On using the elementary row operation `R_(1) to R_(1)-3R_(2)` in the matrix equation `[{:(4,2),(3,3):}]=[{:(1,2),(0,3):}] [{:(2,0),(1,1):}]` we get

Options

A. `[{:(-5,-7),(3,3):}]=[{:(1,-7),(0,3):}] [{:(2,0),(1,1):}]`
B. `[{:(-5,-7),(3,3):}]=[{:(1,2),(0,3):}] [{:(-1,-3),(1,1):}]`
C. `[{:(-5,-7),(3,3):}]=[{:(1,2),(1,-7):}] [{:(2,0),(1,1):}]`
D. `[{:(4,2),(-5-,7):}]=[{:(1,2),(-3,-3):}] [{:(2,0),(1,1):}]`

Answer

To solve the given matrix equation using the elementary row operation \( R_1 \to R_1 - 3R_2 \), we will follow these steps:

### Step 1: Write down the matrix equation
The given matrix equation is:

\[ \begin{pmatrix} 4 & 2 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} \]

### Step 2: Compute the product of the matrices on the right-hand side
First, we need to compute the product of the two matrices on the right-hand side:

\[ \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 1 \cdot 2 + 2 \cdot 1 = 2 + 2 = 4 \)
- First row, second column: \( 1 \cdot 0 + 2 \cdot 1 = 0 + 2 = 2 \)
- Second row, first column: \( 0 \cdot 2 + 3 \cdot 1 = 0 + 3 = 3 \)
- Second row, second column: \( 0 \cdot 0 + 3 \cdot 1 = 0 + 3 = 3 \)

Thus, the product is:

\[ \begin{pmatrix} 4 & 2 \\ 3 & 3 \end{pmatrix} \]

### Step 3: Apply the row operation \( R_1 \to R_1 - 3R_2 \)
Now we apply the row operation \( R_1 \to R_1 - 3R_2 \):

- For the first element of the first row:
\[ 4 - 3 \cdot 3 = 4 - 9 = -5 \]

- For the second element of the first row:
\[ 2 - 3 \cdot 3 = 2 - 9 = -7 \]

The second row remains unchanged:

\[ \begin{pmatrix} 3 & 3 \end{pmatrix} \]

### Step 4: Write the new matrix after the operation
After performing the row operation, the new matrix becomes:

\[ \begin{pmatrix} -5 & -7 \\ 3 & 3 \end{pmatrix} \]

### Step 5: Write down the right-hand side after the operation
Now, we need to apply the same row operation to the right-hand side of the equation:

The original right-hand side is:

\[ \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} \]

Applying \( R_1 \to R_1 - 3R_2 \):

- For the first element of the first row:
\[ 1 - 3 \cdot 0 = 1 \]

- For the second element of the first row:
\[ 2 - 3 \cdot 3 = 2 - 9 = -7 \]

The second row remains unchanged:

\[ \begin{pmatrix} 0 & 3 \end{pmatrix} \]

Thus, the new right-hand side matrix becomes:

\[ \begin{pmatrix} 1 & -7 \\ 0 & 3 \end{pmatrix} \]

### Final Result
After performing the row operation, the new equation is:

\[ \begin{pmatrix} -5 & -7 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 1 & -7 \\ 0 & 3 \end{pmatrix} \]