How do you find the inverse of #A=##((-3, -3, -4), (0, 1, 1), (4, 3, 4))#?

1 Answer
Feb 2, 2017

Answer:

The answer is #=((1,0,1),(4,4,3),(-4,-3,-3))#

Explanation:

We start by calculating the determinant of matrix #A# to see if it's invertible

#det A=|(-3,-3,-4),(0,1,1),(4,3,4)|#

#=-3*|(1,1),(3,4)|+3*|(0,1),(4,4)|-4*|(0,1),(4,3)|#

#=(-3*1)+(3*-4)-4(-4)#

#=-3-12+16=1#

Therefore,

#detA!=0#, the matrix is invertible

We start by calculating the matrix of cofactors

#C=((|(1,1),(3,4)|,-|(0,1),(4,4)|,|(0,1),(4,4)|),(-|(-3,-4),(3,4)|,|(-3,-4),(4,4)|, -|(-3,-3),(4,3)|),(|(-3,-4),(1,1)|,-|(-3,-4),(0,1)|,|(-3,-3),(0,1)|))#

#C=((1,4,-4),(0,4,-3),(1,3,-3))#

Now, we determine the transpose of #C#

#C^T=((1,0,1),(4,4,3),(-4,-3,-3))#

So,

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

#A^-1=C=((1,0,1),(4,4,3),(-4,-3,-3))#

Verification

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

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