Answer
$\{\begin{bmatrix}
1 & -1\\
2 & 1
\end{bmatrix},\frac{1}{8}\begin{bmatrix}
1 & -9\\
2 & -7
\end{bmatrix}\}$
Work Step by Step
According to Gram-Schmidt process, we obtain:
$v_1=A_1=\begin{bmatrix}
1 & -1\\
2 & 1
\end{bmatrix}
v_2=A_2-\frac{(A_2,v_1)}{||v_1||^2}v_1\\
=\begin{bmatrix}
2 & 3\\
4 & 1
\end{bmatrix}-\frac{5.1.1+2.(-3).(-1)+3.4.2+5.1.1}{5.1.1+2.(-1).(-1).+3.2.2+5.1.1}\begin{bmatrix}
1 & -1\\
2 & 1
\end{bmatrix}\\
=\begin{bmatrix}
2 & -3\\
4 & 1
\end{bmatrix} - \frac{15}{8}\begin{bmatrix}
1 & -1\\
2 & 1
\end{bmatrix} \\
=\frac{1}{8}\begin{bmatrix}
1 & -9\\
2 & -7
\end{bmatrix}$
Hence, a corresponding orthonormal set of vector is:
$\{\begin{bmatrix}
1 & -1\\
2 & 1
\end{bmatrix},\frac{1}{8}\begin{bmatrix}
1 & -9\\
2 & -7
\end{bmatrix}\}$
