Amit, Biraj and Chirag were given the task of creating a square matrux of order 2.
Below are the matrices created by them. A, B , C are the matrices created by Amit, Biraj and Chirag respectively.
`A=[(1,2),(-1,3)]B=[(4,0),(1,5)]C=[(2,0),(1,-2)]`
If a=4 and b=-2 based on the above information answer the following:
`AC-BC` is equal to

Options

A. `[(-4,-6),(-4,4)]`
B. `[(-4,-4),(4,-6)]`
C. `[(-4,-4),(-6,4)]`
D. `[(-6,4),(-4,-4)]`

Answer

To solve the problem \( AC - BC \) where \( A \), \( B \), and \( C \) are given matrices, we will follow these steps:

### Step 1: Define the matrices
The matrices are defined as follows:
\[ A = \begin{pmatrix} 1 & 2 \\ -1 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix}, \quad C = \begin{pmatrix} 2 & 0 \\ 1 & -2 \end{pmatrix} \]

### Step 2: Calculate the product \( AC \)
To find \( AC \), we multiply matrix \( A \) by matrix \( C \):
\[ AC = A \cdot C = \begin{pmatrix} 1 & 2 \\ -1 & 3 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \\ 1 & -2 \end{pmatrix} \]

Calculating the elements of \( AC \):
- First row, first column: \( 1 \cdot 2 + 2 \cdot 1 = 2 + 2 = 4 \)
- First row, second column: \( 1 \cdot 0 + 2 \cdot (-2) = 0 - 4 = -4 \)
- Second row, first column: \( -1 \cdot 2 + 3 \cdot 1 = -2 + 3 = 1 \)
- Second row, second column: \( -1 \cdot 0 + 3 \cdot (-2) = 0 - 6 = -6 \)

Thus,
\[ AC = \begin{pmatrix} 4 & -4 \\ 1 & -6 \end{pmatrix} \]

### Step 3: Calculate the product \( BC \)
Next, we calculate \( BC \):
\[ BC = B \cdot C = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \\ 1 & -2 \end{pmatrix} \]

Calculating the elements of \( BC \):
- First row, first column: \( 4 \cdot 2 + 0 \cdot 1 = 8 + 0 = 8 \)
- First row, second column: \( 4 \cdot 0 + 0 \cdot (-2) = 0 + 0 = 0 \)
- Second row, first column: \( 1 \cdot 2 + 5 \cdot 1 = 2 + 5 = 7 \)
- Second row, second column: \( 1 \cdot 0 + 5 \cdot (-2) = 0 - 10 = -10 \)

Thus,
\[ BC = \begin{pmatrix} 8 & 0 \\ 7 & -10 \end{pmatrix} \]

### Step 4: Calculate \( AC - BC \)
Now we find \( AC - BC \):
\[ AC - BC = \begin{pmatrix} 4 & -4 \\ 1 & -6 \end{pmatrix} - \begin{pmatrix} 8 & 0 \\ 7 & -10 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 4 - 8 = -4 \)
- First row, second column: \( -4 - 0 = -4 \)
- Second row, first column: \( 1 - 7 = -6 \)
- Second row, second column: \( -6 - (-10) = -6 + 10 = 4 \)

Thus,
\[ AC - BC = \begin{pmatrix} -4 & -4 \\ -6 & 4 \end{pmatrix} \]

### Final Answer
The result of \( AC - BC \) is:
\[ \begin{pmatrix} -4 & -4 \\ -6 & 4 \end{pmatrix} \]