 Solution

For $A$, the triangular matrix formula gives the characteristic polynomialThe three eigenvalues are distinct, so part (d) already proves the existence of an eigenvector basis. Explicit choices, ordered by eigenvalues $1,3,2$, areTheir column matrix has determinant one, and direct multiplication gives respectively one, three and two times the indicated vector.

For $B$, expansion along the last row givesAgain the three eigenvalues are distinct. An explicit eigenvector basis, ordered by eigenvalues $4,2,3$, isIts column matrix has determinant $-2\ne 0$. In particular, $B$'s diagonal entries are not its eigenvalues: it is not a triangular matrix.

Finally $C=I_3$, so every nonzero vector is an eigenvector with eigenvalue one. Choose the standard basis $e_1,e_2,e_3$. These are linearly independent even though their eigenvalues coincide. Distinct eigenvalues are sufficient for an eigenvector basis, and are not necessary.