Answer
See below
Work Step by Step
Determine basis for rowspace $(A)$:
$A=\begin{bmatrix}
1 & -3 & 2 & 0 & 1\\
4 & -9 & -1 & 1 & 2
\end{bmatrix} \approx \begin{bmatrix}
1 & -3 & 2 & 0 & 1\\
0 & 3 & -9 & 1 & 6
\end{bmatrix}$
Basic for rowspace$(A)$ is $\{(1,-3,2,0,-1),(4,-9,-1,1,2)\}$
According to Gram-Schmidt process, we have:
$v_1=x_1=(1,-3,2,0,-1)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(4,-9,-1,1,2)-\frac{)}{||(1,-3,2,0,-1)||^2}(1,-3,2,0,-1)\\
=(4,-9,-1,1,2)-\frac{4.1+(-9).(-3)+(-1).2+1.0+2.(-1)}{1^2+(-3)^2+2^2+0^2+(-1)^2}(1,-3,2,0,-1)\\
=(4,-9,-1,1,2)-\frac{4+27-2+0-2}{1+9+4+0+1}(1,-3,2,0,-1)\\
=(4,-9,-1,1,2)+\frac{9}{5}(1,-3,2,0,-1)\\
=\frac{1}{5}(11,-18,-23,5,19)$
Hence, an orthogonal basis for rowspace $(A)$ is:
$\{(1,-3,2,0,-1);\frac{1}{5}(11,-18,-23,5,19)\}$
Let $y_1=(1,4)\\
y_2=(0,1)$
Apply Gram- Schmidt
$v_1=y_1=(1,4)\\
v_2=y_2-\frac{}{||v_1||^2}v_1\\
=(0,1)-\frac{}{||(1,4)||^2}(1,4)\\
=(0,1)-\frac{0.1+1.4}{1^2+4^2}(1,4)\\
=(0,1)-\frac{1+4}{17}(1,4)\\
=\frac{1}{17}(-4,1)$
Consequently, an orthogonal basis for colspace $(A)$ is
$\{(1,4),\frac{1}{17}(-4,1)\}$
