How do you find the inverse of A=((2, 4, 1),(-1, 1, -1), (1, 4, 0))?

1 Answer
Jul 11, 2016

The inverted matrix is: ((-4,-4,5),(1,1,-1),(5,4,-6))

Explanation:

There are many ways in invert matrices, but for this problem I used the cofactor transpose method.

If we imagine that
A = ((vecA),(vecB),(vecC))

So that:
vecA = (2,4,1)
vecB = (-1,1,-1)
vecC = (1,4,0)

Then we can define reciprocal vectors:
vecA_R = vecB xx vecC
vecB_R = vecC xx vecA
vecC_R = vecA xx vecB

Each is easily calculated using the determinant rule for cross products:
vecA_R = |(hati,hatj,hatk),(-1,1,-1),(1,4,0)| = (4,-1,-5)
vecB_R = |(hati,hatj,hatk),(-1,4,0),(2,4,1)| = (4,-1,-4)
vecC_R = |(hati,hatj,hatk),(2,4,1),(-1,1,-1)| = (-5,1,6)

We can use these to construct the cofactor transpose of M, barM, as such:
barM = ((vecA_R^T,vecB_R^T,vecC_R^T)) = ((4,4,-5),(-1,-1,1),(-5,-4,6))

The reciprocal vectors and the cofactor transpose matrix have two interesting properties:
vecA*vecA_R=vecB*vecB_R=vecC*vecC_R = det(M)
and
M^-1 = barM/detM

So we can determine that:
det(M) = vecC*vecC_R = (1,4,0)*(-5,1,6) = -1

This means that:
M^-1 = -barM/1 = -((4,4,-5),(-1,-1,1),(-5,-4,6)) = ((-4,-4,5),(1,1,-1),(5,4,-6))