express the following equation in matrix equation and solve them byfinding the inverse of coefficient matrix by elementary transformation method (1) x+y+z=6 x-y+2z=5 2x+y-z=1 (2) 2x+3y-z=11 x+2y+z = 8 3x-y-2z = 5 Share with your friends Share 0 Anuradha Sharma answered this We know that A= IAHere the coefficient matrix is, A=1111-1221-1So we get1111-1221-1=100010001AApply R2 = R2 - R1, we get1110-2121-1 = 100-110001AApply R3 = R3 - 2R1, we get1110-210-1-3 = 100-110-201AApply R1 = R1 + R3, we get10-20-210-1-3 = -101-110-201AApply R2 = R2 - 3R3, we get10-201100-1-3 = -10151-3-201AApply R3 = R3 + R2, we get10-20110007 = -10151-331-2AApply R3 = 17R3, we get10-20110001 = -10151-33/71/7-2/7AApply R1 = R1 + 2R3, we get1000110001 = -1/72/73/751-33/71/7-2/7AApply R2 = R2 - 10R3, we get100010001 = -1/72/73/75/7-3/7-1/73/71/7-2/7Ahence A-1 = -1/72/73/75/7-3/7-1/73/71/7-2/7Now, the given equation in matrix form is,AX=B ............1where A = 1111-1221-1, X = xyz and B = 651premultiplying 1 by A-1 we get, A-1AX=A-1B⇒IX=A-1B⇒X =A-1B = -1/72/73/75/7-3/7-1/73/71/7-2/7 651 = 17-1235-3-131-2 651 =17 -6+10+330-15-118+5-2 = 1771421 =123 So x = 1, y = 2, z = 3With the similar approach try to solve second question on your own. 0 View Full Answer
