Question

If `A=[[1,-1,2],[3,0,-2],[1,0,3]]` verify that A (adj A) = |A| I.

Answer 1

`A=[[1,-1,2],[3,0,-2],[1,0,3]]`

`|A|=|[1,-1,2],[3,0,-2],[1,0,3]|`

`=1(0)+1(9+2)+2(0)`

`=0+11+0`

`therefore |A|=11`

`A_11=(-1)^(1+1)M_11=1|[0,-2],[0,3]|=1(0-0)=1xx0=0`

`A_12=(-1)^(1+2)M_12=-1|[3,-2],[1,3]|=-1(9+2)=-11`

`A_13=(-1)^(1+3)M_13=1|[3,0],[1,0]|=1(0-0)=1xx0=0`

`A_21=(-1)^(2+1)M_21=-1|[-1,2],[0,3]|=-1(-3-0)=1xx0=3`

`A_22=(-1)^(2+2)M_22=1|[1,2],[1,3]|=1(3-2)=1`

`A_23=(-1)^(2+3)M_23=-1|[1,-1],[1,0]|=-1(0+1)=-1`

`A_31=(-1)^(3+1)M_31=1|[-1,2],[0,-2]|=1(2-0)=1xx2=2`

`A_32=(-1)^(3+2)M_32=-1|[1,2],[3,-2]|=-1(-2-6)=8`

`A_33=(-1)^(3+3)M_33=1|[1,-1],[3,0]|=1(0+3)=1xx3=3`

Hence, matrix of the co-factors is

`[[A_11,A_12,A_13],[A_21,A_22,A_23],[A_31,A_32,A_33]]=[[0,-11,0],[3,1,-1],[2,8,3]]=[A_(ij)]_(3xx3)`

`Now , adjA=[A_(ij)]_(3xx3)^T=[[0,3,2],[-11,1,8],[0,-1,3]]`

`A(adjA)=[[1,-1,2],[3,0,-2],[1,0,3]][[0,3,2],[-11,1,8],[0,-1,3]]`

            `=[[0+11+0,3-1-2,2-8+6],[0+0+0,9+0+2,6+0-6],[0+0+0,3+0-3,2+0+9]]`

            `=[[11,0,0],[0,11,0],[0,0,11]]`......................(1)

`|A|.I=11[[1,0,0],[0,1,0],[0,0,1]]`

     `=[[11,0,0],[0,11,0],[0,0,11]]`......................(2)

From equations (1) and (2), we get A(adj A) = |A|. I