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To find the characteristic equation of the given matrix \( B \), we need to follow these steps:

The characteristic equation of a matrix \( B \) is given by:

\[
\text{det}(B - \lambda I) = 0
\]

Where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix of the same order as \( B \).

### Step 1: Matrix \( B \)
Given matrix \( B \) is:

\[
B = \begin{pmatrix} 
1 & 2 & 1 \\
0 & 0 & -1 \\
1 & -1 & -1 
\end{pmatrix}
\]

### Step 2: Subtract \( \lambda I \) from \( B \)

The identity matrix \( I \) of order 3 is:

\[
I = \begin{pmatrix} 
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 
\end{pmatrix}
\]

Now, calculate \( B - \lambda I \):

\[
B - \lambda I = 
\begin{pmatrix} 
1 & 2 & 1 \\
0 & 0 & -1 \\
1 & -1 & -1 
\end{pmatrix}
-
\begin{pmatrix} 
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda 
\end{pmatrix}
=
\begin{pmatrix} 
1 - \lambda & 2 & 1 \\
0 & 0 - \lambda & -1 \\
1 & -1 & -1 - \lambda 
\end{pmatrix}
=
\begin{pmatrix} 
1 - \lambda & 2 & 1 \\
0 & -\lambda & -1 \\
1 & -1 & -1 - \lambda 
\end{pmatrix}
\]

### Step 3: Compute the determinant of \( B - \lambda I \)

Now, we compute the determinant of the resulting matrix:

\[
\text{det}(B - \lambda I) = 
\text{det} \begin{pmatrix} 
1 - \lambda & 2 & 1 \\
0 & -\lambda & -1 \\
1 & -1 & -1 - \lambda 
\end{pmatrix}
\]

The determinant of a 3x3 matrix \( \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is given by:

\[
\text{det} = a(ei - fh) - b(di - fg) + c(dh - eg)
\]

Applying this to our matrix:

\[
\text{det}(B - \lambda I) = (1 - \lambda)[(-\lambda)(-1 - \lambda) - (-1)(-1)] - 2[0 \cdot (-1 - \lambda) - (-1)(1)] + 1[0 \cdot (-1) - (-\lambda)(1)]
\]

Simplifying:

\[
= (1 - \lambda)[\lambda^2 + \lambda - 1] - 2[1] + 1[\lambda]
\]

\[
= (1 - \lambda)(\lambda^2 + \lambda - 1) - 2 + \lambda
\]

Expanding:

\[
= (1 - \lambda)(\lambda^2 + \lambda - 1) = \lambda^2 + \lambda - 1 - \lambda^3 - \lambda^2 + \lambda
\]

\[
= -\lambda^3 + 2\lambda - 1
\]

Now, combining all terms:

\[
-\lambda^3 + 2\lambda - 1 - 2 + \lambda = -\lambda^3 + 3\lambda - 3
\]

Thus, the characteristic equation is:

\[
-\lambda^3 + 3\lambda - 3 = 0
\]

Or equivalently:

\[
\lambda^3 - 3\lambda + 3 = 0
\]

Would you like more details on any step or to verify further calculations?

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Here are five related questions to explore further:
1. How do you solve the characteristic equation to find eigenvalues?
2. What is the significance of eigenvalues in matrix theory?
3. How can the eigenvectors be calculated from the characteristic equation?
4. How does the trace of a matrix relate to its eigenvalues?
5. Can a matrix have complex eigenvalues? How would the characteristic equation look in that case?

**Tip:** The trace of a matrix (the sum of diagonal elements) gives the sum of the eigenvalues, and the determinant of the matrix equals the product of its eigenvalues.

Find the characteristic equation of the matrix \( B = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & -1 \\ 1 & -1 & -1 \end{pmatrix} \).

Learn how to determine the characteristic equation of a 3x3 matrix B = [[1, 2, 1], [0, 0, -1], [1, -1, -1]]. This solution covers key linear algebra concepts, including determinants, characteristic polynomials, and eigenvalues. Ideal for undergraduate students studying linear algebra.

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