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

Related questions
See all questions in Inverse Matrix
Impact of this question
2727 views around the world
You can reuse this answer
Creative Commons License