In the truss shown above, applied loads and reaction in the x direction are positive to the right and in the y direction are positive upwards. Analysis of the truss results in the following system of equations (where members in compression will be positive and members in tension will be negative): 110000 0 0 0 0 A 0 0 1-10 0 0 0 0 0 AD 00 0 0 1 0,6 0 0 0 A, 0 0 0 1 0 0.8 0 0 AB 0 0 0 1 0 -0.6 0 BC 0 0 0.8 0 BD 0.6 0 0 CE 000 0 0 0 1 10 0 0 0 000 ooooo 0 0 000 0 0-0,8 ...TO 0 10 00 0 1-0.667 1 0.667 0 0-0.667 1 0.667 00 0.500 0 00 00 0.833 0 0.500 0.833 0 0.833 0.833 0.500 0 0-0.500 0 0 0-0.667 0 0.667 00 0.667 0 -0.667 0 0 0 1 0 0 -1 0 DE 0 0.6 0 0 CD 000000 -0.8 0 0 1 E 0 0 -1 0 10 The inverse of the matrix [A] is: -1 0 1 -1 0 1 OTO i 0 OOKOO 0 0 20000 0 1010 10 0 10 0.500 0 0.500 0 0 0.500 0 0.500 00 0.375 0 0.375 0 0 -0.625 0 -0.625 0 0 0.625 0 0.675 0 0 -0.375 0 -0.375 1 0 0.500 0 0.500 0 0 0.500 0 0.500 01 If F1 = 42 and F2 F -4, find the force in member CD.