If `A[(1, 0, 0),(x, 1, 0),(x, x, 1)] and I=[(1, 0, 0),(0,1,0),(0,0,1)]`, then `A^(3)-3A^(2)+3A` is equal to

Options

A. 3I
B. I
C. `-I`
D. `-2I`

Answer

To solve the problem \( A^3 - 3A^2 + 3A \) where \( A = \begin{pmatrix} 1 & 0 & 0 \\ x & 1 & 0 \\ x & x & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate \( A - \lambda I \)
We start by calculating \( A - \lambda I \), where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Thus,
\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 & 0 \\ x & 1 - \lambda & 0 \\ x & x & 1 - \lambda \end{pmatrix} \]

### Step 2: Calculate the determinant of \( A - \lambda I \)
Next, we find the determinant of \( A - \lambda I \):
\[ \text{det}(A - \lambda I) = (1 - \lambda) \begin{vmatrix} 1 - \lambda & 0 \\ x & 1 - \lambda \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ = (1 - \lambda)((1 - \lambda)(1 - \lambda) - 0) = (1 - \lambda)^3 \]
Thus, we have:
\[ \text{det}(A - \lambda I) = (1 - \lambda)^3 \]

### Step 3: Set the characteristic polynomial to zero
The characteristic polynomial is given by:
\[ (1 - \lambda)^3 = 0 \]
This implies that \( \lambda = 1 \) is a root of multiplicity 3.

### Step 4: Use the Cayley-Hamilton theorem
According to the Cayley-Hamilton theorem, a matrix satisfies its own characteristic equation. Therefore, we can substitute \( A \) into the characteristic polynomial:
\[ A^3 - 3A^2 + 3A - I = 0 \]
Rearranging gives:
\[ A^3 - 3A^2 + 3A = I \]

### Conclusion
Thus, we conclude that:
\[ A^3 - 3A^2 + 3A = I \]

### Final Answer
The result is:
\[ A^3 - 3A^2 + 3A = I \]