On usign elementry column operation `C_(2)rArrC_(2)-2C_(1)` in the following matrix equation `[{:(1,-3),(2,4):}]=[{:(1,01),(0,1):}][{:(3,1),(2,4):}]` , we have

Options

A. `[{:(1,-5),(0,4):}]=[{:(1,-1),(-2,2):}][{:(3,-5),(2,0):}]`
B. `[{:(1,-5),(0,4):}]=[{:(1,-1),(0,1):}][{:(3,-5),(-0,2):}]`
C. `[{:(1,-5),(2,0):}]=[{:(1,-3),(0,1):}][{:(3,1),(2,4):}]`
D. `[{:(1,-5),(2,0):}] =[{:(1,-1),(0,1):}][{:(3,-5),(2,0):}]`

Explanation

To solve the given matrix equation using the elementary column operation \( C_2 \rightarrow C_2 - 2C_1 \), we will follow these steps:

### Step 1: Write down the initial matrices
The given matrix equation is:
\[ \begin{pmatrix} 1 & -3 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} \]

### Step 2: Identify the columns to operate on
We will apply the operation \( C_2 \rightarrow C_2 - 2C_1 \) on the left matrix \( \begin{pmatrix} 1 & -3 \\ 2 & 4 \end{pmatrix} \).

### Step 3: Perform the elementary column operation
1. **For the first row:**
- The first column remains the same: \( 1 \)
- The second column will be modified:
\[ -3 - 2(1) = -3 - 2 = -5 \]

2. **For the second row:**
- The first column remains the same: \( 2 \)
- The second column will be modified:
\[ 4 - 2(2) = 4 - 4 = 0 \]

### Step 4: Write the resulting matrix
After applying the operation, the new matrix becomes:
\[ \begin{pmatrix} 1 & -5 \\ 2 & 0 \end{pmatrix} \]

### Final Result
Thus, after applying the elementary column operation \( C_2 \rightarrow C_2 - 2C_1 \), the resulting matrix is:
\[ \begin{pmatrix} 1 & -5 \\ 2 & 0 \end{pmatrix} \]

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