How do you find the inverse of #[(4,2) (1,2)]#?

1 Answer
Dec 8, 2016

Answer:

The inverse is #=((1/3,-1/3),(-1/6,2/3))#

Explanation:

If the matrix is

#A=((a,b),(c,d))#

the inverse is

#A^(-1)=1/(ad-bc)((d,-b),(-c,a))#

First we calculate the determinant of the matrix

#| (4,2), (1,2) | =8-2=6!=0#

The determinant is #!=0#, so the matrix is invertible

The inverse is #=1/6((2,-2),(-1,4))#

#=((1/3,-1/3),(-1/6,2/3))#

Verification,

We calculate #A*A^(-1)=((4,2),(1,2))*((1/3,-1/3),(-1/6,2/3))#

#=((4/3-2/6,-4/3+4/3),(1/3-2/6,-1/3+4/3))#

#=((1,0),(0,1))=I#