\( | \left. \begin{array} { c c c | c } { 3 } & { 2 } & { - 1 } \\ { 1 } & { 2 } & { 9 } & { 0 } \\ { 1 } & { 1 } & { 2 } & { 0 } \end{array} \right. \)

It looks like you have provided a matrix in augmented form, which typically represents a system of linear equations. The matrix you provided is: \[ \begin{bmatrix} 3 & 2 & -1 & | & 0 \\ 1 & 2 & 9 & | & 0 \\ 1 & 1 & 2 & | & 0 \end{bmatrix} \] This corresponds to the following system of equations: 1. \( 3x_1 + 2x_2 - x_3 = 0 \) 2. \( x_1 + 2x_2 + 9x_3 = 0 \) 3. \( x_1 + x_2 + 2x_3 = 0 \) To solve this system, we can use the method of Gaussian elimination. Let's perform row operations to bring the matrix to row echelon form. Starting with the original matrix: \[ \begin{bmatrix} 3 & 2 & -1 & 0 \\ 1 & 2 & 9 & 0 \\ 1 & 1 & 2 & 0 \end{bmatrix} \] 1. We can make the first row the pivot row. We can scale the second and third rows to eliminate \(x_1\) from them. - Replace Row 2 with Row 2 - (1/3) * Row 1: \[ R_2 \leftarrow R_2 - \frac{1}{3}R_1 \implies R_2 = \begin{bmatrix} 0 & \frac{5}{3} & \frac{28}{3} & 0 \end{bmatrix} \] - Replace Row 3 with Row 3 - (1/3) * Row 1: \[ R_3 \leftarrow R_3 - \frac{1}{3}R_1 \implies R_3 = \begin{bmatrix} 0 & \frac{1}{3} & \frac{7}{3} & 0 \end{bmatrix} \] The matrix now looks like this: \[ \begin{bmatrix} 3 & 2 & -1 & 0 \\ 0 & \frac{5}{3} & \frac{28}{3} & 0 \\ 0 & \frac{1}{3} & \frac{7}{3} & 0 \end{bmatrix} \] 2. Next, we can scale Row 2 to make the leading coefficient 1: - Multiply Row 2 by \( \frac{3}{5} \): \[ R_2 \leftarrow \frac{3}{5}R_2 \implies R_2 = \begin{bmatrix} 0 & 1 & \frac{28}{5} & 0 \end{bmatrix} \] The matrix now looks like this: \[ \begin{bmatrix} 3 & 2 & -1 & 0 \\ 0 & 1 & \frac{28}{5} & 0 \\ 0 & \frac{1}{3} & \frac{7}{3} & 0 \end{bmatrix} \] 3. Now we can eliminate \(x_2\) from Row 3: - Replace Row 3 with Row 3 - (1/3) * Row 2: \[ R_3 \leftarrow R_3 - \frac{1}{3}R_2 \implies R_3 = \begin{bmatrix} 0 & 0 & -\frac{7}{5} & 0 \end{bmatrix} \] The matrix now looks like this: \[ \begin{bmatrix} 3 & 2 & -1 & 0 \\ 0 & 1 & \frac{28}{5} & 0 \\ 0 & 0 & -\frac{7}{5} & 0 \end{bmatrix} \] 4. Finally, we can scale Row 3 to make the leading coefficient 1: - Multiply Row 3 by \(-\frac{5}{7}\): \[ R_3 \leftarrow -\frac{5}{7}R_3 \implies R_3 = \begin{bmatrix} 0 & 0 & 1 & 0 \end{bmatrix} \] The final row echelon form of the matrix is: \[ \begin{bmatrix} 3 & 2 & -1 & 0 \\ 0 & 1 & \frac{28