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

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 following system of equations: \[ \begin{bmatrix} 3 & -1 & -1 \\ 1 & -3 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] This leads to the following equations: 1. \( 3x_1 - x_2 - x_3 = 0 \) 2. \( x_1 - 3x_2 + 3x_3 = 0 \) You can solve this system of equations to find the values of \( x_1, x_2, \) and \( x_3 \). To do this, you can express one variable in terms of the others. For example, from the first equation, you can express \( x_2 \) in terms of \( x_1 \) and \( x_3 \): \[ x_2 = 3x_1 - x_3 \] Substituting this expression for \( x_2 \) into the second equation gives: \[ x_1 - 3(3x_1 - x_3) + 3x_3 = 0 \] Simplifying this: \[ x_1 - 9x_1 + 3x_3 + 3x_3 = 0 \] \[ -8x_1 + 6x_3 = 0 \] \[ 4x_1 = 3x_3 \quad \Rightarrow \quad x_3 = \frac{4}{3}x_1 \] Now substituting \( x_3 \) back into the expression for \( x_2 \): \[ x_2 = 3x_1 - \frac{4}{3}x_1 = \frac{9}{3}x_1 - \frac{4}{3}x_1 = \frac{5}{3}x_1 \] Thus, we can express the solution in terms of \( x_1 \): \[ x = \begin{bmatrix} x_1 \\ \frac{5}{3}x_1 \\ \frac{4}{3}x_1 \end{bmatrix} = x_1 \begin{bmatrix} 1 \\ \frac{5}{3} \\ \frac{4}{3} \end{bmatrix} \] This means that the solution set is spanned by the vector: \[ \begin{bmatrix} 1 \\ \frac{5}{3} \\ \frac{4}{3} \end{bmatrix} \] where \( x_1 \) can take any real value. Thus, the null space of \( A \) is one-dimensional.