Word Count : 769

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We are asked to find the minimal polynomial of $$ A = \begin{bmatrix} 2 & 1 & 0 & 1\\ -1 & 0 & 0 & 1\\ -2 & -2 & -1 & 3\\ 0 & 0 & 0 & 1 \end{bmatrix} $$ Step 1: Check eigenvalues (roots of characteristic polynomial) The minimal polynomial divides the characteristic polynomial. Let's first compute the characteristic polynomial $\det(A - \lambda I)$. $$ A - \lambda I = \begin{bmatrix} 2-\lambda & 1 & 0 & 1\\ -1 & -\lambda & 0 & 1\\ -2 & -2 & -1-\lambda & 3\\ 0 & 0 & 0 & 1-\lambda \end{bmatrix} $$ We notice that the last row is $[0,0,0,1-\lambda]$, so the determinant can be computed by expanding along the last row: $$ \det(A-\lambda I) _______ _______ ____ _________ __________.
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