answer：To determine the value of ( x ) that makes vectors ( vec{a} ) and ( vec{b} ) perpendicular, I start by recalling that two vectors are perpendicular if their dot product is zero.The dot product of ( vec{a} = (1, 2) ) and ( vec{b} = (x, -2) ) is calculated as:[1 times x + 2 times (-2) = x - 4]Setting this equal to zero for perpendicularity:[x - 4 = 0]Solving for ( x ) gives:[x = 4]

answer：Alright, so I have this problem here, and I need to find the values of a, b, and c. Then, I also need to find the arithmetic square root of a + 6b - c. Hmm, okay, let me break this down step by step.First, the problem says that the square root of (2a + 3) is plus or minus 3. So, mathematically, that should mean that sqrt(2a + 3) = ±3. Wait, but square roots can be both positive and negative, right? So, does that mean I have two equations here? Let me think.If sqrt(2a + 3) = 3, then squaring both sides would give me 2a + 3 = 9. Similarly, if sqrt(2a + 3) = -3, squaring both sides would also give me 2a + 3 = 9 because (-3)^2 is also 9. So, actually, both cases lead to the same equation. That simplifies things a bit. So, I can just solve 2a + 3 = 9.Let me do that. Subtract 3 from both sides: 2a = 6. Then, divide both sides by 2: a = 3. Okay, so a is 3. That was straightforward.Next, the problem mentions the cube root of (3b - 2c) is 2. So, in equation form, that would be cube_root(3b - 2c) = 2. To solve for b, I need to get rid of the cube root. I can do this by cubing both sides of the equation. So, (cube_root(3b - 2c))^3 = 2^3. That simplifies to 3b - 2c = 8. Hmm, okay, so 3b - 2c = 8. I need another equation to solve for both b and c, but I don't have that yet. Wait, the problem also says that c is the integer part of sqrt(6). So, let me figure out what c is.The integer part of sqrt(6) is the greatest integer less than or equal to sqrt(6). I know that sqrt(4) is 2 and sqrt(9) is 3, so sqrt(6) is somewhere between 2 and 3. Specifically, sqrt(6) is approximately 2.449. So, the integer part is 2. Therefore, c = 2. Got it.Now, going back to the equation 3b - 2c = 8. Since c is 2, substitute that in: 3b - 2*2 = 8. So, 3b - 4 = 8. Adding 4 to both sides: 3b = 12. Then, divide both sides by 3: b = 4. Perfect, so b is 4.So, to recap, a is 3, b is 4, and c is 2.Now, the last part of the problem asks for the arithmetic square root of (a + 6b - c). Let me compute that expression first.Substituting the values: a + 6b - c = 3 + 6*4 - 2. Let's compute each term. 6*4 is 24. So, 3 + 24 is 27. Then, subtract 2: 27 - 2 = 25. So, the expression simplifies to 25.Now, the arithmetic square root of 25 is the non-negative square root, which is 5. Because square roots can be both positive and negative, but the arithmetic square root specifically refers to the principal (non-negative) root.Let me just double-check my steps to make sure I didn't make any mistakes.1. Square root of (2a + 3) is ±3. Squared both sides to get 2a + 3 = 9, solved for a = 3. That seems correct.2. Cube root of (3b - 2c) is 2. Cubed both sides to get 3b - 2c = 8. Then, c is the integer part of sqrt(6), which is 2. Substituted c = 2 into the equation, solved for b = 4. That also seems correct.3. Then, computed a + 6b - c: 3 + 24 - 2 = 25. Square root of 25 is 5. All steps check out.I think I've got all the values right. So, a is 3, b is 4, c is 2, and the arithmetic square root of (a + 6b - c) is 5.**Final Answer**The values are ( a = boxed{3} ), ( b = boxed{4} ), ( c = boxed{2} ), and the arithmetic square root is ( boxed{5} ).

answer：Alright, so I have this problem here: I need to find the largest six-digit number by removing nine digits from the number 778157260669103. The catch is that I can't change the order of the digits. The options given are (a) 778152, (b) 781569, (c) 879103, (d) 986103, and (e) 987776. Hmm, okay, let me think about how to approach this.First, I remember that when you want to form the largest possible number by removing digits without changing their order, you should try to keep the leftmost digits as large as possible. That makes sense because the leftmost digits have a higher place value. So, for a six-digit number, I need to pick six digits from the original number in order, making sure each subsequent digit is as large as possible.The original number is 778157260669103. Let me write that out to visualize it better: 7 7 8 1 5 7 2 6 0 6 6 9 1 0 3. So, that's 15 digits in total. I need to remove nine of them, leaving six. Let me count the digits to confirm: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15. Yep, 15 digits.My goal is to select six digits in order, starting from the left, choosing the largest possible digits each time. But I have to be careful because sometimes choosing a slightly smaller digit early on might allow for larger digits later. Hmm, tricky.Let me break it down step by step.1. The first digit: I need the largest possible digit from the first few digits. The original number starts with 7,7,8,1,5,... So, the first three digits are 7,7,8. The largest among these is 8. But wait, 8 is the third digit. So, if I choose 8 as the first digit, I have to skip the first two 7s. But then, I have to make sure that there are enough digits left after that to form the remaining five digits. Let me check: if I take the third digit (8), the remaining digits are 1,5,7,2,6,0,6,6,9,1,0,3. That's 12 digits left, which is more than enough for five more digits. So, starting with 8 seems good.2. The second digit: Now, starting from the digit after 8, which is 1. The remaining digits are 1,5,7,2,6,0,6,6,9,1,0,3. I need the largest possible digit here. The digits are 1,5,7,2,6,0,6,6,9,1,0,3. The largest digit here is 9, but it's quite far to the right. If I choose 9 as the second digit, I might have to skip a lot of digits, but let me see if that's possible.Wait, but if I choose 9 as the second digit, I have to make sure that there are still four digits left after it for the remaining four positions. Let's see: the 9 is the 12th digit in the original number. So, after 9, there are only three digits left: 1,0,3. That's not enough for the remaining four digits. So, choosing 9 as the second digit would leave me with only three digits, which isn't enough. So, I can't choose 9 here.Next, the next largest digit after 1 is 7. The 7 is the fifth digit in the original number. Let me count: after 8, the digits are 1,5,7,2,6,0,6,6,9,1,0,3. So, the first 7 is at position 5. If I choose 7 as the second digit, then the remaining digits after it are 2,6,0,6,6,9,1,0,3. That's nine digits left, which is more than enough for the remaining four digits. So, 7 is a good choice.Wait, but hold on, is there a larger digit than 7 between 1 and 7? The digits are 1,5,7. So, 7 is the largest. So, second digit is 7.3. The third digit: Now, starting from the digit after 7, which is 2. The remaining digits are 2,6,0,6,6,9,1,0,3. I need the largest possible digit here. The digits are 2,6,0,6,6,9,1,0,3. The largest digit here is 9. Let's see where 9 is. It's the sixth digit in this remaining sequence, which is the 11th digit in the original number. After 9, there are three digits left: 1,0,3. Again, that's only three digits, but I need three more digits for the remaining three positions. So, if I choose 9 here, I can still get three more digits after it. So, that's okay.Wait, but if I choose 9 as the third digit, I have to make sure that after 9, there are enough digits for the remaining three positions. Let me check: after 9, the digits are 1,0,3. That's three digits, which is exactly what I need. So, choosing 9 as the third digit is possible.But hold on, is 9 the best choice? Let me see. If I choose 9, then the next digits would be 1,0,3. Alternatively, if I don't choose 9 and instead choose a larger digit later, but 9 is the largest, so I think 9 is the best choice here.So, third digit is 9.4. The fourth digit: Now, starting from the digit after 9, which is 1. The remaining digits are 1,0,3. I need the largest possible digit here. The digits are 1,0,3. The largest is 3, but 3 is the last digit. If I choose 3, then I don't have any digits left for the remaining two positions. Wait, no, I need four digits in total, and I've already chosen three. Wait, no, let me clarify.Wait, I'm confused. Let me recount. The original number is 15 digits. I have already chosen three digits: 8,7,9. So, I need to choose three more digits from the remaining digits after 9, which are 1,0,3. So, I need to choose the next three digits from 1,0,3. So, the fourth digit should be the largest possible from these. The digits are 1,0,3. The largest is 3, but it's the last digit. If I choose 3 as the fourth digit, then I have no digits left for the fifth and sixth positions. So, that's not possible.Wait, so I can't choose 3 as the fourth digit because I need two more digits after it. So, the next largest digit is 1. So, the fourth digit is 1.5. The fifth digit: Now, starting from the digit after 1, which is 0. The remaining digits are 0,3. I need the largest possible digit here. The digits are 0 and 3. The largest is 3, but it's the last digit. If I choose 3 as the fifth digit, then I have no digits left for the sixth position. So, that's not possible. Therefore, I have to choose 0 as the fifth digit.6. The sixth digit: Now, starting from the digit after 0, which is 3. There are no digits left after 3, so I can't choose anything. Wait, that can't be. I must have made a mistake somewhere.Wait, let me go back. Maybe choosing 9 as the third digit was a mistake because it left me with only 1,0,3, which are not enough for the remaining three digits. So, perhaps I should not have chosen 9 as the third digit.Let me try a different approach. Instead of choosing 9 as the third digit, maybe I should choose a different digit earlier on so that I have more digits left for the remaining positions.So, going back to step 3: after choosing 8 and 7, the remaining digits are 2,6,0,6,6,9,1,0,3. I was considering choosing 9 as the third digit, but that left me with only three digits, which isn't enough. So, maybe I should choose the next largest digit that allows me to have enough digits left.Looking at the remaining digits: 2,6,0,6,6,9,1,0,3. The largest digit is 9, but as we saw, choosing 9 leaves only three digits. The next largest digit is 6. Let's see where the first 6 is. It's the second digit in this remaining sequence, which is the 7th digit in the original number. If I choose 6 as the third digit, then the remaining digits after it are 0,6,6,9,1,0,3. That's seven digits left, which is more than enough for the remaining three digits.So, choosing 6 as the third digit instead of 9 might be better because it allows more flexibility for the remaining digits.So, third digit is 6.4. The fourth digit: Now, starting from the digit after 6, which is 0. The remaining digits are 0,6,6,9,1,0,3. I need the largest possible digit here. The digits are 0,6,6,9,1,0,3. The largest is 9. Let's see where 9 is. It's the fourth digit in this remaining sequence, which is the 10th digit in the original number. After 9, the remaining digits are 1,0,3. That's three digits, which is exactly what I need for the remaining three positions.So, choosing 9 as the fourth digit is possible.5. The fifth digit: Now, starting from the digit after 9, which is 1. The remaining digits are 1,0,3. The largest digit here is 3, but it's the last digit. If I choose 3, I can't get a sixth digit. So, I have to choose the next largest, which is 1.6. The sixth digit: Now, starting from the digit after 1, which is 0. The remaining digits are 0,3. The largest digit is 3, but it's the last digit. If I choose 3, I can't get a seventh digit, but I only need one more digit. Wait, no, I need six digits in total. Wait, I've already chosen five digits: 8,7,6,9,1. Now, I need one more digit. The remaining digits are 0 and 3. The largest is 3, so I can choose 3 as the sixth digit.Wait, but hold on. If I choose 3 as the sixth digit, I have to make sure that it's after 1. So, the digits after 1 are 0 and 3. So, 3 is the last digit. So, yes, I can choose 3 as the sixth digit.So, putting it all together, the number would be 8,7,6,9,1,3. Wait, but that's 876913. Hmm, but looking at the options, I don't see that. The options are 778152, 781569, 879103, 986103, 987776. So, 876913 isn't one of them. Did I do something wrong?Wait, maybe I made a mistake in my selection. Let me try another approach. Maybe instead of choosing 6 as the third digit, I should choose a different digit.Alternatively, perhaps I should not have chosen 7 as the second digit. Let me go back to step 2.After choosing 8 as the first digit, the remaining digits are 1,5,7,2,6,0,6,6,9,1,0,3. I chose 7 as the second digit because it was the next largest. But maybe choosing a different digit could lead to a better result.Looking at the remaining digits after 8: 1,5,7,2,6,0,6,6,9,1,0,3. The largest digit is 9, but as before, choosing 9 as the second digit would leave only three digits, which isn't enough. So, the next largest is 7. So, I think choosing 7 as the second digit is still the best option.Wait, but maybe if I choose 5 instead of 7, I can get a larger number later on? Let me see.If I choose 5 as the second digit, then the remaining digits after 5 are 7,2,6,0,6,6,9,1,0,3. The largest digit here is 9, but again, choosing 9 would leave only three digits. So, the next largest is 7. So, choosing 7 as the third digit. Then, the remaining digits after 7 are 2,6,0,6,6,9,1,0,3. The largest is 9, but again, choosing 9 leaves only three digits. So, the next largest is 6. Choosing 6 as the fourth digit, then the remaining digits are 0,6,6,9,1,0,3. The largest is 9, but choosing 9 leaves only three digits. So, the next largest is 6. Choosing 6 as the fifth digit, then the remaining digits are 6,9,1,0,3. The largest is 9, but choosing 9 leaves only two digits. So, the next largest is 6. Choosing 6 as the sixth digit, but then I have only one digit left, which is 9,1,0,3. Wait, this seems messy.Alternatively, maybe choosing 5 as the second digit isn't the best idea because 5 is smaller than 7, which would make the overall number smaller. So, perhaps sticking with 7 as the second digit is better.Wait, let me try another approach. Maybe instead of choosing 8 as the first digit, I should choose a different digit. Let me see.The original number starts with 7,7,8,1,5,7,2,6,0,6,6,9,1,0,3. If I choose the first 7 as the first digit, then I have more digits left, but 7 is smaller than 8. So, 8 is better as the first digit because it's larger. So, I think choosing 8 as the first digit is still the best option.Wait, but let me check the options given. The options are (a) 778152, (b) 781569, (c) 879103, (d) 986103, (e) 987776. So, the largest number among these is 987776, which is option (e). But I'm not sure if that's possible.Wait, let me see. To get 987776, I need to have 9,8,7,7,7,6 in order. Let me check if that's possible in the original number.Looking at the original number: 7,7,8,1,5,7,2,6,0,6,6,9,1,0,3. So, the digits are: positions 1:7, 2:7, 3:8, 4:1, 5:5, 6:7, 7:2, 8:6, 9:0, 10:6, 11:6, 12:9, 13:1, 14:0, 15:3.So, to get 9, I need to pick the 12th digit. To get 8, I need to pick the 3rd digit. But 3rd digit is before the 12th digit, so I can't have 8 after 9. So, 987776 is not possible because 8 comes before 9 in the original number. Therefore, option (e) is not possible.Similarly, option (d) is 986103. Let's see if that's possible. To get 9,8,6,1,0,3. Wait, 9 is at position 12, 8 is at position 3, which is before 9, so again, 8 can't come after 9. So, 986103 is also not possible.Option (c) is 879103. Let's see. 8 is at position 3, 7 is at position 6, 9 is at position 12, 1 is at position 13, 0 is at position 14, 3 is at position 15. So, the order is 8 (3), 7 (6), 9 (12), 1 (13), 0 (14), 3 (15). That seems possible because each subsequent digit is after the previous one. So, 879103 is a valid selection.Option (b) is 781569. Let's check. 7 is at position 1, 8 is at position 3, 1 is at position 4, 5 is at position 5, 6 is at position 8, 9 is at position 12. So, the order is 7 (1), 8 (3), 1 (4), 5 (5), 6 (8), 9 (12). That's also valid.Option (a) is 778152. Let's check. 7 (1), 7 (2), 8 (3), 1 (4), 5 (5), 2 (7). So, that's also valid.So, among the options, (c) 879103, (b) 781569, (a) 778152 are all valid. Now, which one is the largest?Comparing the options:- (a) 778152- (b) 781569- (c) 879103- (d) 986103 (invalid)- (e) 987776 (invalid)So, among the valid ones, (c) 879103 is the largest because it starts with 8, which is larger than 7 in the other options. So, (c) is the largest.Wait, but earlier when I tried to construct the number, I ended up with 876913, which isn't one of the options. So, maybe I made a mistake in my selection.Let me try again, more carefully.Starting with the first digit: 7,7,8,1,5,7,2,6,0,6,6,9,1,0,3.I want the largest first digit. The first three digits are 7,7,8. So, 8 is the largest. So, first digit is 8 (position 3).Now, remaining digits: 1,5,7,2,6,0,6,6,9,1,0,3.I need the largest possible second digit. The digits are 1,5,7,2,6,0,6,6,9,1,0,3. The largest is 9, but choosing 9 would leave only three digits. So, the next largest is 7 (position 6). So, second digit is 7.Now, remaining digits after 7: 2,6,0,6,6,9,1,0,3.Third digit: the largest is 9 (position 12). Choosing 9 leaves three digits: 1,0,3. So, third digit is 9.Fourth digit: from 1,0,3, the largest is 3, but that leaves nothing for fifth and sixth digits. So, next largest is 1.Fifth digit: from 0,3, the largest is 3, but that leaves nothing for sixth digit. So, next largest is 0.Sixth digit: only 3 left. So, sixth digit is 3.So, the number is 8,7,9,1,0,3, which is 879103. That's option (c). So, that's correct.Wait, so earlier when I tried to choose 6 as the third digit, I ended up with a different number, but it seems that choosing 9 as the third digit is better because it leads to a larger number overall, even though it leaves fewer digits for the end. So, 879103 is indeed the largest possible.Let me double-check if there's a way to get a larger number than 879103.Is there a way to get a number starting with 9? Well, 9 is at position 12. If I choose 9 as the first digit, I have to remove the first 11 digits, leaving only four digits: 1,0,3. That's not enough for six digits. So, I can't have 9 as the first digit.What about starting with 8? Yes, as we did, 879103.Is there a way to get a number starting with 8 and then a larger digit than 7? Let's see. After 8, the next digits are 1,5,7,2,6,0,6,6,9,1,0,3. The largest digit after 8 is 9, but choosing 9 as the second digit would leave only three digits, which isn't enough. So, the next largest is 7. So, 8 followed by 7 is the best we can do.Then, after 7, the largest digit is 9, which we can choose, leaving three digits. Then, we have to choose 1,0,3, which gives us 879103.Alternatively, if after 8 and 7, we choose 6 instead of 9, we might get a different number, but 876... would be smaller than 879...So, 879103 is indeed the largest possible.Therefore, the answer is (c) 879103.

answer：First, I recognize that each of the radicals can be simplified by expressing them with exponents.For the fourth root of 81, since 81 is 3 raised to the fourth power, the fourth root of 81 is 3.Next, the cube root of 27 is 3 because 3 cubed equals 27.Lastly, the square root of 9 is 3 since 3 squared is 9.Multiplying these simplified values together: 3 multiplied by 3 multiplied by 3 equals 27.Therefore, the expression simplifies to the positive integer 27.

answer：Okay, so I have this matrix A, which is a 2x2 matrix: A = [2 3] [0 1]And I need to find A^20 minus 2 times A^19. Hmm, that sounds like a lot of matrix multiplication if I try to compute each power separately. There must be a smarter way to do this without multiplying matrices 20 times.Let me think. Maybe I can find a pattern or a formula for A^n, where n is any positive integer. If I can express A^n in terms of A and the identity matrix, that would make it easier to compute A^20 and A^19.First, let me compute the first few powers of A to see if I can spot a pattern.A^1 is just A:A^1 = [2 3] [0 1]Now, let's compute A^2:A^2 = A * A = [2 3] [2 3] = [2*2 + 3*0, 2*3 + 3*1] = [4 + 0, 6 + 3] = [4 9] [0 1] [0 1] [0*2 + 1*0, 0*3 + 1*1] [0 + 0, 0 + 1] [0 1]So, A^2 = [4 9] [0 1]Hmm, interesting. Let's compute A^3:A^3 = A^2 * A = [4 9] [2 3] = [4*2 + 9*0, 4*3 + 9*1] = [8 + 0, 12 + 9] = [8 21] [0 1] [0 1] [0*2 + 1*0, 0*3 + 1*1] [0 + 0, 0 + 1] [0 1]So, A^3 = [8 21] [0 1]Wait a second, I see a pattern here. The bottom right entry is always 1, and the bottom left is always 0. The top right entry seems to be increasing in a certain way. Let me note down the top right entries:A^1: 3A^2: 9A^3: 21Hmm, 3, 9, 21... Let me see if I can find a pattern or a recurrence relation here.Looking at the top right entries:From A^1 to A^2: 3 to 9, which is multiplied by 3.From A^2 to A^3: 9 to 21, which is multiplied by roughly 2.333... Hmm, that doesn't seem consistent.Wait, maybe it's not a multiplicative pattern. Let me compute A^4 to see if I can figure it out.A^4 = A^3 * A = [8 21] [2 3] = [8*2 + 21*0, 8*3 + 21*1] = [16 + 0, 24 + 21] = [16 45] [0 1] [0 1] [0*2 + 1*0, 0*3 + 1*1] [0 + 0, 0 + 1] [0 1]So, A^4 = [16 45] [0 1]Top right entry is 45. So the sequence is 3, 9, 21, 45...Looking at the differences between terms:9 - 3 = 621 - 9 = 1245 - 21 = 24Hmm, the differences are 6, 12, 24... which are doubling each time. So, the difference between A^(n+1) and A^n in the top right entry is doubling each time.So, starting from 3:3, then 3 + 6 = 9,9 + 12 = 21,21 + 24 = 45,45 + 48 = 93,and so on.So, it seems like the top right entry follows the pattern where each term is the previous term plus 6*(2)^(n-2). Wait, let me check:For n=1, top right is 3.For n=2, it's 3 + 6 = 9.For n=3, it's 9 + 12 = 21.For n=4, it's 21 + 24 = 45.So, the increment for each step is 6, 12, 24, which is 6*2^(n-2) where n starts at 2.So, the top right entry for A^n is 3 + 6*(2^(n-2) - 1). Wait, let me see:Wait, the increments are 6, 12, 24,... which is a geometric series with first term 6 and ratio 2.So, the total increment after n-1 steps (since starting from n=1) would be 6*(2^(n-1) - 1)/(2 - 1) = 6*(2^(n-1) - 1).Wait, let me test that.For n=2: 6*(2^(2-1) - 1) = 6*(2 - 1) = 6*1 = 6. So, 3 + 6 = 9. Correct.For n=3: 6*(2^(3-1) - 1) = 6*(4 - 1) = 6*3 = 18. So, 3 + 18 = 21. Correct.For n=4: 6*(2^(4-1) - 1) = 6*(8 - 1) = 6*7 = 42. So, 3 + 42 = 45. Correct.So, the top right entry for A^n is 3 + 6*(2^(n-1) - 1) = 3 + 6*2^(n-1) - 6 = 6*2^(n-1) - 3.Simplify that: 6*2^(n-1) - 3 = 3*(2^n - 1).Wait, let me check:6*2^(n-1) = 3*2^n, so 3*2^n - 3 = 3*(2^n - 1). Yes, that's correct.So, the top right entry is 3*(2^n - 1).So, putting it all together, A^n is:[2^n 3*(2^n - 1)][0 1 ]Wait, let me verify this formula with the previous results.For n=1: [2 3*(2 - 1)] = [2 3], which matches A^1.For n=2: [4 3*(4 - 1)] = [4 9], which matches A^2.For n=3: [8 3*(8 - 1)] = [8 21], which matches A^3.For n=4: [16 3*(16 - 1)] = [16 45], which matches A^4.Great, so the formula seems to hold.So, generalizing, A^n is:[2^n 3*(2^n - 1)][0 1 ]Therefore, A^20 is:[2^20 3*(2^20 - 1)][0 1 ]Similarly, A^19 is:[2^19 3*(2^19 - 1)][0 1 ]So, now, we need to compute A^20 - 2*A^19.Let me write that out:A^20 - 2*A^19 = [2^20 3*(2^20 - 1)] - 2*[2^19 3*(2^19 - 1)] [0 1 ] [0 1 ]Let me compute each entry:Top left entry: 2^20 - 2*2^19 = 2^20 - 2^20 = 0.Top right entry: 3*(2^20 - 1) - 2*3*(2^19 - 1) = 3*(2^20 - 1 - 2*2^19 + 2).Simplify the expression inside the parentheses:2^20 - 1 - 2*2^19 + 2 = 2^20 - 2*2^19 + ( -1 + 2 ) = 2^20 - 2^20 + 1 = 0 + 1 = 1.So, top right entry is 3*1 = 3.Bottom left entry: 0 - 2*0 = 0.Bottom right entry: 1 - 2*1 = 1 - 2 = -1.Wait, hold on. Let me double-check the bottom right entry.Wait, A^20 has 1 in the bottom right, and A^19 also has 1 in the bottom right. So, 1 - 2*1 = -1. That seems correct.Wait, but in our formula, A^n has 1 in the bottom right, so A^20 - 2*A^19 would have 1 - 2*1 = -1.But let me think about this. Is that correct?Wait, let me re-examine the computation step by step.Compute A^20 - 2*A^19:Top left: 2^20 - 2*2^19 = 2^19*(2 - 2) = 0.Top right: 3*(2^20 - 1) - 2*3*(2^19 - 1) = 3*(2^20 - 1 - 2*2^19 + 2).Simplify inside:2^20 - 2*2^19 = 2^19*(2 - 2) = 0.Then, -1 + 2 = 1.So, total inside is 0 + 1 = 1.Multiply by 3: 3*1 = 3.Bottom left: 0 - 2*0 = 0.Bottom right: 1 - 2*1 = -1.So, putting it all together, A^20 - 2*A^19 is:[0 3][0 -1]Wait, that seems consistent.But let me verify this result another way. Maybe by using the formula for A^n and plugging in n=20 and n=19, then subtracting.But before that, let me think if there's another approach, maybe using eigenvalues or diagonalization.Given that A is a 2x2 matrix, perhaps we can diagonalize it or find its eigenvalues and eigenvectors to compute A^n more easily.Let me try that approach as a cross-check.First, find the eigenvalues of A.The characteristic equation is det(A - λI) = 0.Compute determinant:|2 - λ 3 ||0 1 - λ |Determinant = (2 - λ)(1 - λ) - 0 = (2 - λ)(1 - λ).Set to zero: (2 - λ)(1 - λ) = 0.So, eigenvalues are λ = 2 and λ = 1.So, A has eigenvalues 2 and 1.Since A is upper triangular and has distinct eigenvalues, it is diagonalizable.Therefore, we can write A = PDP^{-1}, where D is the diagonal matrix of eigenvalues.Then, A^n = PD^nP^{-1}.So, let me find the eigenvectors to construct P.For λ = 2:Solve (A - 2I)v = 0.A - 2I = [0 3] [0 -1]So, the equations are:0*v1 + 3*v2 = 00*v1 -1*v2 = 0From the second equation: -v2 = 0 => v2 = 0.From the first equation: 3*v2 = 0, which is also satisfied.So, eigenvectors for λ=2 are scalar multiples of [1; 0].For λ = 1:Solve (A - I)v = 0.A - I = [1 3] [0 0]So, the equations are:1*v1 + 3*v2 = 00*v1 + 0*v2 = 0From the first equation: v1 = -3*v2.So, eigenvectors are scalar multiples of [-3; 1].Therefore, matrix P can be formed by the eigenvectors:P = [1 -3] [0 1]And P^{-1} is the inverse of P.Since P is upper triangular with 1s on the diagonal, its inverse is also upper triangular with 1s on the diagonal.Compute P^{-1}:P = [1 -3] [0 1]Inverse of P is [1 3] [0 1]Because:[1 -3] [1 3] = [1*1 + (-3)*0, 1*3 + (-3)*1] = [1, 0][0 1] [0 1] [0*1 + 1*0, 0*3 + 1*1] [0, 1]Yes, that's correct.So, P^{-1} = [1 3] [0 1]Therefore, A^n = P D^n P^{-1}, where D = diag(2, 1).Compute D^n:D^n = [2^n 0] [0 1^n] = [2^n 0] [0 1]So, now compute A^n = P D^n P^{-1}:A^n = [1 -3] [2^n 0] [1 3] [0 1] [0 1] [0 1]First, compute the product [2^n 0; 0 1] [1 3; 0 1]:= [2^n*1 + 0*0, 2^n*3 + 0*1] [0*1 + 1*0, 0*3 + 1*1]= [2^n 3*2^n] [0 1 ]Then, multiply by [1 -3; 0 1]:= [1*2^n + (-3)*0, 1*3*2^n + (-3)*1] [0*2^n + 1*0, 0*3*2^n + 1*1]= [2^n, 3*2^n - 3] [0, 1 ]Which matches our earlier formula for A^n.So, A^n = [2^n 3*(2^n - 1)] [0 1 ]Therefore, the previous computation is correct.So, going back, A^20 - 2*A^19 is:[0 3][0 -1]Wait, but let me think again about the bottom right entry. When I computed it earlier, I got -1, but is that correct?Wait, let me re-examine:A^20 has 1 in the bottom right, and A^19 also has 1 in the bottom right. So, 1 - 2*1 = -1. That seems correct.But let me think about the structure of the matrix. The bottom right entry is 1 for all A^n, so subtracting 2*A^19 would subtract 2 from the 1, giving -1. That makes sense.Similarly, the top right entry is 3*(2^20 - 1) - 2*3*(2^19 - 1) = 3*(2^20 - 1 - 2*2^19 + 2) = 3*(2^20 - 2^20 +1) = 3*1 = 3.Top left is 2^20 - 2*2^19 = 0.So, the resulting matrix is:[0 3][0 -1]Is there another way to see this? Maybe by considering the recurrence relation for A^n.Alternatively, perhaps we can find a recurrence relation for the powers of A.Let me denote A^n as:[ a_n b_n ][ 0 1 ]From the earlier computations, we saw that a_n = 2^n and b_n = 3*(2^n - 1).But let's try to find a recurrence relation for a_n and b_n.From the multiplication A^(n+1) = A^n * A.So,A^(n+1) = [a_n b_n] [2 3] = [a_n*2 + b_n*0, a_n*3 + b_n*1] [0 1] [0 1] [0*2 + 1*0, 0*3 + 1*1]So,a_{n+1} = 2*a_n + 0*b_n = 2*a_nb_{n+1} = 3*a_n + 1*b_nSo, we have the recurrence relations:a_{n+1} = 2*a_nb_{n+1} = 3*a_n + b_nWith initial conditions:a_1 = 2b_1 = 3From a_{n+1} = 2*a_n, we can see that a_n = 2^n, since each term is double the previous.Then, for b_n, we have:b_{n+1} = 3*2^n + b_nWith b_1 = 3.This is a linear recurrence relation. Let's solve it.The recurrence is:b_{n+1} - b_n = 3*2^nThis is a nonhomogeneous linear recurrence. We can solve it by finding the homogeneous solution and a particular solution.First, solve the homogeneous equation:b_{n+1} - b_n = 0Characteristic equation: r - 1 = 0 => r = 1So, homogeneous solution is b_n^{(h)} = C*1^n = C.Now, find a particular solution. Since the nonhomogeneous term is 3*2^n, we can try a particular solution of the form b_n^{(p)} = D*2^n.Plug into the recurrence:b_{n+1}^{(p)} - b_n^{(p)} = 3*2^nD*2^{n+1} - D*2^n = 3*2^nD*2^n*(2 - 1) = 3*2^nD*2^n = 3*2^nSo, D = 3.Therefore, the general solution is:b_n = b_n^{(h)} + b_n^{(p)} = C + 3*2^nNow, apply the initial condition. For n=1, b_1 = 3.So,3 = C + 3*2^1 = C + 6Thus, C = 3 - 6 = -3.Therefore, the solution is:b_n = -3 + 3*2^n = 3*(2^n - 1)Which matches our earlier result.So, that's consistent.Therefore, A^n = [2^n 3*(2^n - 1)] [0 1 ]So, A^20 - 2*A^19 is:[2^20 3*(2^20 - 1)] - 2*[2^19 3*(2^19 - 1)][0 1 ] [0 1 ]Compute each entry:Top left: 2^20 - 2*2^19 = 2^20 - 2^20 = 0Top right: 3*(2^20 - 1) - 2*3*(2^19 - 1) = 3*(2^20 -1 - 2*2^19 + 2) = 3*(2^20 - 2^20 +1) = 3*1 = 3Bottom left: 0 - 2*0 = 0Bottom right: 1 - 2*1 = -1So, the resulting matrix is:[0 3][0 -1]Wait, that seems consistent across both methods: direct computation using the formula and using eigenvalues and diagonalization.Just to be thorough, let me compute A^20 - 2*A^19 using the recurrence relations.From the recurrence relations:a_{n} = 2^nb_{n} = 3*(2^n - 1)So, for n=20:a_20 = 2^20b_20 = 3*(2^20 - 1)For n=19:a_19 = 2^19b_19 = 3*(2^19 - 1)So, A^20 - 2*A^19:Top left: 2^20 - 2*2^19 = 0Top right: 3*(2^20 -1) - 2*3*(2^19 -1) = 3*(2^20 -1 - 2*2^19 + 2) = 3*(2^20 - 2^20 +1) = 3*1 = 3Bottom left: 0 - 2*0 = 0Bottom right: 1 - 2*1 = -1Same result.Therefore, I'm confident that the result is:[0 3][0 -1]So, the final answer is:boxed{begin{pmatrix} 0 & 3 0 & -1 end{pmatrix}}

answer：Okay, so I have this problem where there are five points on a line: P, Q, R, S, T, in that order. I know that the sum of the distances from P to the other four points is 67, and the sum from Q to the other four points is 34. I need to find the length of segment PQ. Hmm, let me try to visualize this.First, since all the points are on a line in the order P, Q, R, S, T, I can represent them as points on a number line. Let me assign coordinates to each point to make it easier. Let's say the coordinate of P is 0. Then, the coordinate of Q would be some positive number, say x. Similarly, R would be at x + y, S at x + y + z, and T at x + y + z + w, where x, y, z, w are the distances between consecutive points. But maybe assigning coordinates like this will complicate things. Alternatively, I can assign coordinates such that P is at 0, Q is at a, R is at b, S is at c, and T is at d, where 0 < a < b < c < d. That might be a better approach because it keeps the order clear.So, let me denote the coordinates as follows:- P: 0- Q: a- R: b- S: c- T: dGiven that, the sum of distances from P to the other points is 67. That would be the distance from P to Q, plus P to R, plus P to S, plus P to T. So, mathematically, that's:a + b + c + d = 67.Similarly, the sum of distances from Q to the other points is 34. That would be the distance from Q to P, plus Q to R, plus Q to S, plus Q to T. Since distance is absolute, but since all points are in order, these distances are just:(a - 0) + (b - a) + (c - a) + (d - a) = 34.Wait, hold on. Let me clarify that. The distance from Q to P is |a - 0| = a. The distance from Q to R is |b - a| = b - a, since b > a. Similarly, Q to S is c - a, and Q to T is d - a. So, adding these up:a + (b - a) + (c - a) + (d - a) = 34.Simplify that:a + b - a + c - a + d - a = 34.So, the a terms cancel out: a - a - a - a = -2a. Wait, no:Wait, let's do it step by step:First term: aSecond term: b - aThird term: c - aFourth term: d - aAdding them together:a + (b - a) + (c - a) + (d - a) = a + b - a + c - a + d - a.Now, combine like terms:a - a - a - a + b + c + d.Which simplifies to:(-2a) + b + c + d = 34.So, we have two equations:1. a + b + c + d = 672. -2a + b + c + d = 34Hmm, interesting. So, if I subtract the second equation from the first, I can eliminate b, c, d.So, equation 1 minus equation 2:(a + b + c + d) - (-2a + b + c + d) = 67 - 34Simplify:a + b + c + d + 2a - b - c - d = 33So, that's 3a = 33Therefore, a = 11.So, the length of segment PQ is 11.Wait, that seems straightforward. Let me just verify.So, if a = 11, then from equation 1:11 + b + c + d = 67 => b + c + d = 56.From equation 2:-22 + b + c + d = 34 => b + c + d = 56.Which is consistent. So, that checks out.Therefore, the length of PQ is 11.I think that's the answer. It seems a bit too straightforward, but the equations worked out. Let me just think again.We had two equations:1. Sum from P: a + b + c + d = 672. Sum from Q: a + (b - a) + (c - a) + (d - a) = 34, which simplifies to -2a + b + c + d = 34.Subtracting equation 2 from equation 1 gives 3a = 33, so a = 11.Yes, that seems correct. So, PQ is 11 units long.**Final Answer**The length of segment ( PQ ) is boxed{11}.