In a), the columns are pairwise orthogonal; I can check that all the dot products are zero.
\begin{gather*}
\begin{pmatrix}
1 \\ -1 \\ 0
\end{pmatrix} \cdot
\begin{pmatrix}
1 \\ 1 \\ 1
\end{pmatrix} = 0 \\
\begin{pmatrix}
1 \\ -1 \\ 0
\end{pmatrix} \cdot
\begin{pmatrix}
-1 \\ -1 \\ 2
\end{pmatrix} = 0 \\
\begin{pmatrix}
1 \\ 1 \\ 1
\end{pmatrix} \cdot
\begin{pmatrix}
-1 \\ -1 \\ 2
\end{pmatrix} = 0
\end{gather*}
However, the columns are not unit vectors. The length of the first column is \(\sqrt{2}\text{,}\) the second is \(\sqrt{3} \) and the third if \(\sqrt{6}\text{.}\) An orthogonal matrix has to have columns of length one, so this is not an orthogonal matrix.
In b), the columns are pairwise orthogonal; I can check that all the dot products are zero. (Not that for these checks, I can scale the vectors as I wish, since whether or not the dot product is zero doesn’t depend on the scaling. I’ve scaled the columns to remove the square roots to make these calculations easier.)
\begin{gather*}
\begin{pmatrix}
1 \\ 1 \\ -6
\end{pmatrix} \cdot
\begin{pmatrix}
3 \\ 3 \\ 1
\end{pmatrix} = 0 \\
\begin{pmatrix}
1 \\ 1 \\ -6
\end{pmatrix} \cdot
\begin{pmatrix}
1 \\ -1 \\ 0
\end{pmatrix} = 0 \\
\begin{pmatrix}
3 \\ 3 \\ 1
\end{pmatrix} \cdot
\begin{pmatrix}
1 \\ -1 \\ 0
\end{pmatrix} = 0
\end{gather*}
In addition, the column vectors all have length one.
\begin{align*}
\begin{vmatrix}
\dfrac{1}{\sqrt{38}} \\[0.5em]
\dfrac{1}{\sqrt{38}} \\[0.5em]
\dfrac{-6}{\sqrt{38}}
\end{vmatrix} \amp =
\dfrac{1}{\sqrt{38}} \begin{vmatrix}
1 \\
1 \\
-6
\end{vmatrix} = \dfrac{1}{\sqrt{38}} \sqrt{1^2 + 1^2 +
(-6)^2} = \dfrac{\sqrt{38}}{\sqrt{38}} = 1 \\
\begin{vmatrix}
\dfrac{3}{\sqrt{19}} \\[0.5em]
\dfrac{3}{\sqrt{19}} \\[0.5em]
\dfrac{1}{\sqrt{19}}
\end{vmatrix} \amp = \dfrac{1}{\sqrt{19}}
\begin{vmatrix}
3 \\
3 \\
1
\end{vmatrix} = \dfrac{1}{\sqrt{19}} \sqrt{3^2 + 3^2 +
1^2} = \dfrac{\sqrt{19}}{\sqrt{19}} = 1\\
\begin{vmatrix}
\dfrac{1}{\sqrt{2}} \\[0.5em]
\dfrac{-1}{\sqrt{2}} \\[0.5em]
0
\end{vmatrix} \amp =
\dfrac{1}{\sqrt{2}} \begin{vmatrix}
1 \\
-1 \\
0
\end{vmatrix} = \dfrac{1}{\sqrt{2}} \sqrt{1^2 + (-1)^2 +
0^2} = \dfrac{\sqrt{2}}{\sqrt{2}} = 1
\end{align*}
In c), the columns are pairwise orthogonal; I can check that all the dot products are zero. (Note that for these checks, I can scale the vectors as I wish, since whether or not the dot product is zero doesn’t depend on the scaling. I’ve scaled the columns to remove the square roots to make these calculations easier.)
\begin{gather*}
\begin{pmatrix}
2 \\ -1 \\ 1
\end{pmatrix} \cdot
\begin{pmatrix}
3 \\ 4 \\ -2
\end{pmatrix} = 0 \\
\begin{pmatrix}
2 \\ -1 \\ 1
\end{pmatrix} \cdot
\begin{pmatrix}
-2 \\ 7 \\ 11
\end{pmatrix} = 0 \\
\begin{pmatrix}
3 \\ 4 \\ -2
\end{pmatrix} \cdot
\begin{pmatrix}
-2 \\ 7 \\ 11
\end{pmatrix} = 0
\end{gather*}
In addition, the column vectors all have length one.
\begin{align*}
\begin{vmatrix}
\dfrac{\sqrt{2}}{\sqrt{6}} \\[0.5em]
\dfrac{-1}{\sqrt{6}} \\[0.5em]
\dfrac{1}{\sqrt{6}}
\end{vmatrix} \amp =
\dfrac{1}{\sqrt{6}} \begin{vmatrix}
2 \\
-1 \\
1
\end{vmatrix} = \dfrac{1}{\sqrt{6}} \sqrt{2^2 + (-1)^2 +
1^2} = \dfrac{\sqrt{6}}{\sqrt{6}} = 1\\
\begin{vmatrix}
\dfrac{3}{\sqrt{29}} \\[0.5em]
\dfrac{4}{\sqrt{29}} \\[0.5em]
\dfrac{-2}{\sqrt{29}}
\end{vmatrix} \amp =
\dfrac{1}{\sqrt{29}} \begin{vmatrix}
3 \\
4 \\
-2
\end{vmatrix} = \dfrac{1}{\sqrt{29}} \sqrt{3^2 + 4^2 +
(-2)^2} = \dfrac{\sqrt{29}}{\sqrt{29}} = 1\\
\begin{vmatrix}
\dfrac{-2}{\sqrt{174}} \\[0.5em]
\dfrac{7}{\sqrt{174}} \\[0.5em]
\dfrac{11}{\sqrt{174}}
\end{vmatrix} \amp =
\dfrac{1}{\sqrt{174}} \begin{vmatrix}
-2 \\
7 \\
11
\end{vmatrix} = \dfrac{1}{\sqrt{174}} \sqrt{(-2)^2 + 7^2
+ (11)^2} = \dfrac{\sqrt{174}}{\sqrt{174}} = 1
\end{align*}
In d), the columns are not pairwise orthogonal; I can check to see that this fails with the first and third columns.
\begin{align*}
\amp \begin{pmatrix}
3 \\ -2 \\ -2
\end{pmatrix} \cdot
\begin{pmatrix}
-6 \\ -11 \\ 4
\end{pmatrix} \\
\amp = (3)(-6) + (-2)(-11) + (-2)(-4) = -18 + 22 + 8 =
12 \neq 0
\end{align*}
Since the columns are not orthogonal, this cannot be an orthogonal matrix. (I could check that the columns are unit vectors, which indeed they are, but I don’t need to since I’ve already shown that this cannot match the criteria for othogonality).
