If`A==[[1, -1, 3], [2, 1, 0], [3, 3, 1]]`, then first `3R_3` and then `C_3 to C_3 +2C_2` on A gives

Options

A. `[[1, -1, 1], [2, 1, 2], [9, 9, 21]]`
B. `[[1,-1, 3], [2, 1, 0], [9, 9, 3]]`
C. `[[1, -1, 1], [2, 1, 2], [3, 3, 7]]`
D. `[[1, -1, 6], [2, 1, 3], [3, 3, 12]]`

Answer

To solve the problem step by step, we will perform the operations on the matrix \( A \) as described.

Given:
\[ A = \begin{bmatrix} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix} \]

### Step 1: Perform the operation \( R_3 \to 3R_3 \)

This means we will multiply every element in the third row \( R_3 \) by 3.

Calculating \( 3R_3 \):
\[ R_3 = [3, 3, 1] \implies 3R_3 = [3 \cdot 3, 3 \cdot 3, 3 \cdot 1] = [9, 9, 3] \]

Now, we replace the third row in matrix \( A \) with \( 3R_3 \):
\[ A = \begin{bmatrix} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 9 & 9 & 3 \end{bmatrix} \]

### Step 2: Perform the operation \( C_3 \to C_3 + 2C_2 \)

This means we will add 2 times the second column \( C_2 \) to the third column \( C_3 \).

Calculating \( 2C_2 \):
\[ C_2 = \begin{bmatrix} -1 \\ 1 \\ 9 \end{bmatrix} \implies 2C_2 = \begin{bmatrix} 2 \cdot -1 \\ 2 \cdot 1 \\ 2 \cdot 9 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \\ 18 \end{bmatrix} \]

Now, we add \( 2C_2 \) to \( C_3 \):
\[ C_3 = \begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix} \implies C_3 + 2C_2 = \begin{bmatrix} 3 + (-2) \\ 0 + 2 \\ 3 + 18 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 21 \end{bmatrix} \]

Now, we replace the third column in the matrix with the new values:
\[ A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 9 & 9 & 21 \end{bmatrix} \]

### Final Result

The final matrix after performing the operations is:
\[ \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 9 & 9 & 21 \end{bmatrix} \]