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).