You can choose any $\hat{x}_{1} , \ldots , \hat{x}_{5}$, and use the system $ \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{bmatrix} = \begin{bmatrix} \hat{x}_{1} \\ \hat{x}_{2} \\ \hat{x}_{3} \\ \hat{x}_{4} \\ \hat{x}_{5} \\ \end{bmatrix}. $ If you want something that is less obvious, you can start multiplying the rows by nonzero constants: $ \begin{bmatrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{bmatrix} = \begin{bmatrix} 2 \hat{x}_{1} \\ \hat{x}_{2} \\ \hat{x}_{3} \\ \hat{x}_{4} \\ \hat{x}_{5} \\ \end{bmatrix} $ (in this example, we multiply the first row by $2$), or adding multiples of one row to another row: $ \begin{bmatrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{bmatrix} = \begin{bmatrix} 2 \hat{x}_{1} \\ \hat{x}_{2} \\ \hat{x}_{3} - 3 \hat{x}_{5} \\ \hat{x}_{4} \\ \hat{x}_{5} \\ \end{bmatrix} $ (in this example, we add $-3$ times the fifth row to the third row).