How do you find the inverse of #A=##((3, 0), (4, 17))#?

1 Answer
Nov 10, 2016

Answer:

The inverse is #A^(-1)=((1/3,0),(-4/51,1/17))#

Explanation:

The inverse of a matrix #((a,b),(c,d))# is #=1/(ad-bc)((d,-b),(-c,a))#

Here #A=((3,0),(4,17))#

The determinant of the matrix is #=(3*17-4*0)=51#

The determinant #!=0#, therefore it is invertible.

So, #A^(-1)=1/51((17,0),(-4,3))=((1/3,0),(-4/51,1/17))#

Verification
#A*A^(-1)=((3,0),(4,17))*((1/3,0),(-4/51,1/17)))=((1,0),(0,1))=I#
#I=# Identity Matrix