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

1 Answer
Jul 11, 2016

Answer:

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))#