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

Question

1683 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-16T11:30:55

#((1, 0, 0, 0),(0,0,0,1),(0,1,0,0),(0,0,1,0))#

#A = ((1, 0, 0, 0),(0,0,1,0),(0,0,0,1),(0,1,0,0))#

Notice what this matrix does as a linear transformation:

#((1, 0, 0, 0),(0,0,1,0),(0,0,0,1),(0,1,0,0))((a),(b),(c),(d)) = ((a),(c),(d),(b))#

It cyclically permutes three ordinates.

Therefore if applied #3# times it results in the identity:

#A^3 = I#

So the inverse of #A# is:

#A^2=((1, 0, 0, 0),(0,0,1,0),(0,0,0,1),(0,1,0,0))((1, 0, 0, 0),(0,0,1,0),(0,0,0,1),(0,1,0,0)) = ((1, 0, 0, 0),(0,0,0,1),(0,1,0,0),(0,0,1,0))#