a) The following row manipulations on the identity matrix produce the inverse matrix:
- R1←R2-R1
- R2←R2-R1
- R3←R3-R2
- R2←R2-R1
- R3←R3-R2
- R2←R2-R3
- R1←R1-4R3
(It would take too long to show the transformation step by step, but you can perform the 7 steps above to see what happens to the original matrix and the identity matrix. R1 means row 1, etc., and the arrow shows how a particular row is altered by carrying out the operation on the right.)
When these operations are performed on the original matrix they reduce it to the identity matrix. When applied to the identity matrix, we get the inverse matrix:
( 11 -3 -4 )
( 5 -2 -1 )
( -3 1 1 )
b) When this is multiplied by the original matrix we get the identity matrix:
( 11 -3 -4 )( 1 1 5 )
( 5 -2 -1 )( 2 1 9 )
( -3 1 1 )( 1 2 7 )=
( 11-6-4=1 11-3-8=0 55-27-28=0 )
( 5-4-1=0 5-2-2=1 25-18-7=0 )
( -3+2+1=0 -3+1+2=0 -15+9+7=1 )
c) Apply the inverse matrix to the constants:
( 11 -3 -4 )( 0 )
( 5 -2 -1 )( 30 )
( -3 1 1 )( 10 )=
( 0-90-40=-130 )
( 0-60-10=-70 )
( 0+30+10=40 )
So (x,y,z)=(-130,-70,40).