Discussion
Matrix
\[ A = \begin{pmatrix} 1 & 3 & -2 \\ 0 & 4 & -2 \\ 0 & 3 & -1 \end{pmatrix} \]
is similar to :
Answer: B
A is similar to the diagonal matrix diag(1, 1, 2).
Expanding along the first column, the characteristic polynomial is \((1 - \lambda)\) times that of the block \(\begin{pmatrix} 4 & -2 \\ 3 & -1 \end{pmatrix}\), whose trace is 3 and determinant is \(-4 + 6 = 2\). So the block has eigenvalues 1 and 2, and A has eigenvalues 1, 1, 2.
To decide between the diagonal form and a Jordan block for 1, compute
\[ A - I = \begin{pmatrix} 0 & 3 & -2 \\ 0 & 3 & -2 \\ 0 & 3 & -2 \end{pmatrix} \]
This has rank 1, so the eigenspace for 1 has dimension \(3 - 1 = 2\), equal to the algebraic multiplicity. A is diagonalisable, so it is similar to diag(1, 1, 2).
The matrix with a 1 above the diagonal in the eigenvalue-1 block would need a one-dimensional eigenspace. The options with eigenvalues 1, 2, 2 have the wrong trace: trace A = 1 + 4 - 1 = 4, not 5.
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