if `A=[{:(1,6),(2,4),(-3,5):}]B=[{:(3,4),(1,-2),(2,-1):}],` then find a matrix C such that 2A-B+c=0

Explanation

To solve the problem, we need to find a matrix \( C \) such that \( 2A - B + C = 0 \). This can be rearranged to find \( C \):

\[ C = B - 2A \]

### Step 1: Define Matrices A and B

Given:
\[ A = \begin{pmatrix} 1 & 6 \\ 2 & 4 \\ -3 & 5 \end{pmatrix}, \quad B = \begin{pmatrix} 3 & 4 \\ 1 & -2 \\ 2 & -1 \end{pmatrix} \]

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

We need to multiply matrix \( A \) by 2:

\[ 2A = 2 \times \begin{pmatrix} 1 & 6 \\ 2 & 4 \\ -3 & 5 \end{pmatrix} = \begin{pmatrix} 2 \times 1 & 2 \times 6 \\ 2 \times 2 & 2 \times 4 \\ 2 \times -3 & 2 \times 5 \end{pmatrix} = \begin{pmatrix} 2 & 12 \\ 4 & 8 \\ -6 & 10 \end{pmatrix} \]

### Step 3: Calculate \( B - 2A \)

Now we will subtract \( 2A \) from \( B \):

\[ B - 2A = \begin{pmatrix} 3 & 4 \\ 1 & -2 \\ 2 & -1 \end{pmatrix} - \begin{pmatrix} 2 & 12 \\ 4 & 8 \\ -6 & 10 \end{pmatrix} \]

Calculating element-wise:

1. First row:
- \( 3 - 2 = 1 \)
- \( 4 - 12 = -8 \)

2. Second row:
- \( 1 - 4 = -3 \)
- \( -2 - 8 = -10 \)

3. Third row:
- \( 2 - (-6) = 2 + 6 = 8 \)
- \( -1 - 10 = -11 \)

Putting it all together:

\[ B - 2A = \begin{pmatrix} 1 & -8 \\ -3 & -10 \\ 8 & -11 \end{pmatrix} \]

### Step 4: Conclusion

Thus, the matrix \( C \) is:

\[ C = \begin{pmatrix} 1 & -8 \\ -3 & -10 \\ 8 & -11 \end{pmatrix} \]