If `A=[(1,2),(3,4)],B=[(2,3),(4,5)]`, and `4A-3B+C=O`, then C=

Options

A. `[(2,-1),(0,1)]`
B. `[(2,1),(0,-1)]`
C. `[(-2,1),(0,-1)]`
D. None

Answer

To solve the equation \( 4A - 3B + C = O \) for the matrix \( C \), we will follow these steps:

### Step 1: Define the matrices A and B
Given:
\[ A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} \]

### Step 2: Calculate \( 4A \)
To find \( 4A \), we multiply each element of matrix \( A \) by 4:
\[ 4A = 4 \times \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 4 \cdot 1 & 4 \cdot 2 \\ 4 \cdot 3 & 4 \cdot 4 \end{pmatrix} = \begin{pmatrix} 4 & 8 \\ 12 & 16 \end{pmatrix} \]

### Step 3: Calculate \( 3B \)
Next, we find \( 3B \) by multiplying each element of matrix \( B \) by 3:
\[ 3B = 3 \times \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} 3 \cdot 2 & 3 \cdot 3 \\ 3 \cdot 4 & 3 \cdot 5 \end{pmatrix} = \begin{pmatrix} 6 & 9 \\ 12 & 15 \end{pmatrix} \]

### Step 4: Substitute \( 4A \) and \( 3B \) into the equation
Now, we substitute \( 4A \) and \( 3B \) into the equation \( 4A - 3B + C = O \):
\[ C = O - 4A + 3B \]
Since \( O \) is the zero matrix:
\[ O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]
Thus,
\[ C = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 4 & 8 \\ 12 & 16 \end{pmatrix} + \begin{pmatrix} 6 & 9 \\ 12 & 15 \end{pmatrix} \]

### Step 5: Perform the matrix operations
Now we perform the operations step by step:
1. Calculate \( O - 4A \):
\[ O - 4A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 4 & 8 \\ 12 & 16 \end{pmatrix} = \begin{pmatrix} -4 & -8 \\ -12 & -16 \end{pmatrix} \]
2. Now add \( 3B \):
\[ C = \begin{pmatrix} -4 & -8 \\ -12 & -16 \end{pmatrix} + \begin{pmatrix} 6 & 9 \\ 12 & 15 \end{pmatrix} = \begin{pmatrix} -4 + 6 & -8 + 9 \\ -12 + 12 & -16 + 15 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 0 & -1 \end{pmatrix} \]

### Final Result
Thus, the matrix \( C \) is:
\[ C = \begin{pmatrix} 2 & 1 \\ 0 & -1 \end{pmatrix} \]