The (x, y) coordinates of the hand at the top of a robot arm are given by: x = L, cos(8,) + L2 cos(8, + 0,) y = L, sin(8,) + L,sin(8, + 02) Where L1 and L2 are the distances of the parts of the arm from the shoulder to he elbow and from the elbow to the hand, Đ1 and 62 are the angles that these arts make with the x-axis. These angles can be expressed as: 0,(t) = a,t³ + a,t“ + azt³ 02(t) = b,t5 + bzt“ + bąt³ vhere t is the time. We assume that the arm starts to move at t = 0s and stops t tf=2s. The conditions that the angular position 61(t), angular velocity and ngular acceleration should meet at the final time tf= 2 (in sec) are expressed n the following matrix form and allow us to obtain the coefficients a1, az and 3: [a 3x22||a2 园- [01f – 010] 24 4x23 20x23 12x22 6x2 Jlaz] 25 23 5x24 vith 0lf= 437/180, 010 = -197/180. a) Use MATLAB left division to solve the above system and get the oefficients a1, az and az ) Similarly the coefficients b1, b2 and b; can be obtained by solving a similar ystem: 25 24 23 [82f – 0201 5x24 20x23 12x22 6x2 3x22 ||b2 = 4x23 [b3. vith 02f=1517/180, 020 = 447/180. Obtain the coefficients bi bɔ and b; using MATLAB matrix inversion