Consider the system of linear equations in x, y, z : x + 2y + tz = 0 , 6x + y + 5tz = 0 , 3x + t^2 y + f(t) z = 0 , where f: R R is a differentiable function. If this system has infinitely many solutions for all t R , then f

Step-by-step solution

For a homogeneous system of linear equations to have infinitely many solutions, the determinant of its coefficient matrix must be zero. The coefficient matrix is: = vmatrix 1 & 2 & t 6 & 1 & 5t 3 & t^2 & f(t) vmatrix Expanding the determinant along the first row, we get: = 1(1 f(t) - 5t t^2) - 2(6 f(t) - 3 5t) + t(6 t^2 - 3 1) = f(t) - 5t^3 - 12f(t) + 30t + 6t^3 - 3t = -11f(t) + t^3 + 27t Since the system has infinitely many solutions for all t R , we must have = 0 for all t R . -11f(t) + t^3 + 27t = 0 f(t) = t^3 +