How do you find the inverse of #A=##((1, 2, 1), (2, 5, 4), (1, 4, 9)) #?

1 Answer
Mar 2, 2017

Answer:

The inverse is #=((29/4,-7/2,3/4),(-7/2,2,-1/2),(3/4,-1/2,1/4))#

Explanation:

For a matrix #A# to be invertible, the determinant #detA!=0#

Let's calculate

#detA=|(1,2,1),(2,5,4),(1,4,9)|#

#=1*|(5,4),(4,9)|-2*|(2,4),(1,9)|+1*|(2,5),(1,4)|#

#=1*(45-16)-2*(18-4)+1*(8-5)#

#=29-28+3#

#=4#

As #detA!=0#, matrix #A# is invertible

We calculate the matrix of cofactors

#C=((|(5,4),(4,9)|,-|(2,4),(1,9)|,|(2,5),(1,4)|),(-|(2,1),(4,9)|,|(1,1),(1,9)|,-|(1,2),(1,4)|),(|(2,1),(5,4)|,-|(1,1),(2,4)|,|(1,2),(2,5)|))#

#=((29,-14,3),(-14,8,-2),(3,-2,1))#

The transpose of #C# is

#C^T=((29,-14,3),(-14,8,-2),(3,-7,1))#

The inverse is

#A^-1=1/detA*C^T#

#=1/4((29,-14,3),(-14,8,-2),(3,-2,1))#

#=((29/4,-14/4,3/4),(-14/4,8/4,-2/4),(3/4,-2/4,1/4))#

Verification

#A*A^-1=((1,2,1),(2,5,4),(1,4,9))*((29/4,-14/4,3/4),(-14/4,8/4,-2/4),(3/4,-2/4,1/4))#

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

#=I#