Obtain the characteristic equation of the matrix `A=[(1, 0,2),(0,2,1),(2,0,3)]` and verify that it is satisfied y A and hence find its inverse.

Answer

To solve the problem of obtaining the characteristic equation of the matrix \( A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} \) and verifying it, as well as finding its inverse, we will follow these steps:

### Step 1: Find the characteristic polynomial

The characteristic polynomial of a matrix \( A \) is given by the determinant of \( A - \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is a scalar.

1. **Formulate \( A - \lambda I \)**:

\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 & 2 \\ 0 & 2 - \lambda & 1 \\ 2 & 0 & 3 - \lambda \end{pmatrix} \]

2. **Calculate the determinant**:

We need to find \( \det(A - \lambda I) \):

\[ \det(A - \lambda I) = (1 - \lambda) \det\begin{pmatrix} 2 - \lambda & 1 \\ 0 & 3 - \lambda \end{pmatrix} - 2 \det\begin{pmatrix} 0 & 1 \\ 2 & 3 - \lambda \end{pmatrix} \]

First, calculate the two 2x2 determinants:

- For \( \begin{pmatrix} 2 - \lambda & 1 \\ 0 & 3 - \lambda \end{pmatrix} \):

\[ \det = (2 - \lambda)(3 - \lambda) - 0 = (2 - \lambda)(3 - \lambda) \]

- For \( \begin{pmatrix} 0 & 1 \\ 2 & 3 - \lambda \end{pmatrix} \):

\[ \det = 0 \cdot (3 - \lambda) - 1 \cdot 2 = -2 \]

Now substituting back:

\[ \det(A - \lambda I) = (1 - \lambda)((2 - \lambda)(3 - \lambda)) + 4 \]

Expanding this:

\[ = (1 - \lambda)(6 - 5\lambda + \lambda^2) + 4 \]

\[ = 6 - 5\lambda + \lambda^2 - 6\lambda + 5\lambda^2 - \lambda^3 + 4 \]

\[ = -\lambda^3 + 6\lambda^2 - 11\lambda + 10 \]

Thus, the characteristic polynomial is:

\[ \lambda^3 - 6\lambda^2 + 11\lambda - 10 = 0 \]

### Step 2: Verify the characteristic equation is satisfied by \( A \)

To verify, we substitute \( A \) into the characteristic polynomial:

1. **Calculate \( A^2 \) and \( A^3 \)**:

- First, calculate \( A^2 \):

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 8 \\ 4 & 4 & 7 \\ 8 & 0 & 13 \end{pmatrix} \]

- Then, calculate \( A^3 \):

\[ A^3 = A^2 \cdot A = \begin{pmatrix} 5 & 0 & 8 \\ 4 & 4 & 7 \\ 8 & 0 & 13 \end{pmatrix} \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 21 & 16 & 38 \\ 14 & 8 & 27 \\ 26 & 0 & 46 \end{pmatrix} \]

2. **Substitute into the polynomial**:

We need to check if:

\[ A^3 - 6A^2 + 11A - 10I = 0 \]

After substituting \( A, A^2, A^3 \) and simplifying, we should arrive at the zero matrix.

### Step 3: Find the inverse of \( A \)

To find the inverse of \( A \), we can use the formula:

\[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \]

1. **Calculate \( \det(A) \)**:

\[ \det(A) = 1(2 \cdot 3 - 1 \cdot 0) - 0 + 2(0 - 2 \cdot 1) = 6 - 4 = 2 \]

2. **Calculate the adjugate of \( A \)**:

The adjugate is the transpose of the cofactor matrix. Calculate the cofactors and then transpose:

\[ \text{adj}(A) = \begin{pmatrix} 2 & -2 & 0 \\ -1 & 3 & -2 \\ 0 & -2 & 2 \end{pmatrix} \]

3. **Compute the inverse**:

\[ A^{-1} = \frac{1}{2} \begin{pmatrix} 2 & -2 & 0 \\ -1 & 3 & -2 \\ 0 & -2 & 2 \end{pmatrix} = \begin{pmatrix} 1 & -1 & 0 \\ -0.5 & 1.5 & -1 \\ 0 & -1 & 1 \end{pmatrix} \]

### Summary of Steps:
1. Calculate \( A - \lambda I \).
2. Find the determinant to get the characteristic polynomial.
3. Verify the polynomial is satisfied by substituting \( A \).
4. Calculate the inverse using the adjugate and determinant.