Find the characteristic polynomial of the matrix

\[A= \begin{bmatrix} 3 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 2 & 0 \\ \end{bmatrix}\]

Solution:

For a square matrix, the solution is \(P_{a}(x) = det(A - \lambda I)\) so in this case we have:

\[ \begin{bmatrix} 3 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 2 & 0 \\ \end{bmatrix}-\lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}= \begin{bmatrix} 3-\lambda & 2 & 1 \\ 0 & 1-\lambda & 1 \\ 1 & 2 & -\lambda \\ \end{bmatrix} \]

Then:

\[A= det \begin{bmatrix} 3-\lambda & 0 & 0 \\ 0 & 1-\lambda & 0 \\ 0 & 0 & -\lambda \\ \end{bmatrix} \] \[A= det \begin{bmatrix} 3-\lambda & 0 & 0 \\ 0 & 1-\lambda & 0 \\ 0 & 0 & -\lambda \\ \end{bmatrix} \]

Calculate the determinant:

\[ P_{a}(x)= -(3-\lambda)\cdot(1-\lambda)\cdot\lambda+2-(1-\lambda)-2\cdot(3-\lambda)-(3-\lambda)\cdot(1-\lambda)\cdot\lambda+2-(1-\lambda)-2\cdot(3-\lambda) \] \[ P_{a}(x)=-5-x^3+4x^2 \]