Exorcise 3.11 . Let \( \mathrm{A}=\left(\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 6 & -5\end{array}\right) \). Find the matrix obtained from \( A \) by applying the following elementary row operation. \( \begin{array}{lll}\text { a. } R_{1} \leftrightarrow R_{3} & \text { b. } R_{2} \rightarrow R_{2}-4 R_{1} & \text { c. } R_{3} \rightarrow \frac{1}{3} R_{3}\end{array} \)

Let’s tackle each of the elementary row operations step by step on the matrix \( A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 6 & -5 \end{pmatrix} \). **a.** For the operation \( R_{1} \leftrightarrow R_{3} \), we will swap the first row and the third row: \[ A' = \begin{pmatrix} 3 & 6 & -5 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{pmatrix} \] **b.** Next, we perform \( R_{2} \rightarrow R_{2} - 4 R_{1} \). We need to replace \( R_2 \) with the result of \( R_2 \) minus 4 times \( R_1 \): \[ R_{2} = R_{2} - 4 R_{1} = \begin{pmatrix} 4 & 5 & 6 \end{pmatrix} - 4 \begin{pmatrix} 3 & 6 & -5 \end{pmatrix} = \begin{pmatrix} 4 - 12 & 5 - 24 & 6 + 20 \end{pmatrix} = \begin{pmatrix} -8 & -19 & 26 \end{pmatrix} \] The updated matrix becomes: \[ A'' = \begin{pmatrix} 3 & 6 & -5 \\ -8 & -19 & 26 \\ 1 & 2 & 3 \end{pmatrix} \] **c.** Finally, for the operation \( R_{3} \rightarrow \frac{1}{3} R_{3} \), we multiply the third row by \(\frac{1}{3}\): \[ R_{3} = \frac{1}{3} \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{2}{3} & 1 \end{pmatrix} \] Thus, the final matrix is: \[ A''' = \begin{pmatrix} 3 & 6 & -5 \\ -8 & -19 & 26 \\ \frac{1}{3} & \frac{2}{3} & 1 \end{pmatrix} \] In summary, after applying all the specified row operations, the resulting matrix is: \[ \begin{pmatrix} 3 & 6 & -5 \\ -8 & -19 & 26 \\ \frac{1}{3} & \frac{2}{3} & 1 \end{pmatrix} \]