If `A=[(1,2,3),(2,1,2),(2,2,3)]B=[(1,2,2),(-2,-1,-2),(2,2,3)]` and `C=[(-1,-2,-2),(2,1,2),(2,2,3)]` then find the value of tr. `(A+B^(T)+3C)`.

Explanation

To find the value of \( \text{tr}(A + B^T + 3C) \), we will follow these steps:

### Step 1: Write down the matrices
Given:
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 2 & 2 \\ -2 & -1 & -2 \\ 2 & 2 & 3 \end{pmatrix}, \quad C = \begin{pmatrix} -1 & -2 & -2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{pmatrix} \]

### Step 2: Calculate the transpose of matrix \( B \)
The transpose of matrix \( B \), denoted \( B^T \), is obtained by swapping rows and columns:
\[ B^T = \begin{pmatrix} 1 & -2 & 2 \\ 2 & -1 & 2 \\ 2 & -2 & 3 \end{pmatrix} \]

### Step 3: Calculate \( 3C \)
To find \( 3C \), we multiply each element of matrix \( C \) by 3:
\[ 3C = 3 \times \begin{pmatrix} -1 & -2 & -2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{pmatrix} = \begin{pmatrix} -3 & -6 & -6 \\ 6 & 3 & 6 \\ 6 & 6 & 9 \end{pmatrix} \]

### Step 4: Add the matrices \( A \), \( B^T \), and \( 3C \)
Now we will add the matrices \( A \), \( B^T \), and \( 3C \):
\[ A + B^T + 3C = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{pmatrix} + \begin{pmatrix} 1 & -2 & 2 \\ 2 & -1 & 2 \\ 2 & -2 & 3 \end{pmatrix} + \begin{pmatrix} -3 & -6 & -6 \\ 6 & 3 & 6 \\ 6 & 6 & 9 \end{pmatrix} \]

Calculating the sum element-wise:
\[ = \begin{pmatrix} 1 + 1 - 3 & 2 - 2 - 6 & 3 + 2 - 6 \\ 2 + 2 + 6 & 1 - 1 + 3 & 2 + 2 + 6 \\ 2 + 2 + 6 & 2 - 2 + 6 & 3 + 3 + 9 \end{pmatrix} \]
\[ = \begin{pmatrix} -1 & -6 & -1 \\ 10 & 3 & 10 \\ 10 & 6 & 15 \end{pmatrix} \]

### Step 5: Calculate the trace of the resulting matrix
The trace of a matrix is the sum of the diagonal elements. For the matrix we obtained:
\[ \text{tr}(A + B^T + 3C) = -1 + 3 + 15 = 17 \]

### Final Answer
Thus, the value of \( \text{tr}(A + B^T + 3C) \) is:
\[ \boxed{17} \]