Please solve in English

Here's the continuation of the solution for Question 4, now in English. Question 4: Matrix and Curve Length a) Given the matrix M = 3 & 1 \\ 0 & 2 , prove by induction that M^n = 3^n & 3^n-2^n \\ 0 & 2^n for all positive integers n. Step 1: Base Case (n=1). We need to show that the formula holds for n=1. M^1 = 3^1 & 3^1-2^1 \\ 0 & 2^1 = 3 & 3-2 \\ 0 & 2 = 3 & 1 \\ 0 & 2 . This matches the given matrix M. So, the base case holds. Step 2: Inductive Hypothesis. Assume that the formula holds for some positive integer k. That is, assume: M^k = 3^k & 3^k-2^k \\ 0 & 2^k Step 3: Inductive Step (n=k+1). We need to prove that the formula holds for n=k+1, i.e., M^k+1 = 3^k+1 & 3^k+1-2^k+1 \\ 0 & 2^k+1 . We know that M^k+1 = M^k · M. Substitute the inductive hypothesis for M^k and the given matrix M: M^k+1 = 3^k & 3^k-2^k \\ 0 & 2^k 3 & 1 \\ 0 & 2 Perform the matrix multiplication: The element in the first row, first column is: (3^k)(3) + (3^k-2^k)(0) = 3^k+1 The element in the first row, second column is: (3^k)(1) + (3^k-2^k)(2) = 3^k + 2 · 3^k - 2 · 2^k = 3^k + 2 · 3^k - 2^k+1 = 3 · 3^k - 2^k+1 = 3^k+1 - 2^k+1 The element in the second row, first column is: (0)(3) + (2^k)(0) = 0 The element in the second row, second column is: (0)(1) + (2^k)(2) = 2^k+1 So, we get: M^k+1 = 3^k+1 & 3^k+1-2^k+1 \\ 0 & 2^k+1 This matches the formula for n=k+1. By the principle of mathematical induction, the formula is true for all positive integers n. M^n = 3^n & 3^n-2^n \\ 0 & 2^n b) Find the length of the curve y = (x^3)/(6) + (1)/(2x) from x=1 to x=2. Step 1: Find the derivative (dy)/(dx). Given y = (x^3)/(6) + (1)/(2x) = (1)/(6)x^3 + (1)/(2)x^-1. Differentiate with respect to x: (dy)/(dx) = (1)/(6)(3x^2) + (1)/(2)(-1x^-2) = (1)/(2)x^2 - (1)/(2x^2) Step 2: Calculate 1 + ((dy)/(dx))^2. ((dy)/(dx))^2 = ((1)/(2)x^2 - (1)/(2x^2))^2 = (1)/(4)x^4 - 2((1)/(2)x^2)((1)/(2x^2)) + (1)/(4x^4) = (1)/(4)x^4 - (1)/(2) + (1)/(4x^4) Now, add 1: 1 + ((dy)/(dx))^2 = 1 + (1)/(4)x^4 - (1)/(2) + (1)/(4x^4) = (1)/(4)x^4 + (1)/(2) + (1)/(4x^4) Notice that this expression is a perfect square: (1)/(4)x^4 + (1)/(2) + (1)/(4x^4) = ((1)/(2)x^2 + (1)/(2x^2))^2 Step 3: Set up the arc length integral. The arc length L of a curve y=f(x) from x=a to x=b is given by the formula: L = _a^b sqrt(1 + ((dy)/(dx))^2)\,dx Substitute the expression from Step 2: L = _1^2 sqrt(((1)/(2)x^2 + (1)/(2x^2))^2)\,dx L = _1^2 ((1)/(2)x^2 + (1)/(2x^2))\,dx Step 4: Evaluate the integral. L = _1^2 ((1)/(2)x^2 + (1)/(2)x^-2)\,dx L = [ (1)/(2)(x^3)/(3) + (1)/(2)x^-1-1 ]_1^2 L = [ (x^3)/(6) - (1)/(2x) ]_1^2 Evaluate at the limits: L = ( (2^3)/(6) - (1)/(2(2)) ) - ( (1^3)/(6) - (1)/(2(1)) ) L = ( (8)/(6) - (1)/(4) ) - ( (1)/(6) - (1)/(2) ) L = ( (4)/(3) - (1)/(4) ) - ( (1)/(6) - (3)/(6) ) L = ( (16)/(12) - (3)/(12) ) - ( -(2)/(6) ) L = (13)/(12) - ( -(1)/(3) ) L = (13)/(12) + (1)/(3) = (13)/(12) + (4)/(12) = (17)/(12) The length of the curve is (17)/(12). That's 2 down. 3 left today — send the next one.