\( A = \left[ \begin{array} { r r r } { 3 } & { - 1 } & { - 1 } \\ { 1 } & { - 3 } & { 3 } \end{array} \right] , \quad x = \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right] , \quad 0 = \left[ \begin{array} { l } { 0 } \\ { 0 } \end{array} \right] \)

It looks like you have a matrix \( A \) and a vector \( x \), along with a zero vector. The matrix \( A \) is a \( 2 \times 3 \) matrix, and the vector \( x \) is a \( 3 \times 1 \) vector. The matrix \( A \) is given by: \[ A = \begin{bmatrix} 3 & -1 & -1 \\ 1 & -3 & 3 \end{bmatrix} \] The vector \( x \) is: \[ x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \] And the zero vector is: \[ 0 = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] If you are looking to solve the equation \( A x = 0 \), you would set up the equation as follows: \[ \begin{bmatrix} 3 & -1 & -1 \\ 1 & -3 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] This leads to the following system of equations: 1. \( 3x_1 - x_2 - x_3 = 0 \) 2. \( x_1 - 3x_2 + 3x_3 = 0 \) To solve this system, we can express it in augmented matrix form and then perform row reduction: \[ \begin{bmatrix} 3 & -1 & -1 & | & 0 \\ 1 & -3 & 3 & | & 0 \end{bmatrix} \] We can perform row operations to simplify this matrix. For example, we can multiply the second row by 3 and subtract it from the first row: 1. Multiply the second row by 3: \[ 3 \cdot \begin{bmatrix} 1 & -3 & 3 & | & 0 \end{bmatrix} = \begin{bmatrix} 3 & -9 & 9 & | & 0 \end{bmatrix} \] 2. Subtract this from the first row: \[ \begin{bmatrix} 3 & -1 & -1 & | & 0 \\ 3 & -9 & 9 & | & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 0 & 8 & -10 & | & 0 \\ 1 & -3 & 3 & | & 0 \end{bmatrix} \] Now we can express the second row in terms of the first row: From the second row, we have: \[ x_1 - 3x_2 + 3x_3 = 0 \implies x_1 = 3x_2 - 3x_3 \] From the first row, we can express \( x_2 \) in terms of \( x_3 \): \[ 8x_2 - 10x_3 = 0 \implies 8x_2 = 10x_3 \implies x_2 = \frac{10}{8}x_3 = \frac{5}{4}x_3 \] Substituting \( x_2 \) back into the equation for \( x_1 \): \[ x_1 = 3\left(\frac{5}{4}x_3\right) - 3x_3 = \frac{15}{4}x_3 - 3x_3 = \frac{15}{4}x_3 - \frac{12}{4}x_3 = \frac{3}{4}x_3 \] Thus, we can express the solution in terms of \( x_3 \): \[ x_1 = \frac{3}{4}x_3, \quad x_2 = \frac{5}{4}x_3, \quad x_3 = x_3 \] Letting \( x_3 = t \) (a free variable), we have the general solution: \[ x = t \begin{bmatrix} \frac{3}{4} \\ \frac{5}{4} \\ 1 \end{bmatrix} \] where \( t \) is any real number. This represents the solution set to the homogeneous system \( Ax = 0 \).