Given `A=[{:(,-3,6),(,0,-9):}] and A^t` is it transpose matrix. Find :
(i) `2A+3A^(t)` (ii) `2A^(t)-3A`

Answer

To solve the given problem, we need to find the expressions \(2A + 3A^t\) and \(2A^t - 3A\) where \(A\) is given as:

\[ A = \begin{pmatrix} -3 & 6 \\ 0 & -9 \end{pmatrix} \]

### Step 1: Find the Transpose of Matrix A

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

\[ A^t = \begin{pmatrix} -3 & 0 \\ 6 & -9 \end{pmatrix} \]

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

Now, we will multiply matrix \(A\) by 2:

\[ 2A = 2 \times \begin{pmatrix} -3 & 6 \\ 0 & -9 \end{pmatrix} = \begin{pmatrix} 2 \times -3 & 2 \times 6 \\ 2 \times 0 & 2 \times -9 \end{pmatrix} = \begin{pmatrix} -6 & 12 \\ 0 & -18 \end{pmatrix} \]

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

Next, we will multiply the transpose \(A^t\) by 3:

\[ 3A^t = 3 \times \begin{pmatrix} -3 & 0 \\ 6 & -9 \end{pmatrix} = \begin{pmatrix} 3 \times -3 & 3 \times 0 \\ 3 \times 6 & 3 \times -9 \end{pmatrix} = \begin{pmatrix} -9 & 0 \\ 18 & -27 \end{pmatrix} \]

### Step 4: Calculate \(2A + 3A^t\)

Now we can add \(2A\) and \(3A^t\):

\[ 2A + 3A^t = \begin{pmatrix} -6 & 12 \\ 0 & -18 \end{pmatrix} + \begin{pmatrix} -9 & 0 \\ 18 & -27 \end{pmatrix} = \begin{pmatrix} -6 + (-9) & 12 + 0 \\ 0 + 18 & -18 + (-27) \end{pmatrix} \]

Calculating the elements:

\[ = \begin{pmatrix} -15 & 12 \\ 18 & -45 \end{pmatrix} \]

### Step 5: Calculate \(2A^t\)

Next, we will multiply the transpose \(A^t\) by 2:

\[ 2A^t = 2 \times \begin{pmatrix} -3 & 0 \\ 6 & -9 \end{pmatrix} = \begin{pmatrix} 2 \times -3 & 2 \times 0 \\ 2 \times 6 & 2 \times -9 \end{pmatrix} = \begin{pmatrix} -6 & 0 \\ 12 & -18 \end{pmatrix} \]

### Step 6: Calculate \(3A\) Again

We already calculated \(3A\) earlier, but for clarity, we will do it again:

\[ 3A = 3 \times \begin{pmatrix} -3 & 6 \\ 0 & -9 \end{pmatrix} = \begin{pmatrix} 3 \times -3 & 3 \times 6 \\ 3 \times 0 & 3 \times -9 \end{pmatrix} = \begin{pmatrix} -9 & 18 \\ 0 & -27 \end{pmatrix} \]

### Step 7: Calculate \(2A^t - 3A\)

Now we can subtract \(3A\) from \(2A^t\):

\[ 2A^t - 3A = \begin{pmatrix} -6 & 0 \\ 12 & -18 \end{pmatrix} - \begin{pmatrix} -9 & 18 \\ 0 & -27 \end{pmatrix} = \begin{pmatrix} -6 - (-9) & 0 - 18 \\ 12 - 0 & -18 - (-27) \end{pmatrix} \]

Calculating the elements:

\[ = \begin{pmatrix} 3 & -18 \\ 12 & 9 \end{pmatrix} \]

### Final Answers

Thus, the final answers are:

(i) \(2A + 3A^t = \begin{pmatrix} -15 & 12 \\ 18 & -45 \end{pmatrix}\)

(ii) \(2A^t - 3A = \begin{pmatrix} 3 & -18 \\ 12 & 9 \end{pmatrix}\)