For the matrix `A=[{:(1,1,0),(1,2,1),(2,1,0):}]`, which of the following is correct ?

Options

A. `A^3+3A^2 -I=0`
B. `A^3-3A^2 -I=0`
C. `A^3+2A^2 -I=0`
D. `A^3 -A^2 + I=0`

Answer

To solve the problem, we need to analyze the given matrix \( A \) and determine the correct statement among the provided options. The matrix \( A \) is given as:

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

We will find the characteristic polynomial of the matrix \( A \) by calculating the determinant of \( A - \lambda I \), where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix.

### Step 1: Set up \( A - \lambda I \)

The identity matrix \( I \) for a \( 3 \times 3 \) matrix is:

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

Thus, \( \lambda I \) is:

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

Now, we can compute \( A - \lambda I \):

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

### Step 2: Calculate the determinant of \( A - \lambda I \)

To find the determinant, we can use the formula for the determinant of a \( 3 \times 3 \) matrix:

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

For our matrix \( A - \lambda I \):

\[ \text{det}(A - \lambda I) = (1 - \lambda) \left( (2 - \lambda)(-\lambda) - (1)(1) \right) - (1) \left( (1)(-\lambda) - (1)(2) \right) + (0) \left( (1)(1) - (2 - \lambda)(2) \right) \]

Calculating the determinant step-by-step:

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

2. Calculate \( (1)(-\lambda) - (1)(2) \):
\[ = -\lambda - 2 \]

Now substituting back into the determinant formula:

\[ \text{det}(A - \lambda I) = (1 - \lambda)(\lambda^2 - 2\lambda - 1) + \lambda + 2 \]

### Step 3: Expand and simplify

Expanding \( (1 - \lambda)(\lambda^2 - 2\lambda - 1) \):

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

Combining like terms:

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

### Step 4: Set the characteristic polynomial to zero

The characteristic polynomial is:

\[ -\lambda^3 + 3\lambda^2 + 1 = 0 \]

This can be rewritten as:

\[ \lambda^3 - 3\lambda^2 - 1 = 0 \]

### Conclusion

The characteristic polynomial is \( \lambda^3 - 3\lambda^2 - 1 = 0 \). Based on the video transcript, the correct option is:

**Option 2 is correct.**