Answer
$Q=\begin{bmatrix}
1/2&-1/\sqrt{8}&1/2
\\-1/2&1/\sqrt{8}&1/2
\\0&2/\sqrt{8}&0
\\1/2&1/\sqrt{8}&-1/2
\\1/2&1/\sqrt{8}&1/2
\end{bmatrix}$ and $R=\begin{bmatrix}
2&8&7\\
0&2\sqrt{2}&3\sqrt{2}\\
0&0&6
\end{bmatrix}$
Work Step by Step
The three orthogonal columns that form the orthogonal basis for the columns space of the given matrix was found in exercise 12.
$u_1=\begin{bmatrix}
1\\-1\\0\\1\\1
\end{bmatrix}$, $u_2=\begin{bmatrix}
-1\\1\\2\\1\\1
\end{bmatrix}$, and $u_3=\begin{bmatrix}
1\\1\\0\\-1\\1
\end{bmatrix}$.
To form the matrix Q, we need to normalize these vectors.
$Q=\begin{bmatrix}
1/2&-1/\sqrt{8}&1/2
\\-1/2&1/\sqrt{8}&1/2
\\0&2/\sqrt{8}&0
\\1/2&1/\sqrt{8}&-1/2
\\1/2&1/\sqrt{8}&1/2
\end{bmatrix}$
$A=QR$
$R=Q^TA=\begin{bmatrix}
2&8&7\\
0&2\sqrt{2}&3\sqrt{2}\\
0&0&6
\end{bmatrix}$
