If `A=[{:(,2,1,-1),(,0,1,-2):}]` find:
(i) `A^(t).A` (ii) `A.A^(t)`
where `A^t` is the transpose of matrix A.

Answer

To solve the problem, we need to find \( A^t \cdot A \) and \( A \cdot A^t \) where \( A \) is given as:

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

### Step 1: Find the transpose of matrix \( A \)

The transpose of a matrix is obtained by swapping its rows and columns. Therefore, the transpose \( A^t \) will be:

\[ A^t = \begin{pmatrix} 2 & 0 \\ 1 & 1 \\ -1 & -2 \end{pmatrix} \]

### Step 2: Calculate \( A^t \cdot A \)

Now, we will multiply \( A^t \) by \( A \):

\[ A^t \cdot A = \begin{pmatrix} 2 & 0 \\ 1 & 1 \\ -1 & -2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 1 & -1 \\ 0 & 1 & -2 \end{pmatrix} \]

To perform the multiplication, we will calculate each element of the resulting matrix:

1. **First row, first column**:
\[ 2 \cdot 2 + 0 \cdot 0 = 4 \]

2. **First row, second column**:
\[ 2 \cdot 1 + 0 \cdot 1 = 2 \]

3. **First row, third column**:
\[ 2 \cdot (-1) + 0 \cdot (-2) = -2 \]

4. **Second row, first column**:
\[ 1 \cdot 2 + 1 \cdot 0 = 2 \]

5. **Second row, second column**:
\[ 1 \cdot 1 + 1 \cdot 1 = 2 \]

6. **Second row, third column**:
\[ 1 \cdot (-1) + 1 \cdot (-2) = -3 \]

7. **Third row, first column**:
\[ -1 \cdot 2 + (-2) \cdot 0 = -2 \]

8. **Third row, second column**:
\[ -1 \cdot 1 + (-2) \cdot 1 = -3 \]

9. **Third row, third column**:
\[ -1 \cdot (-1) + (-2) \cdot (-2) = 1 + 4 = 5 \]

Putting these together, we get:

\[ A^t \cdot A = \begin{pmatrix} 4 & 2 & -2 \\ 2 & 2 & -3 \\ -2 & -3 & 5 \end{pmatrix} \]

### Step 3: Calculate \( A \cdot A^t \)

Next, we will multiply \( A \) by \( A^t \):

\[ A \cdot A^t = \begin{pmatrix} 2 & 1 & -1 \\ 0 & 1 & -2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \\ 1 & 1 \\ -1 & -2 \end{pmatrix} \]

Calculating each element:

1. **First row, first column**:
\[ 2 \cdot 2 + 1 \cdot 1 + (-1) \cdot (-1) = 4 + 1 + 1 = 6 \]

2. **First row, second column**:
\[ 2 \cdot 0 + 1 \cdot 1 + (-1) \cdot (-2) = 0 + 1 + 2 = 3 \]

3. **Second row, first column**:
\[ 0 \cdot 2 + 1 \cdot 1 + (-2) \cdot (-1) = 0 + 1 + 2 = 3 \]

4. **Second row, second column**:
\[ 0 \cdot 0 + 1 \cdot 1 + (-2) \cdot (-2) = 0 + 1 + 4 = 5 \]

Putting these together, we get:

\[ A \cdot A^t = \begin{pmatrix} 6 & 3 \\ 3 & 5 \end{pmatrix} \]

### Final Results

Thus, the results are:

(i) \( A^t \cdot A = \begin{pmatrix} 4 & 2 & -2 \\ 2 & 2 & -3 \\ -2 & -3 & 5 \end{pmatrix} \)

(ii) \( A \cdot A^t = \begin{pmatrix} 6 & 3 \\ 3 & 5 \end{pmatrix} \)