R(A)= R(A ~= 2,
Entonces hay
(k 2)(k-1)=(k-1)=0,
Entonces k=1
②Solución: Porque A=
.3 2 1
2 1 0
1 0 0
Por lo tanto
a 11 = 1×0-0× 0 = 0,
A21=-(2×0-1×0)=0,
a 31 = 2×0-1×1 =-1;
A12=-(2×0-1×0)=0,
a22 = 3×0-1×1 =-1,
a32 =-( 3×0-2×1)= 2;
a 13 = 2×0-1×1 =-1,
A23=-(3×0-2×1 )= 2,
a33 = 3×1-2×2 =-1;
Entonces A*=
0 0 -1
0 -1 2
-1 2 -1
Y | a =-(1×1×1)=-1
Entonces -1 =
0 0 1
0 1 -2
1 -2 1