How do you find the inverse of #[(2,-3), (-2,-2)]#?

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Answer 1 · 2017-01-11T08:25:26

The answer is #=((1/5,-3/10),(-1/5,-1/5))#

The inverse of the matrix #((a,b),(c,d))# is

#1/(ad-bc)*((d,-b),(-c,a))#

Let #A=((2,-3),(-2,-2))#

#DetA= | (a,b), (c,d) |=ad-bc=-4-6=-10#

As #DetA!=0#, the matrix is invertible

#A^-1=-1/10((-2,3),(2,2))#

#=((1/5,-3/10),(-1/5,-1/5))#

Verification

#A*A^-1=((2,-3),(-2,-2))*((1/5,-3/10),(-1/5,-1/5))#

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