If `A=[{:(2,-1),(-1,2):}]`, verify `A^(2)-4A+3I=0`, where `I=[{:(1,0),(0,1):}]` and `O=[{:(0,0),(0,0):}]`. Hence find `A^(-1)`.

Answer

To solve the problem, we need to verify the equation \( A^2 - 4A + 3I = 0 \) and find the inverse of matrix \( A \). Let's break this down step by step.

### Step 1: Define the matrices
Given:
\[ A = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}, \quad I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \cdot \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 2 \cdot 2 + (-1) \cdot (-1) = 4 + 1 = 5 \)
- First row, second column: \( 2 \cdot (-1) + (-1) \cdot 2 = -2 - 2 = -4 \)
- Second row, first column: \( -1 \cdot 2 + 2 \cdot (-1) = -2 - 2 = -4 \)
- Second row, second column: \( -1 \cdot (-1) + 2 \cdot 2 = 1 + 4 = 5 \)

Thus,
\[ A^2 = \begin{pmatrix} 5 & -4 \\ -4 & 5 \end{pmatrix} \]

### Step 3: Calculate \( 4A \)
Next, we calculate \( 4A \):
\[ 4A = 4 \cdot \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 8 & -4 \\ -4 & 8 \end{pmatrix} \]

### Step 4: Calculate \( 3I \)
Now, we calculate \( 3I \):
\[ 3I = 3 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]

### Step 5: Substitute into the equation
Now we substitute \( A^2 \), \( 4A \), and \( 3I \) into the equation \( A^2 - 4A + 3I \):
\[ A^2 - 4A + 3I = \begin{pmatrix} 5 & -4 \\ -4 & 5 \end{pmatrix} - \begin{pmatrix} 8 & -4 \\ -4 & 8 \end{pmatrix} + \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]

Calculating this step by step:
1. \( A^2 - 4A \):
\[ \begin{pmatrix} 5 & -4 \\ -4 & 5 \end{pmatrix} - \begin{pmatrix} 8 & -4 \\ -4 & 8 \end{pmatrix} = \begin{pmatrix} 5 - 8 & -4 + 4 \\ -4 + 4 & 5 - 8 \end{pmatrix} = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} \]

2. Now add \( 3I \):
\[ \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} + \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} -3 + 3 & 0 + 0 \\ 0 + 0 & -3 + 3 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

Thus, we have verified that:
\[ A^2 - 4A + 3I = 0 \]

### Step 6: Finding \( A^{-1} \)
From the equation \( A^2 - 4A + 3I = 0 \), we can rearrange it to find \( A^{-1} \):
\[ A^2 - 4A + 3I = 0 \implies A^2 = 4A - 3I \]
Multiplying both sides by \( A^{-1} \):
\[ A = 4I - 3A^{-1} \]
Rearranging gives:
\[ 3A^{-1} = 4I - A \]
Thus,
\[ A^{-1} = \frac{1}{3}(4I - A) \]

Substituting \( A \) and \( I \):
\[ A^{-1} = \frac{1}{3}\left(4 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\right) = \frac{1}{3}\left(\begin{pmatrix} 4 & 0 \\ 0 & 4 \end{pmatrix} - \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\right) \]
Calculating:
\[ A^{-1} = \frac{1}{3}\begin{pmatrix} 4 - 2 & 0 + 1 \\ 0 + 1 & 4 - 2 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \]
So,
\[ A^{-1} = \begin{pmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} \end{pmatrix} \]

### Final Answer
Thus, we have verified the equation and found:
\[ A^{-1} = \begin{pmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} \end{pmatrix} \]