Part A

(a) A is a 2 x2 matrix and 5(A + 1) = I. Enter det (A + I). (b) A is a 4 x4 matrix and 2 A + 3 1 = 0. Enter det (A + I). (C) A is a 3 x3 matrix and A- - 4 A - 5 1 = 0. If det ( A - 2 1 ) > 0, enter det (A - 2 1).You will integrate the following rational function: 7x-5 (r+1) (x-2) To do this, start by writing 7x-5 B (x+1) (x-2) I- (a) Put the right-hand side over a common denominator. Enter the numerator of the result. Expand the numerator. It is a linear function of , which must equal the numerator of the original function. By comparing coefficients you should get linear equations in B and C. Enter your equations with the constant on the right hand side, e.g. B+2*C=3. What equation can you deduce from: (b) the coefficients of r: Fi (c) the constants (the coefficients of 1): (d) Solve these two equations, and enter the solution in the form B, C. B, C = (e) Use these values, and the partial fraction form, to calculate the integral. Use a constant of integration k. Don't forget that the integral of _ is In (x| which has syntax In(abs(x)).We will evaluate the following improper integral: I 2 (2 + 2)2 First we will do the indefinite integral I (12 + 2)2 dx (a) Use the substitution u = + 2 to formulate as an integral with respect to u, filling in the integrand below (remember to use proper Mobius syntax in your function) A du (b) Now integrate to get a function of u: (c) Substitute the value of u interms of c to get an antiderivative in terms of r, we will call this F(I): F(I) = (d) Now to do the improper integral we require lim dx (-+00 J2 (2 + 2)2 so for this we need [F(x)15 (the answer here is a function of a) (e) Finally to evaluate the answer take the limit as a - co. Enter the exact answer (e.g. as a fraction if necessary, not a decimal)