Show that ,` if A = [{:(1,3),(2,6) :}] and B = [{:(-1,4), (2,1):}] ` then `: (A +B) ^(2) ne A ^(2) + 2 AB + B ^(2).`

Answer

To show that \( (A + B)^2 \neq A^2 + 2AB + B^2 \) for the given matrices \( A \) and \( B \), we will follow these steps:

### Step 1: Define the matrices
Let
\[ A = \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 4 \\ 2 & 1 \end{pmatrix} \]

### Step 2: Calculate \( A + B \)
We add the matrices \( A \) and \( B \) component-wise:
\[ A + B = \begin{pmatrix} 1 + (-1) & 3 + 4 \\ 2 + 2 & 6 + 1 \end{pmatrix} = \begin{pmatrix} 0 & 7 \\ 4 & 7 \end{pmatrix} \]

### Step 3: Calculate \( (A + B)^2 \)
Now we need to calculate \( (A + B)^2 \):
\[ (A + B)^2 = (A + B)(A + B) = \begin{pmatrix} 0 & 7 \\ 4 & 7 \end{pmatrix} \begin{pmatrix} 0 & 7 \\ 4 & 7 \end{pmatrix} \]
To multiply these matrices, we compute:
- First row, first column: \( 0 \cdot 0 + 7 \cdot 4 = 28 \)
- First row, second column: \( 0 \cdot 7 + 7 \cdot 7 = 49 \)
- Second row, first column: \( 4 \cdot 0 + 7 \cdot 4 = 28 \)
- Second row, second column: \( 4 \cdot 7 + 7 \cdot 7 = 28 + 49 = 77 \)

Thus,
\[ (A + B)^2 = \begin{pmatrix} 28 & 49 \\ 28 & 77 \end{pmatrix} \]

### Step 4: Calculate \( A^2 \)
Next, we calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix} \]
Calculating this gives:
- First row, first column: \( 1 \cdot 1 + 3 \cdot 2 = 1 + 6 = 7 \)
- First row, second column: \( 1 \cdot 3 + 3 \cdot 6 = 3 + 18 = 21 \)
- Second row, first column: \( 2 \cdot 1 + 6 \cdot 2 = 2 + 12 = 14 \)
- Second row, second column: \( 2 \cdot 3 + 6 \cdot 6 = 6 + 36 = 42 \)

Thus,
\[ A^2 = \begin{pmatrix} 7 & 21 \\ 14 & 42 \end{pmatrix} \]

### Step 5: Calculate \( B^2 \)
Now, we calculate \( B^2 \):
\[ B^2 = B \cdot B = \begin{pmatrix} -1 & 4 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} -1 & 4 \\ 2 & 1 \end{pmatrix} \]
Calculating this gives:
- First row, first column: \( -1 \cdot -1 + 4 \cdot 2 = 1 + 8 = 9 \)
- First row, second column: \( -1 \cdot 4 + 4 \cdot 1 = -4 + 4 = 0 \)
- Second row, first column: \( 2 \cdot -1 + 1 \cdot 2 = -2 + 2 = 0 \)
- Second row, second column: \( 2 \cdot 4 + 1 \cdot 1 = 8 + 1 = 9 \)

Thus,
\[ B^2 = \begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix} \]

### Step 6: Calculate \( 2AB \)
Next, we calculate \( 2AB \):
\[ AB = A \cdot B = \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix} \begin{pmatrix} -1 & 4 \\ 2 & 1 \end{pmatrix} \]
Calculating this gives:
- First row, first column: \( 1 \cdot -1 + 3 \cdot 2 = -1 + 6 = 5 \)
- First row, second column: \( 1 \cdot 4 + 3 \cdot 1 = 4 + 3 = 7 \)
- Second row, first column: \( 2 \cdot -1 + 6 \cdot 2 = -2 + 12 = 10 \)
- Second row, second column: \( 2 \cdot 4 + 6 \cdot 1 = 8 + 6 = 14 \)

Thus,
\[ AB = \begin{pmatrix} 5 & 7 \\ 10 & 14 \end{pmatrix} \]
Now, multiplying by 2:
\[ 2AB = 2 \begin{pmatrix} 5 & 7 \\ 10 & 14 \end{pmatrix} = \begin{pmatrix} 10 & 14 \\ 20 & 28 \end{pmatrix} \]

### Step 7: Calculate \( A^2 + 2AB + B^2 \)
Now we can add \( A^2 \), \( 2AB \), and \( B^2 \):
\[ A^2 + 2AB + B^2 = \begin{pmatrix} 7 & 21 \\ 14 & 42 \end{pmatrix} + \begin{pmatrix} 10 & 14 \\ 20 & 28 \end{pmatrix} + \begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix} \]
Calculating this gives:
- First row, first column: \( 7 + 10 + 9 = 26 \)
- First row, second column: \( 21 + 14 + 0 = 35 \)
- Second row, first column: \( 14 + 20 + 0 = 34 \)
- Second row, second column: \( 42 + 28 + 9 = 79 \)

Thus,
\[ A^2 + 2AB + B^2 = \begin{pmatrix} 26 & 35 \\ 34 & 79 \end{pmatrix} \]

### Step 8: Compare \( (A + B)^2 \) and \( A^2 + 2AB + B^2 \)
Now we compare:
\[ (A + B)^2 = \begin{pmatrix} 28 & 49 \\ 28 & 77 \end{pmatrix} \]
\[ A^2 + 2AB + B^2 = \begin{pmatrix} 26 & 35 \\ 34 & 79 \end{pmatrix} \]
Clearly, \( (A + B)^2 \neq A^2 + 2AB + B^2 \).

### Conclusion
Thus, we have shown that:
\[ (A + B)^2 \neq A^2 + 2AB + B^2 \]