Given `A=[{:(,0,4,6),(,3,0,-1):}] and B=[{:(,0,1),(,-1,2),(,-5,-6):}]`, is the following possible
(i) AB (ii) BA (iii) `A^2`.

Answer

To determine whether the matrix multiplications \( AB \), \( BA \), and \( A^2 \) are possible, we first need to examine the dimensions of matrices \( A \) and \( B \).

Given:
- Matrix \( A \) is a \( 2 \times 3 \) matrix:
\[ A = \begin{pmatrix} 0 & 4 & 6 \\ 3 & 0 & -1 \end{pmatrix} \]
- Matrix \( B \) is a \( 3 \times 2 \) matrix:
\[ B = \begin{pmatrix} 0 & 1 \\ -1 & 2 \\ -5 & -6 \end{pmatrix} \]

### Step 1: Check if \( AB \) is possible
To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix.

- \( A \) has 3 columns and \( B \) has 3 rows.
- Therefore, the multiplication \( AB \) is possible.

**Result of \( AB \)**:
\[ AB = \begin{pmatrix} 0 & 4 & 6 \\ 3 & 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 2 \\ -5 & -6 \end{pmatrix} \]

Calculating \( AB \):
- First row, first column: \( 0 \cdot 0 + 4 \cdot (-1) + 6 \cdot (-5) = 0 - 4 - 30 = -34 \)
- First row, second column: \( 0 \cdot 1 + 4 \cdot 2 + 6 \cdot (-6) = 0 + 8 - 36 = -28 \)
- Second row, first column: \( 3 \cdot 0 + 0 \cdot (-1) + (-1) \cdot (-5) = 0 + 0 + 5 = 5 \)
- Second row, second column: \( 3 \cdot 1 + 0 \cdot 2 + (-1) \cdot (-6) = 3 + 0 + 6 = 9 \)

Thus,
\[ AB = \begin{pmatrix} -34 & -28 \\ 5 & 9 \end{pmatrix} \]

### Step 2: Check if \( BA \) is possible
Now we check the multiplication \( BA \):
- \( B \) has 2 columns and \( A \) has 2 rows.
- Therefore, the multiplication \( BA \) is possible.

**Result of \( BA \)**:
\[ BA = \begin{pmatrix} 0 & 1 \\ -1 & 2 \\ -5 & -6 \end{pmatrix} \begin{pmatrix} 0 & 4 & 6 \\ 3 & 0 & -1 \end{pmatrix} \]

Calculating \( BA \):
- First row, first column: \( 0 \cdot 0 + 1 \cdot 3 = 3 \)
- First row, second column: \( 0 \cdot 4 + 1 \cdot 0 = 0 \)
- First row, third column: \( 0 \cdot 6 + 1 \cdot (-1) = -1 \)
- Second row, first column: \( -1 \cdot 0 + 2 \cdot 3 = 6 \)
- Second row, second column: \( -1 \cdot 4 + 2 \cdot 0 = -4 \)
- Second row, third column: \( -1 \cdot 6 + 2 \cdot (-1) = -6 - 2 = -8 \)
- Third row, first column: \( -5 \cdot 0 + -6 \cdot 3 = -18 \)
- Third row, second column: \( -5 \cdot 4 + -6 \cdot 0 = -20 \)
- Third row, third column: \( -5 \cdot 6 + -6 \cdot (-1) = -30 + 6 = -24 \)

Thus,
\[ BA = \begin{pmatrix} 3 & 0 & -1 \\ 6 & -4 & -8 \\ -18 & -20 & -24 \end{pmatrix} \]

### Step 3: Check if \( A^2 \) is possible
For \( A^2 \) (which means \( A \times A \)):
- \( A \) is a \( 2 \times 3 \) matrix.
- The multiplication \( A \times A \) is not possible since the number of columns in the first \( A \) (3) does not equal the number of rows in the second \( A \) (2).

### Final Conclusion:
- \( AB \) is possible.
- \( BA \) is possible.
- \( A^2 \) is not possible.