You're absolutely correct. The cofactor expansion for the first element \( 1 - \lambda \) was not computed correctly in my previous solution. Let me go through the computation of the characteristic polynomial carefully again.

### Given Matrix \( A \):
\[
A = \begin{bmatrix} 
1 & 0 & 1 \\
2 & -1 & -1 \\
0 & 2 & -2
\end{bmatrix}
\]

### Step 1: **Find the characteristic polynomial**

The characteristic polynomial is determined by \( \det(A - \lambda I) \), where \( I \) is the identity matrix and \( \lambda \) is the eigenvalue. The matrix \( A - \lambda I \) is:

\[
A - \lambda I = \begin{bmatrix}
1 - \lambda & 0 & 1 \\
2 & -1 - \lambda & -1 \\
0 & 2 & -2 - \lambda
\end{bmatrix}.
\]

Now we compute the determinant of this matrix:
\[
\det(A - \lambda I) = \begin{vmatrix}
1 - \lambda & 0 & 1 \\
2 & -1 - \lambda & -1 \\
0 & 2 & -2 - \lambda
\end{vmatrix}.
\]

### Step 2: **Expand the determinant**

We will perform a cofactor expansion along the first row:
\[
\det(A - \lambda I) = (1 - \lambda) \det \begin{vmatrix} 
-1 - \lambda & -1 \\
2 & -2 - \lambda
\end{vmatrix} - 0 + 1 \cdot \det \begin{vmatrix}
2 & -1 \\
0 & -2 - \lambda
\end{vmatrix}.
\]

### Step 3: **Compute the \( 2 \times 2 \) determinants**

#### First determinant:
\[
\det \begin{vmatrix} 
-1 - \lambda & -1 \\
2 & -2 - \lambda
\end{vmatrix}
= (-1 - \lambda)(-2 - \lambda) - (-1)(2)
= (\lambda^2 + 3\lambda + 2) + 2
= \lambda^2 + 3\lambda + 4.
\]

#### Second determinant:
\[
\det \begin{vmatrix} 
2 & -1 \\
0 & -2 - \lambda

This solution details the correction of a cofactor expansion error in the characteristic polynomial of a matrix. It includes the use of cofactor expansion and determinant computation for eigenvalue problems in linear algebra, suitable for undergraduate students.

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