be 90 degrees. But just to be thorough, let me compute the magnitudes of n1 and n2 to confirm. The magnitude of n1 is sqrt(3² + (-1)² + 2²) = sqrt(9 + 1 + 4) = sqrt(14). The magnitude of n2 is sqrt(5² + 9² + (-3)²) = sqrt(25 + 81 + 9) = sqrt(115). So, plugging into the formula: cosθ = 0 / (sqrt(14) * sqrt(115)) = 0. Therefore, θ = arccos(0) = 90 degrees. Wait, but sometimes angles between planes are considered as the acute angle. So, if the normals are perpendicular, the planes are perpendicular, which is 90 degrees, which is already acute. So, I think that's the answer. But just to make sure, let me visualize. If two planes have perpendicular normals, they intersect at a right angle. So, yes, the angle between them is 90 degrees. I don't think I made any calculation mistakes. The dot product was 15 - 9 - 6, which is indeed 0. The magnitudes squared are 14 and 115, so their square roots are fine. So, yeah, the angle is 90 degrees. **Final Answer**The angle between the planes is boxed{90^circ}.

answer：Okay, I need to write a Python function that translates a list of sentences from English to French using a given dictionary. Hmm, let's think about how to approach this.So, the input is a list of sentences, each being a string. And a dictionary where the keys are English words and the values are their French translations. The output should be a list where each sentence is translated word by word. If a word isn't in the dictionary, it stays the same.First, I should process each sentence individually. For each sentence, I need to split it into words. Oh right, using the split() method will split on whitespace by default, which should handle spaces between words.Once I have the list of words for a sentence, I need to translate each word. So for each word in the sentence, I check if it's a key in the translation_dict. If it is, replace it with the corresponding value. If not, leave it as is.After translating all the words in a sentence, I need to join them back into a string. So I'll use the join() method with a space to combine the translated words.Let me outline the steps:1. Initialize an empty list to hold the translated sentences.2. Loop through each sentence in the input list.3. For each sentence, split into words.4. For each word, check if it exists in the translation_dict. - If yes, replace it with the French word. - If no, keep the original word.5. Join the translated words back into a sentence.6. Add this translated sentence to the result list.7. After processing all sentences, return the result list.Wait, what about case sensitivity? Like, if a word is capitalized, does the dictionary have it in lowercase? The problem statement says the input is a list of sentences, but it's not specified whether the words are case-sensitive. The example shows all lowercase, so I think the function should treat words as case-sensitive. So if a word is "Hello" and the dictionary has "hello", it won't translate. So we don't need to handle case; the function treats words as they are.Another thing: punctuation. The problem statement doesn't mention handling punctuation. So I assume that the sentences are split into words correctly, and any punctuation is part of the word. For example, if a word is "world!", and the dictionary has "world", it won't translate. So the function doesn't handle that, as per the problem's example.So, for each word in the split sentence, we look it up in the dictionary. If found, replace; else, leave.Let me think about the example:Input sentences: ["hello world", "good morning"]Dictionary: {"hello": "bonjour", "world": "monde", "good": "bon", "morning": "matin"}Processing first sentence:"hello" → "bonjour""world" → "monde"So joined as "bonjour monde"Second sentence:"good" → "bon""morning" → "matin"So joined as "bon matin"Which matches the expected output.So the function should handle this correctly.Now, how to implement this in Python.Function steps:def translate_sentences(sentences, translation_dict): translated = [] for sentence in sentences: words = sentence.split() translated_words = [] for word in words: if word in translation_dict: translated_words.append(translation_dict[word]) else: translated_words.append(word) translated_sentence = ' '.join(translated_words) translated.append(translated_sentence) return translatedYes, that seems right.Testing the example:sentences = ["hello world", "good morning"]translation_dict = {"hello": "bonjour", "world": "monde", "good": "bon", "morning": "matin"}The function should return ["bonjour monde", "bon matin"]Another test case: what if a word is not in the dictionary?E.g., sentences = ["hello world today", "good morning"]translation_dict has "hello", "world", "good", "morning", but not "today".So the output should be ["bonjour monde today", "bon matin"]Yes, the function handles that.What about empty sentences? Like an empty string in the list. Splitting an empty string gives an empty list, so the translated sentence is an empty string.What about multiple spaces between words? The split() method will handle that, as it treats any number of whitespace as a separator, and the resulting list will have the words without empty strings.So the function should handle all these cases.I think this should work.

answer：First, I identify the number of gifts each person gave to Rosalina.Emilio gave 11 gifts, Jorge gave 6 gifts, and Pedro gave 4 gifts.To find the total number of gifts Rosalina received, I add the gifts from each person together.So, 11 (from Emilio) plus 6 (from Jorge) plus 4 (from Pedro) equals 21 gifts in total.

answer：Okay, so I need to figure out the ticket price that will maximize the number of tickets sold for the community events in Hermitage. The number of tickets sold is given by the quadratic function ( t(p) = -50p^2 + 300p ). Hmm, quadratic functions have a parabola shape, and since the coefficient of ( p^2 ) is negative (-50), the parabola opens downward. That means the vertex of the parabola will be the maximum point. So, the vertex will give me the ticket price that maximizes the number of tickets sold.I remember that for a quadratic function in the form ( f(p) = ap^2 + bp + c ), the vertex occurs at ( p = -frac{b}{2a} ). In this case, ( a = -50 ) and ( b = 300 ). Let me plug those values into the formula.So, ( p = -frac{300}{2 times -50} ). Calculating the denominator first: 2 times -50 is -100. Then, ( p = -frac{300}{-100} ). The negatives cancel out, so ( p = 3 ). Wait, so the ticket price should be set at 3 to maximize the number of tickets sold. Let me just verify that. If I plug p = 3 back into the function ( t(p) ), I should get the maximum number of tickets.Calculating ( t(3) = -50(3)^2 + 300(3) ). That's ( -50 times 9 + 900 ), which is ( -450 + 900 = 450 ). So, at 3 per ticket, 450 tickets will be sold. That seems right.Now, moving on to the second part. The cost of organizing the events is 1,000, and we need to ensure that the revenue from ticket sales covers this cost. The revenue function is given by ( R(p) = p cdot t(p) ). So, first, I need to write out the revenue function explicitly.Given ( t(p) = -50p^2 + 300p ), then ( R(p) = p times (-50p^2 + 300p) ). Let me multiply that out. ( R(p) = -50p^3 + 300p^2 ). Hmm, so it's a cubic function. To find the break-even point, we need to set the revenue equal to the cost, which is 1,000. So, we set up the equation:( -50p^3 + 300p^2 = 1000 ).Let me rearrange this equation to bring all terms to one side:( -50p^3 + 300p^2 - 1000 = 0 ).Hmm, solving a cubic equation can be tricky. Maybe I can factor out a common term first. Let's see, all coefficients are divisible by -50. Let me factor that out:( -50(p^3 - 6p^2 + 20) = 0 ).Wait, actually, if I factor out -50, it becomes:( -50(p^3 - 6p^2 + 20) = 0 ).But that might not help much. Alternatively, maybe I can factor out a 50:( 50(-p^3 + 6p^2 - 20) = 0 ).But that still leaves me with a cubic inside. Maybe I can factor the cubic equation. Let me write it as:( -p^3 + 6p^2 - 20 = 0 ).Alternatively, multiply both sides by -1 to make it positive leading coefficient:( p^3 - 6p^2 + 20 = 0 ).Now, trying to factor this cubic. Maybe using rational root theorem. Possible rational roots are factors of 20 over factors of 1, so ±1, ±2, ±4, ±5, ±10, ±20.Let me test p=2: ( 8 - 24 + 20 = 4 neq 0 ).p=4: ( 64 - 96 + 20 = -12 neq 0 ).p=5: ( 125 - 150 + 20 = -5 neq 0 ).p=10: ( 1000 - 600 + 20 = 420 neq 0 ).p=1: ( 1 - 6 + 20 = 15 neq 0 ).p=-1: ( -1 - 6 + 20 = 13 neq 0 ).Hmm, none of these seem to work. Maybe I made a mistake earlier.Wait, let me double-check my revenue function. ( R(p) = p times t(p) = p(-50p^2 + 300p) = -50p^3 + 300p^2 ). That seems correct.Setting that equal to 1000: ( -50p^3 + 300p^2 = 1000 ). Then, moving 1000 to the left: ( -50p^3 + 300p^2 - 1000 = 0 ). Factoring out -50: ( -50(p^3 - 6p^2 + 20) = 0 ). So, ( p^3 - 6p^2 + 20 = 0 ). Hmm, still not factorable with integer roots.Maybe I need to use another method, like the rational root theorem didn't work, so perhaps I can use the method of depressed cubic or numerical methods.Alternatively, maybe I can graph the function or use trial and error to approximate the roots.Alternatively, perhaps I can rewrite the equation:( -50p^3 + 300p^2 - 1000 = 0 ).Divide both sides by -50 to simplify:( p^3 - 6p^2 + 20 = 0 ).Wait, same as before. Maybe I can write it as ( p^3 - 6p^2 = -20 ).Alternatively, perhaps I can factor by grouping, but I don't see an obvious grouping.Alternatively, maybe I can use the cubic formula, but that's quite complicated.Alternatively, perhaps I can use substitution. Let me set ( q = p - frac{b}{3a} ) to eliminate the quadratic term. For a cubic equation ( p^3 + ap^2 + bp + c = 0 ), the substitution is ( p = q - frac{a}{3} ). In this case, the equation is ( p^3 - 6p^2 + 20 = 0 ). So, a = -6, b = 0, c = 20.Wait, actually, the standard form is ( p^3 + Ap^2 + Bp + C = 0 ). So, here, A = -6, B = 0, C = 20.So, the substitution is ( p = q - frac{A}{3} = q - (-6)/3 = q + 2 ).So, substituting ( p = q + 2 ) into the equation:( (q + 2)^3 - 6(q + 2)^2 + 20 = 0 ).Let me expand this:First, ( (q + 2)^3 = q^3 + 6q^2 + 12q + 8 ).Then, ( -6(q + 2)^2 = -6(q^2 + 4q + 4) = -6q^2 -24q -24 ).Adding the constant term 20.So, putting it all together:( q^3 + 6q^2 + 12q + 8 -6q^2 -24q -24 + 20 = 0 ).Simplify term by term:- ( q^3 )- ( 6q^2 -6q^2 = 0 )- ( 12q -24q = -12q )- ( 8 -24 +20 = 4 )So, the equation becomes:( q^3 -12q + 4 = 0 ).Hmm, that's a depressed cubic (no ( q^2 ) term). Now, I can use the depressed cubic formula. The general form is ( q^3 + mq + n = 0 ). Here, m = -12, n = 4.The depressed cubic formula is:( q = sqrt[3]{-frac{n}{2} + sqrt{left(frac{n}{2}right)^2 + left(frac{m}{3}right)^3}} + sqrt[3]{-frac{n}{2} - sqrt{left(frac{n}{2}right)^2 + left(frac{m}{3}right)^3}} ).Plugging in m = -12, n = 4:First, compute ( frac{n}{2} = 2 ).Then, compute ( left(frac{n}{2}right)^2 = 4 ).Compute ( frac{m}{3} = -4 ), so ( left(frac{m}{3}right)^3 = (-4)^3 = -64 ).So, the discriminant inside the square root is ( 4 + (-64) = -60 ).Uh-oh, that's negative, which means we have complex numbers involved. So, the equation has three real roots, but they can be expressed using trigonometric substitution.Alternatively, since the discriminant is negative, we can express the roots using cosine.The formula for roots when discriminant is negative is:( q = 2sqrt{-frac{m}{3}} cosleft( frac{1}{3} arccosleft( frac{-n}{2} sqrt{ -frac{27}{m^3} } right) right) ).Wait, let's see. The general formula for depressed cubic with discriminant < 0 is:( q = 2sqrt{ frac{|m|}{3} } cosleft( frac{1}{3} arccosleft( frac{3n}{2m} sqrt{ frac{3}{|m|} } right) right) ).Wait, maybe I should look up the exact formula, but since I can't, I'll try to recall.Alternatively, let me use the identity for solving cubics with three real roots.Given ( q^3 + pq + r = 0 ), when discriminant ( Delta = (r/2)^2 + (p/3)^3 < 0 ), then the roots are:( q = 2sqrt{ -p/3 } cosleft( frac{1}{3} arccosleft( frac{ -r }{ 2 } sqrt{ -27/p^3 } right) - frac{2pi k}{3} right) ), for k=0,1,2.In our case, the equation is ( q^3 -12q + 4 = 0 ). So, p = -12, r = 4.Compute discriminant ( Delta = (4/2)^2 + (-12/3)^3 = (2)^2 + (-4)^3 = 4 - 64 = -60 ). So, discriminant is negative, confirming three real roots.So, using the formula:( q = 2sqrt{ -(-12)/3 } cosleft( frac{1}{3} arccosleft( frac{ -4 }{ 2 } sqrt{ -27/(-12)^3 } right) right) ).Wait, let's compute step by step.First, ( sqrt{ -p/3 } = sqrt{ -(-12)/3 } = sqrt{4} = 2 ).Next, compute the argument inside arccos:( frac{ -r }{ 2 } sqrt{ -27 / p^3 } ).But p = -12, so:( frac{ -4 }{ 2 } sqrt{ -27 / (-12)^3 } = (-2) sqrt{ -27 / (-1728) } ).Simplify inside the square root:( -27 / (-1728) = 27/1728 = 1/64 ).So, ( sqrt{1/64} = 1/8 ).Therefore, the argument is ( (-2) times (1/8) = -1/4 ).So, the expression becomes:( q = 2 times 2 cosleft( frac{1}{3} arccos( -1/4 ) right) ).Simplify:( q = 4 cosleft( frac{1}{3} arccos( -1/4 ) right) ).Let me compute ( arccos(-1/4) ). The arccos of -1/4 is approximately 1.823 radians (since cos(1.823) ≈ -0.25). Then, one-third of that is approximately 0.6077 radians.So, ( cos(0.6077) ≈ 0.823 ).Therefore, ( q ≈ 4 times 0.823 ≈ 3.292 ).But since it's a depressed cubic with three real roots, we need to consider the other two roots as well. The general solution is:( q = 2sqrt{4} cosleft( frac{1}{3} arccos(-1/4) - frac{2pi k}{3} right) ) for k=0,1,2.So, for k=0: ( q ≈ 3.292 ).For k=1: ( q = 4 cosleft( 0.6077 - 2.0944 right) = 4 cos(-1.4867) ≈ 4 times 0.087 ≈ 0.348 ).For k=2: ( q = 4 cosleft( 0.6077 - 4.1888 right) = 4 cos(-3.5811) ≈ 4 times (-0.946) ≈ -3.784 ).So, the three real roots are approximately q ≈ 3.292, q ≈ 0.348, and q ≈ -3.784.But remember, we had substituted ( p = q + 2 ). So, converting back to p:For q ≈ 3.292: p ≈ 3.292 + 2 ≈ 5.292.For q ≈ 0.348: p ≈ 0.348 + 2 ≈ 2.348.For q ≈ -3.784: p ≈ -3.784 + 2 ≈ -1.784.Since ticket prices can't be negative, we discard p ≈ -1.784. So, the relevant roots are p ≈ 5.292 and p ≈ 2.348.So, the revenue function equals 1,000 at approximately p ≈ 2.35 and p ≈ 5.29.Now, to find the range of ticket prices where revenue is at least 1,000, we need to consider the behavior of the revenue function.Since the revenue function is a cubic ( R(p) = -50p^3 + 300p^2 ), which tends to negative infinity as p approaches positive infinity and positive infinity as p approaches negative infinity. However, since p represents ticket price, it must be positive.Looking at the graph of the revenue function, it starts at 0 when p=0, increases to a maximum, then decreases. The revenue function crosses the 1,000 line at two points: approximately p ≈ 2.35 and p ≈ 5.29.Therefore, the revenue is above 1,000 between these two prices. So, ticket prices between approximately 2.35 and 5.29 will result in revenue that covers the 1,000 cost.But let me verify this by testing a value between 2.35 and 5.29, say p=3.Compute R(3) = -50*(27) + 300*(9) = -1350 + 2700 = 1350, which is more than 1000. Good.Testing p=2: R(2) = -50*(8) + 300*(4) = -400 + 1200 = 800, which is less than 1000. So, below 2.35, revenue is less than 1000.Testing p=6: R(6) = -50*(216) + 300*(36) = -10800 + 10800 = 0, which is less than 1000. So, above 5.29, revenue drops below 1000.Therefore, the range of p where revenue is at least 1000 is between approximately 2.35 and 5.29.But let me express these more accurately. Earlier, I approximated the roots as 2.348 and 5.292. So, rounding to two decimal places, p ≈ 2.35 and p ≈ 5.29.Therefore, the ticket prices should be between approximately 2.35 and 5.29 to break even or make a profit.But let me check if these are exact values or if I can express them more precisely. Since the cubic equation didn't factor nicely, these are approximate values. So, in the answer, I can present them as approximate decimals.Alternatively, maybe I can express them as exact expressions, but they involve cube roots and cosines, which might be too complicated. So, probably, the answer expects decimal approximations.So, summarizing:1. The ticket price that maximizes the number of tickets sold is 3, resulting in 450 tickets sold.2. The ticket prices should be between approximately 2.35 and 5.29 to ensure revenue covers the 1,000 cost.Wait, but let me check if the revenue function is indeed positive between these two points. Since it's a cubic with negative leading coefficient, it goes from positive infinity to negative infinity. But in the positive p region, it starts at 0 when p=0, increases to a maximum, then decreases. So, the revenue is above 1000 between the two roots we found.Yes, that makes sense.So, final answers:1. 3, 450 tickets.2. Between approximately 2.35 and 5.29.But let me write the exact values using the cubic solution. Wait, but since they are irrational, it's better to present them as approximate decimals.Alternatively, maybe I can express them as exact roots, but that might not be necessary.Alternatively, perhaps I can write the exact roots in terms of cosines, but that might be overcomplicating.So, I think it's acceptable to present the approximate decimal values.**Final Answer**1. The ticket price should be set to boxed{3} dollars, and the maximum number of tickets sold will be boxed{450}.2. The range of ticket prices that will allow breaking even or making a profit is from boxed{2.35} dollars to boxed{5.29} dollars.

answer：Okay, so I have this matrix problem here. The matrix A is given as:[mathbf{A} = begin{pmatrix} a & 1 -2 & d end{pmatrix}]And it's given that:[mathbf{A} + mathbf{A}^{-1} = mathbf{0}]I need to find the determinant of A, which is denoted as det A. Hmm, okay. Let me think about how to approach this.First, I remember that for a matrix A, if it's invertible, then:[mathbf{A}^{-1} = frac{1}{det mathbf{A}} begin{pmatrix} d & -1 2 & a end{pmatrix}]Wait, is that right? Let me double-check. The inverse of a 2x2 matrix (begin{pmatrix} m & n p & q end{pmatrix}) is (frac{1}{mq - np} begin{pmatrix} q & -n -p & m end{pmatrix}). So, applying that to matrix A:[mathbf{A}^{-1} = frac{1}{ad - (-2)(1)} begin{pmatrix} d & -1 2 & a end{pmatrix}]Simplify the denominator: ad - (-2)(1) = ad + 2. So,[mathbf{A}^{-1} = frac{1}{ad + 2} begin{pmatrix} d & -1 2 & a end{pmatrix}]Alright, so now the equation given is:[mathbf{A} + mathbf{A}^{-1} = mathbf{0}]Which means:[mathbf{A} + mathbf{A}^{-1} = begin{pmatrix} 0 & 0 0 & 0 end{pmatrix}]So, let's write out the equation component-wise. Let me denote the inverse matrix as:[mathbf{A}^{-1} = begin{pmatrix} frac{d}{ad + 2} & frac{-1}{ad + 2} frac{2}{ad + 2} & frac{a}{ad + 2} end{pmatrix}]Therefore, adding A and A^{-1}:First element (1,1):[a + frac{d}{ad + 2} = 0]Element (1,2):[1 + frac{-1}{ad + 2} = 0]Element (2,1):[-2 + frac{2}{ad + 2} = 0]Element (2,2):[d + frac{a}{ad + 2} = 0]So, that gives us four equations. Let me write them down:1. ( a + frac{d}{ad + 2} = 0 ) 2. ( 1 - frac{1}{ad + 2} = 0 ) 3. ( -2 + frac{2}{ad + 2} = 0 ) 4. ( d + frac{a}{ad + 2} = 0 )Hmm, interesting. Let's see if these equations are consistent and if we can solve for a and d.Looking at equations 2 and 3, they both involve the term ( frac{1}{ad + 2} ). Let me solve equation 2 first.Equation 2: ( 1 - frac{1}{ad + 2} = 0 )So,( 1 = frac{1}{ad + 2} )Multiply both sides by (ad + 2):( ad + 2 = 1 )So,( ad = 1 - 2 = -1 )Therefore, ad = -1.Similarly, let's check equation 3:Equation 3: ( -2 + frac{2}{ad + 2} = 0 )So,( -2 + frac{2}{ad + 2} = 0 )Move -2 to the other side:( frac{2}{ad + 2} = 2 )Multiply both sides by (ad + 2):( 2 = 2(ad + 2) )Divide both sides by 2:( 1 = ad + 2 )So,( ad = 1 - 2 = -1 )Same result as equation 2. So, both equations 2 and 3 give us ad = -1. Good, consistent.So, now we know that ad = -1.Now, let's look at equations 1 and 4.Equation 1: ( a + frac{d}{ad + 2} = 0 )But since ad = -1, substitute that in:( a + frac{d}{-1 + 2} = 0 )Simplify denominator: -1 + 2 = 1So,( a + frac{d}{1} = 0 )Which is:( a + d = 0 )Similarly, equation 4: ( d + frac{a}{ad + 2} = 0 )Again, ad = -1, so:( d + frac{a}{-1 + 2} = 0 )Simplify denominator: 1So,( d + a = 0 )Same as equation 1. So, equations 1 and 4 both give us a + d = 0.So, from equations 2 and 3, we have ad = -1.From equations 1 and 4, we have a + d = 0.So, now we have a system of two equations:1. a + d = 0 2. a*d = -1So, we can solve for a and d.From equation 1: d = -aSubstitute into equation 2:a*(-a) = -1Which is:- a^2 = -1Multiply both sides by -1:a^2 = 1Therefore, a = 1 or a = -1.So, if a = 1, then d = -1.If a = -1, then d = 1.So, we have two possibilities:Case 1: a = 1, d = -1Case 2: a = -1, d = 1Now, let's check if these satisfy the original equation.But before that, let's compute det A.det A = a*d - (1)*(-2) = a*d + 2But from earlier, we have ad = -1.So, det A = (-1) + 2 = 1.Wait, so regardless of the values of a and d, as long as ad = -1, det A is 1.So, det A = 1.But let me confirm this.Given that det A = a*d - (1)*(-2) = a*d + 2.But we have ad = -1, so det A = (-1) + 2 = 1.Yes, that's correct.But just to be thorough, let me verify with the two cases.Case 1: a = 1, d = -1Matrix A:[begin{pmatrix} 1 & 1 -2 & -1 end{pmatrix}]Compute det A: (1)(-1) - (1)(-2) = -1 + 2 = 1.Case 2: a = -1, d = 1Matrix A:[begin{pmatrix} -1 & 1 -2 & 1 end{pmatrix}]Compute det A: (-1)(1) - (1)(-2) = -1 + 2 = 1.So, in both cases, determinant is 1.Therefore, det A = 1.But just to make sure, let's also check if A + A^{-1} = 0 in both cases.Starting with Case 1: a=1, d=-1.Compute A^{-1}:We have det A = 1, so inverse is:[frac{1}{1} begin{pmatrix} -1 & -1 2 & 1 end{pmatrix} = begin{pmatrix} -1 & -1 2 & 1 end{pmatrix}]So, A + A^{-1}:[begin{pmatrix} 1 & 1 -2 & -1 end{pmatrix} + begin{pmatrix} -1 & -1 2 & 1 end{pmatrix} = begin{pmatrix} 0 & 0 0 & 0 end{pmatrix}]Perfect, that's the zero matrix.Similarly, Case 2: a=-1, d=1.Compute A^{-1}:det A = 1, so inverse is:[frac{1}{1} begin{pmatrix} 1 & -1 2 & -1 end{pmatrix} = begin{pmatrix} 1 & -1 2 & -1 end{pmatrix}]So, A + A^{-1}:[begin{pmatrix} -1 & 1 -2 & 1 end{pmatrix} + begin{pmatrix} 1 & -1 2 & -1 end{pmatrix} = begin{pmatrix} 0 & 0 0 & 0 end{pmatrix}]Again, zero matrix. So, both cases satisfy the given condition.Therefore, regardless of the values of a and d, as long as the conditions are satisfied, det A is 1.So, the answer is 1.**Final Answer**boxed{1}

answer：Alright, so I'm trying to figure out this problem about the plant-based food company launching new protein bars. They have some conditions on protein and carbohydrate content, and also a cost constraint. Let me break it down step by step.First, the problem says that the protein content, p, in grams, should be at least 1.5 times the carbohydrate content, c, but no more than 3 times. So, I need to translate that into an inequality. Hmm, okay. "At least" means greater than or equal to, and "no more than" means less than or equal to. So, if p is at least 1.5 times c, that would be p ≥ 1.5c. And p should also be no more than 3 times c, so p ≤ 3c. So, combining these two, the inequality should be 1.5c ≤ p ≤ 3c. That seems right.Moving on to the second part. The cost function is given by C(p, c) = 2p + 3c - 5. The company wants the production cost to be no more than 10 per bar. So, that means 2p + 3c - 5 ≤ 10. Let me write that down as an inequality: 2p + 3c - 5 ≤ 10. I think I should solve this for p and c to find the range of possible values.First, let's simplify the cost inequality. Adding 5 to both sides gives 2p + 3c ≤ 15. So, 2p + 3c ≤ 15. Now, I need to consider this along with the earlier condition on p and c, which is 1.5c ≤ p ≤ 3c. Also, p and c are non-negative, so p ≥ 0 and c ≥ 0.So, now I have a system of inequalities:1. 1.5c ≤ p ≤ 3c2. 2p + 3c ≤ 153. p ≥ 04. c ≥ 0I need to find all pairs (p, c) that satisfy these conditions. Maybe I can express p in terms of c from the first inequality and substitute into the cost inequality.From the first inequality, p is between 1.5c and 3c. So, let's substitute the lower and upper bounds of p into the cost inequality to see how c is constrained.First, let's take p = 1.5c and plug it into 2p + 3c ≤ 15:2*(1.5c) + 3c ≤ 153c + 3c ≤ 156c ≤ 15c ≤ 2.5So, when p is at its minimum (1.5c), c can be at most 2.5 grams.Now, let's take p = 3c and plug it into the cost inequality:2*(3c) + 3c ≤ 156c + 3c ≤ 159c ≤ 15c ≤ 15/9c ≤ 5/3 ≈ 1.6667So, when p is at its maximum (3c), c can be at most approximately 1.6667 grams.Therefore, c is constrained between 0 and 2.5 grams, but depending on p, it might be less. Wait, actually, since p can vary between 1.5c and 3c, the maximum c allowed depends on p.Alternatively, maybe I should solve the system graphically or algebraically.Let me try to express c in terms of p from the cost inequality.From 2p + 3c ≤ 15, we can write 3c ≤ 15 - 2p, so c ≤ (15 - 2p)/3.Also, from the protein condition, p ≥ 1.5c and p ≤ 3c.So, let's express c in terms of p for the protein condition:From p ≥ 1.5c, we get c ≤ (2/3)p.From p ≤ 3c, we get c ≥ p/3.So, combining these, p/3 ≤ c ≤ (2/3)p.And from the cost inequality, c ≤ (15 - 2p)/3.So, now we have:p/3 ≤ c ≤ min{(2/3)p, (15 - 2p)/3}Also, since c must be non-negative, p must be such that (15 - 2p)/3 is non-negative.So, (15 - 2p)/3 ≥ 0 => 15 - 2p ≥ 0 => 2p ≤ 15 => p ≤ 7.5.So, p can be at most 7.5 grams.But also, from the protein condition, p must be at least 1.5c. Since c is non-negative, p must be non-negative as well.So, p is between 0 and 7.5.But let's see when (2/3)p is less than or equal to (15 - 2p)/3.Set (2/3)p ≤ (15 - 2p)/3.Multiply both sides by 3: 2p ≤ 15 - 2pAdd 2p to both sides: 4p ≤ 15p ≤ 15/4 = 3.75So, when p ≤ 3.75, (2/3)p ≤ (15 - 2p)/3, so c ≤ (2/3)p.When p > 3.75, then (15 - 2p)/3 < (2/3)p, so c ≤ (15 - 2p)/3.Therefore, the constraints on c depend on the value of p.So, summarizing:For p between 0 and 3.75:p/3 ≤ c ≤ (2/3)pAnd for p between 3.75 and 7.5:p/3 ≤ c ≤ (15 - 2p)/3Also, we need to ensure that c is non-negative, so let's check the lower bounds.From p/3 ≤ c, since p ≥ 0, c must be at least 0.So, putting it all together, the range of possible values for p and c is:- When 0 ≤ p ≤ 3.75, c is between p/3 and (2/3)p.- When 3.75 ≤ p ≤ 7.5, c is between p/3 and (15 - 2p)/3.I think that's the range. Let me verify with some test points.For example, take p = 0. Then c must be 0, since c ≥ p/3 = 0 and c ≤ (2/3)*0 = 0. So, c = 0. That makes sense.Take p = 3. Then, c is between 1 and 2. Let's check the cost: 2*3 + 3*2 = 6 + 6 = 12, which is more than 10. Wait, that's a problem.Wait, hold on. If p = 3, c can be up to 2, but 2p + 3c = 6 + 6 = 12, which is over 10. So, that's not allowed. Hmm, so maybe my earlier conclusion is incorrect.Wait, perhaps I made a mistake in interpreting the inequalities. Let me go back.The cost inequality is 2p + 3c ≤ 15. So, when p = 3, c can be up to (15 - 6)/3 = 3. But from the protein condition, c must be ≤ 2. So, actually, c is limited by the protein condition in this case.Wait, but if p = 3, then c must be ≤ 2 (from 1.5c ≤ p => c ≤ 2). But 2p + 3c = 6 + 6 = 12, which is more than 10. So, that's a problem.Wait, so maybe I need to consider both constraints together.Perhaps I should solve the system of inequalities:1. 1.5c ≤ p ≤ 3c2. 2p + 3c ≤ 153. p ≥ 0, c ≥ 0So, maybe I can graph these inequalities to find the feasible region.Alternatively, I can find the intersection points.Let me try to find the points where the lines intersect.First, the protein condition is p = 1.5c and p = 3c.The cost condition is 2p + 3c = 15.Let me find where p = 1.5c intersects 2p + 3c = 15.Substitute p = 1.5c into 2p + 3c = 15:2*(1.5c) + 3c = 3c + 3c = 6c = 15 => c = 2.5, so p = 1.5*2.5 = 3.75.So, intersection point at (3.75, 2.5).Next, find where p = 3c intersects 2p + 3c = 15.Substitute p = 3c into 2p + 3c = 15:2*(3c) + 3c = 6c + 3c = 9c = 15 => c = 15/9 = 5/3 ≈ 1.6667, so p = 3*(5/3) = 5.So, intersection point at (5, 5/3).Now, let's see the feasible region.From p = 1.5c and p = 3c, and the cost line 2p + 3c = 15.The feasible region is bounded by:- p ≥ 1.5c- p ≤ 3c- 2p + 3c ≤ 15- p ≥ 0, c ≥ 0So, the feasible region is a polygon with vertices at:1. (0, 0): where p = 0 and c = 0.2. (5, 5/3): intersection of p = 3c and 2p + 3c = 15.3. (3.75, 2.5): intersection of p = 1.5c and 2p + 3c = 15.4. (0, 0): Wait, but when c = 0, p must be 0 as well because p ≥ 1.5c = 0 and p ≤ 3c = 0.Wait, actually, the feasible region is a polygon with vertices at (0,0), (5, 5/3), (3.75, 2.5), and back to (0,0). Hmm, but that doesn't seem right because (3.75, 2.5) is also connected to (0,0) via p = 1.5c.Wait, maybe the feasible region is a quadrilateral with vertices at (0,0), (5, 5/3), (3.75, 2.5), and (0,0). Wait, that can't be, because (3.75, 2.5) is above (5, 5/3). Let me plot these points.Wait, (5, 5/3) is approximately (5, 1.6667), and (3.75, 2.5) is approximately (3.75, 2.5). So, these are two distinct points.So, the feasible region is bounded by:- From (0,0) to (5, 5/3) along p = 3c.- From (5, 5/3) to (3.75, 2.5) along 2p + 3c = 15.- From (3.75, 2.5) back to (0,0) along p = 1.5c.Wait, no, because p = 1.5c goes from (0,0) to (3.75, 2.5). So, the feasible region is actually a triangle with vertices at (0,0), (5, 5/3), and (3.75, 2.5).Wait, let me check if (5, 5/3) is on p = 3c: yes, because 5 = 3*(5/3). And (3.75, 2.5) is on p = 1.5c: 3.75 = 1.5*2.5.And both points are on the cost line 2p + 3c = 15.So, the feasible region is a triangle with vertices at (0,0), (5, 5/3), and (3.75, 2.5).Therefore, the range of p and c is all points within this triangle.So, to express the range, we can say:For p between 0 and 5 grams, c is between p/3 and min{(2/3)p, (15 - 2p)/3}.But wait, when p is between 0 and 3.75, (2/3)p is less than (15 - 2p)/3, so c is between p/3 and (2/3)p.When p is between 3.75 and 5, (15 - 2p)/3 is less than (2/3)p, so c is between p/3 and (15 - 2p)/3.Wait, but earlier I thought p could go up to 7.5, but actually, the intersection point is at p = 5, so p can't go beyond 5 because beyond that, c would have to be negative, which is not allowed.Wait, let me check. If p = 5, c = 5/3 ≈ 1.6667. If p increases beyond 5, say p = 6, then from the cost inequality, c ≤ (15 - 12)/3 = 1. But from the protein condition, c must be at least p/3 = 2. So, c would have to be between 2 and 1, which is impossible. Therefore, p cannot exceed 5.So, p is between 0 and 5, and c is between p/3 and min{(2/3)p, (15 - 2p)/3}.Therefore, the range of possible values is:- When 0 ≤ p ≤ 3.75, c is between p/3 and (2/3)p.- When 3.75 ≤ p ≤ 5, c is between p/3 and (15 - 2p)/3.And c must be non-negative, which is already satisfied since p is non-negative.Let me test p = 3.75. Then c is between 3.75/3 = 1.25 and (15 - 7.5)/3 = 7.5/3 = 2.5. Wait, but at p = 3.75, c is exactly 2.5 from the protein condition, so that's consistent.Another test: p = 4. Then c must be between 4/3 ≈ 1.333 and (15 - 8)/3 = 7/3 ≈ 2.333. But wait, from the protein condition, c must be ≤ 4/3 ≈ 1.333? Wait, no, p = 4, so from p ≤ 3c, c ≥ 4/3 ≈ 1.333, and from p ≥ 1.5c, c ≤ 4/1.5 ≈ 2.6667. But from the cost condition, c ≤ (15 - 8)/3 ≈ 2.333. So, c is between 1.333 and 2.333.Wait, but earlier I thought when p is between 3.75 and 5, c is between p/3 and (15 - 2p)/3. For p = 4, p/3 ≈ 1.333, and (15 - 8)/3 ≈ 2.333. So, that's correct.So, to summarize, the range is:- For p from 0 to 3.75 grams, c ranges from p/3 to (2/3)p.- For p from 3.75 to 5 grams, c ranges from p/3 to (15 - 2p)/3.And p cannot exceed 5 grams because beyond that, the constraints conflict.Therefore, the possible values for p and c are all pairs where p is between 0 and 5, and c is between p/3 and the minimum of (2/3)p and (15 - 2p)/3.I think that's the correct range.