Let `A=[{:(,1,0),(,2,1):}], B=[{:(,2,3),(,-1,0):}]`. Find `A^2+AB+B^2`

Answer

To solve the problem, we need to find \( A^2 + AB + B^2 \) for the matrices \( A \) and \( B \).

Given:
\[ A = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 3 \\ -1 & 0 \end{pmatrix} \]

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply matrix \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 0 \cdot 2 = 1 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, first column: \( 2 \cdot 1 + 1 \cdot 2 = 4 \)
- Second row, second column: \( 2 \cdot 0 + 1 \cdot 1 = 1 \)

Thus,
\[ A^2 = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix} \]

### Step 2: Calculate \( B^2 \)

Now, we calculate \( B^2 \):
\[ B^2 = B \cdot B = \begin{pmatrix} 2 & 3 \\ -1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ -1 & 0 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 2 \cdot 2 + 3 \cdot (-1) = 4 - 3 = 1 \)
- First row, second column: \( 2 \cdot 3 + 3 \cdot 0 = 6 + 0 = 6 \)
- Second row, first column: \( -1 \cdot 2 + 0 \cdot (-1) = -2 + 0 = -2 \)
- Second row, second column: \( -1 \cdot 3 + 0 \cdot 0 = -3 + 0 = -3 \)

Thus,
\[ B^2 = \begin{pmatrix} 1 & 6 \\ -2 & -3 \end{pmatrix} \]

### Step 3: Calculate \( AB \)

Next, we calculate \( AB \):
\[ AB = A \cdot B = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ -1 & 0 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 1 \cdot 2 + 0 \cdot (-1) = 2 + 0 = 2 \)
- First row, second column: \( 1 \cdot 3 + 0 \cdot 0 = 3 + 0 = 3 \)
- Second row, first column: \( 2 \cdot 2 + 1 \cdot (-1) = 4 - 1 = 3 \)
- Second row, second column: \( 2 \cdot 3 + 1 \cdot 0 = 6 + 0 = 6 \)

Thus,
\[ AB = \begin{pmatrix} 2 & 3 \\ 3 & 6 \end{pmatrix} \]

### Step 4: Calculate \( A^2 + AB + B^2 \)

Now, we add \( A^2 \), \( AB \), and \( B^2 \):
\[ A^2 + AB + B^2 = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix} + \begin{pmatrix} 2 & 3 \\ 3 & 6 \end{pmatrix} + \begin{pmatrix} 1 & 6 \\ -2 & -3 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 1 + 2 + 1 = 4 \)
- First row, second column: \( 0 + 3 + 6 = 9 \)
- Second row, first column: \( 4 + 3 - 2 = 5 \)
- Second row, second column: \( 1 + 6 - 3 = 4 \)

Thus,
\[ A^2 + AB + B^2 = \begin{pmatrix} 4 & 9 \\ 5 & 4 \end{pmatrix} \]

### Final Answer:
\[ A^2 + AB + B^2 = \begin{pmatrix} 4 & 9 \\ 5 & 4 \end{pmatrix} \]