If `A=[(9,1),(4,3)],B=[(1,5),(6,11)]` and `3A+5B+2C=0` then C=………….

Answer

To solve the equation \(3A + 5B + 2C = 0\) for \(C\), we will follow these steps:

### Step 1: Write down the matrices A and B
Given:
\[ A = \begin{pmatrix} 9 & 1 \\ 4 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 5 \\ 6 & 11 \end{pmatrix} \]

### Step 2: Calculate \(3A\)
To calculate \(3A\), we multiply each element of matrix \(A\) by 3:
\[ 3A = 3 \times \begin{pmatrix} 9 & 1 \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} 3 \times 9 & 3 \times 1 \\ 3 \times 4 & 3 \times 3 \end{pmatrix} = \begin{pmatrix} 27 & 3 \\ 12 & 9 \end{pmatrix} \]

### Step 3: Calculate \(5B\)
Next, we calculate \(5B\) by multiplying each element of matrix \(B\) by 5:
\[ 5B = 5 \times \begin{pmatrix} 1 & 5 \\ 6 & 11 \end{pmatrix} = \begin{pmatrix} 5 \times 1 & 5 \times 5 \\ 5 \times 6 & 5 \times 11 \end{pmatrix} = \begin{pmatrix} 5 & 25 \\ 30 & 55 \end{pmatrix} \]

### Step 4: Substitute \(3A\) and \(5B\) into the equation
Now, we substitute \(3A\) and \(5B\) into the equation \(3A + 5B + 2C = 0\):
\[ 3A + 5B = \begin{pmatrix} 27 & 3 \\ 12 & 9 \end{pmatrix} + \begin{pmatrix} 5 & 25 \\ 30 & 55 \end{pmatrix} = \begin{pmatrix} 27 + 5 & 3 + 25 \\ 12 + 30 & 9 + 55 \end{pmatrix} = \begin{pmatrix} 32 & 28 \\ 42 & 64 \end{pmatrix} \]

### Step 5: Rearranging the equation to find \(2C\)
Now we rearrange the equation:
\[ 2C = - (3A + 5B) = - \begin{pmatrix} 32 & 28 \\ 42 & 64 \end{pmatrix} = \begin{pmatrix} -32 & -28 \\ -42 & -64 \end{pmatrix} \]

### Step 6: Calculate \(C\)
To find \(C\), we divide each element of \(2C\) by 2:
\[ C = \frac{1}{2} \begin{pmatrix} -32 & -28 \\ -42 & -64 \end{pmatrix} = \begin{pmatrix} -16 & -14 \\ -21 & -32 \end{pmatrix} \]

### Final Answer
Thus, the value of \(C\) is:
\[ C = \begin{pmatrix} -16 & -14 \\ -21 & -32 \end{pmatrix} \]