Word Count : 554

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We are to find the eigenvalues and eigenvectors of $$ B = \begin{bmatrix} 1 & 1 & 0 \\ -1 & 3 & 0 \\ 1 & -1 & 1 \end{bmatrix} $$ and check if it is diagonalisable. Let’s solve step by step manually. --- ### Step 1: Characteristic polynomial The eigenvalues $\lambda$ satisfy $$ \det(B - \lambda I) = 0 $$ $$ B - \lambda I = \begin{bmatrix} 1-\lambda & 1 & 0 \\ -1 & 3-\lambda & 0 \\ 1 & -1 & 1-\lambda \end{bmatrix} $$ Determinant: $$ \det(B - \lambda I) = \begin{vmatrix} 1-\lambda & 1 & 0 \\ -1 & 3-\lambda & 0 \\ 1 & -1 & 1-\lambda \end{vmatrix} $$ Use cofactor expansion along 3rd column (many zeros): $$ \det(B-\lambda I) = (1-\lambda) \cdot \begin{vmatrix} 1-\lambda & 1 \\ -1 & 3-\lambda \end{vmatrix} _________ ________ ________ ____ ______ __________ _______ _____ ____ ____ ____ _________.
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