How do you find the inverse of #A=##((1, -1, 0), (1, 0, -1), (-6, 2, 3))#?

1 Answer
Jan 2, 2017

Answer:

The answer is #=((-2,-3,-1),(-3,-3,-1),(-2,-4,-1))#

Explanation:

#A=((1,-1,0),(1,0,-1),(-6,2,3))#

We start by calculating the determinant

#detA= | (1,-1,0), (1,0,-1), (-6,2,3) | #

#=1| (0,-1), (2,3) |+1| (1,-1), (-6,3) |+0| (1,0), (-6,2) |#

#=(2)+(-3)=-1#

As #detA=-1#, the matrix is invertible

To determine the inverse #A^(-1)#, we start by calculating the matrix of cofactors

#C=((| (0,-1), (2,3) |,-| (1,-1), (-6,3) |,| (1,0), (-6,2) |),(-| (-1,0), (2,3) |,| (1,0), (-6,3) |,-| (1,-1), (-6,2) |),(| (-1,0), (0,-1) |,-| (1,0), (1,-1) |,| (1,-1), (1,0) |))#

#=((2,3,2),(3,3,4),(1,1,1))#

Now we calculate the traspose matrix of #C#

#C^T=((2,3,1),(3,3,1),(2,4,1))#

To obtain the inverse of #A#, we divide #C^T# by the determinant

#A^-1=C^T/det(A)#

#=((-2,-3,-1),(-3,-3,-1),(-2,-4,-1))#

Verification

#A*A^-1=((1,-1,0),(1,0,-1),(-6,2,3))*((-2,-3,-1),(-3,-3,-1),(-2,-4,-1))#

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