If `A=[{:(1,5),(7,12):}]` and `B=[{:(9,1),( 7,8):}]` then find a matrix C such that `3A+5B+2C` is a null matrix.

Explanation

To solve the problem, we need to find a matrix \( C \) such that \( 3A + 5B + 2C \) is a null matrix (i.e., a matrix with all elements equal to zero).

Given:
- \( A = \begin{pmatrix} 1 & 5 \\ 7 & 12 \end{pmatrix} \)
- \( B = \begin{pmatrix} 9 & 1 \\ 7 & 8 \end{pmatrix} \)

We will follow these steps:

### Step 1: Calculate \( 3A \)
We multiply matrix \( A \) by 3:
\[ 3A = 3 \times \begin{pmatrix} 1 & 5 \\ 7 & 12 \end{pmatrix} = \begin{pmatrix} 3 \times 1 & 3 \times 5 \\ 3 \times 7 & 3 \times 12 \end{pmatrix} = \begin{pmatrix} 3 & 15 \\ 21 & 36 \end{pmatrix} \]

### Step 2: Calculate \( 5B \)
Next, we multiply matrix \( B \) by 5:
\[ 5B = 5 \times \begin{pmatrix} 9 & 1 \\ 7 & 8 \end{pmatrix} = \begin{pmatrix} 5 \times 9 & 5 \times 1 \\ 5 \times 7 & 5 \times 8 \end{pmatrix} = \begin{pmatrix} 45 & 5 \\ 35 & 40 \end{pmatrix} \]

### Step 3: Add \( 3A \) and \( 5B \)
Now, we add the results from Step 1 and Step 2:
\[ 3A + 5B = \begin{pmatrix} 3 & 15 \\ 21 & 36 \end{pmatrix} + \begin{pmatrix} 45 & 5 \\ 35 & 40 \end{pmatrix} = \begin{pmatrix} 3 + 45 & 15 + 5 \\ 21 + 35 & 36 + 40 \end{pmatrix} = \begin{pmatrix} 48 & 20 \\ 56 & 76 \end{pmatrix} \]

### Step 4: Set up the equation
We need to find \( C \) such that:
\[ 3A + 5B + 2C = 0 \]
This can be rearranged to:
\[ 2C = - (3A + 5B) = -\begin{pmatrix} 48 & 20 \\ 56 & 76 \end{pmatrix} = \begin{pmatrix} -48 & -20 \\ -56 & -76 \end{pmatrix} \]

### Step 5: Solve for \( C \)
Now we divide each element by 2 to find \( C \):
\[ C = \frac{1}{2} \begin{pmatrix} -48 & -20 \\ -56 & -76 \end{pmatrix} = \begin{pmatrix} -24 & -10 \\ -28 & -38 \end{pmatrix} \]

### Final Answer
Thus, the matrix \( C \) is:
\[ C = \begin{pmatrix} -24 & -10 \\ -28 & -38 \end{pmatrix} \]