How do you find the inverse of #A=##((-3, -1), (-7, 8))#?

2 Answers
Cem Sentin
Mar 12, 2018

#A^(-1)=##((-8/31, -1/31), (-7/31, 3/31))#

Explanation:

#A=##((-3, -1), (-7, 8))#

#Cof(A)=##((8, 7), (1, -3))#

#Adj(A)=##((8, 1), (7, -3))#

#Det(A)=(-3)*8-(-1)(-7)=-31#

Thus, #A^(-1)=(Adj(A))/(Det(A))#

#A^(-1)=##((-8/31, -1/31), (-7/31, 3/31))#

Y
Jul 14, 2018

Answer: #[(-8/31,-1/31),(-7/31,3/31)]#

Explanation:

An easy way to find the inverse of a 2x2 square matrix is to apply the following formula*:
Let #A=[(a,b),(c,d)]#, then
#A^(-1)=1/(det(A))[(d,-b),(-c,a)]#

For this problem, we have
#A=[(-3,-1),(-7,8)]#, so #a=-3,b=-1,c=-7,d=8#

We can find the determinant:
#det(A)=ad-bc=-24-7=-31#

Plugging in our values, we have:
#A^(-1)=1/(-31)[(8,1),(7,-3)]=[(-8/31,-1/31),(-7/31,3/31)]#, which is our answer

*Note: This formula only works for 2x2 matrices and does not work for larger matrices

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