How do you find the inverse of #A=##((1, 0, -2), (3, 1, -6), (0, 1, 1)) #?

1 Answer

Answer:

#( (7, -2, 2), (-3, 1, 0), (3, -1, 1))#

Explanation:

#((1,0,-2), (3,1,-6), (0,1,1)) * ((x_1, x_2, x_3), (y_1, y_2, y_3), (z_1, z_2, z_3)) = ((1,0,0), (0,1,0), (0,0,1))#

#L_2 := L_2 - 3 L_1#

#((1,0,-2), (0,1,0), (0,1,1)) * ((x_1, x_2, x_3), (y_1, y_2, y_3), (z_1, z_2, z_3)) = ((1,0,0), (-3,1,0), (0,0,1))#

#L_2: y_1 = -3, y_2 = 1, y_3 = 0#

#L_3: y_1 + z_1 = 0 ; y_2 + z_2 = 0; y_3 + z_3 = 1#

#z_1 = 3 ; z_2 = -1; z_3 = 1#

#L_1: x_1 - 2 z_1 = 1 ; x_2 - 2 z_2 = 0 ; x_3 - 2 z_3 = 0#

#x_1 = 7 ; x_2 = - 2 ; x_3 = 2#