Question

3 -1-4 210.

Answer 1

Consider a matrix A

A=[ 3 −1 −4 2 ]

The expression for the matrix A is,

A=IA

[ 3 −1 −4 2 ]=[ 1 0 0 1 ]A

R 1 → 1 3 R 1 [ 3 3 −1 3 −4 2 ]=[ 1 3 0 3 0 1 ]A [ 1 −1 3 −4 2 ]=[ 1 3 0 0 1 ]A

R 2 → R 2 +4 R 1 [ 1 ( −1 3 ) −4+4( 1 ) 2+4( −1 3 ) ]=[ ( 1 3 ) 0 0+4( 1 3 ) 1+4( 0 ) ]A [ 1 ( −1 3 ) 0 2 3 ]=[ ( 1 3 ) 0 4 3 1 ]A

R 2 → 3 2 R 2 [ 1 −1 3 3 2 ( 0 ) 3 2 ( 2 3 ) ]=[ 1 3 0 3 2 ( 4 3 ) 3 2 ( 1 ) ]A [ 1 −1 3 0 1 ]=[ 1 3 0 2 3 2 ]A

R 1 → R 1 + 1 3 R 2 [ 1+ 1 3 ( 0 ) −1 3 + 1 3 ( 1 ) 0 1 ]=[ 1 3 + 1 3 ( 2 ) 0+ 1 3 ( 3 2 ) 2 3 2 ]A [ 1 0 0 1 ]=[ 1 ( 1 2 ) 2 3 2 ]A I=[ 1 ( 1 2 ) 2 3 2 ]A

Compare the above resultant expression withthe identity,

I= A −1 A

Thus, the inverse of the matrix A is [ 1 ( 1 2 ) 2 3 2 ]