How do you find the inverse of #A=##((-7, 5), (5, -4))#?

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Answer 1 · 2016-10-27T11:58:10

The inverse is #((-4/3,-5/3),(-5/3,-7/3))#

DetA=Det #((-7,5),(5,-4))=(-7*-4)-(5*5)=28-25=3#

The determinant is #!=0#, so the inverse exists

Then we calculate the cofactors

#A_11=-4# #A_12=-5#
#A_21=-5# #A_22=-7#

So the cofactor matrix is #((-4,-5),(-5,-7))#

And the transpose is #((-4,-5),(-5,-7))#

And finally we divide all the components by the determinant
#A^(-1)=((-4/3,-5/3),(-5/3,-7/3))#

We have to check by multiplying #A*A^(-1)#

#((-7,5),(5,-4))*((-4/3,-5/3),(-5/3,-7/3))=((1,0),(0,1))#