Evaluate the given expression. Take
\[
\begin{array}{l}
A=\left[\begin{array}{rrr}
1 & -1 & 0 \\
0 & 6 & -1
\end{array}\right] \text { and } C=\left[\begin{array}{lll}
x & 1 & w \\
z & r & 2
\end{array}\right] \\
6 A^{T}-C^{T}
\end{array}
\]

Step 1 ：Find the transpose of A and C. The transpose of a matrix is obtained by interchanging its rows into columns or columns into rows. The transpose of A, denoted as A^T, is \[A^{T}=\left[\begin{array}{ll}1 & 0 \\-1 & 6 \\0 & -1\end{array}\right]\] The transpose of C, denoted as C^T, is \[C^{T}=\left[\begin{array}{ll}x & z \\1 & r \\w & 2\end{array}\right]\]

Step 2 ：Multiply A^T by 6. \[6A^{T}=\left[\begin{array}{ll}6 & 0 \\-6 & 36 \\0 & -6\end{array}\right]\]

Step 3 ：Subtract C^T from 6A^T. \[6A^{T}-C^{T}=\left[\begin{array}{ll}6-x & -z \\-6-1 & 36-r \\0-w & -6-2\end{array}\right]\]

Step 4 ：So, the final result is \[6A^{T}-C^{T}=\left[\begin{array}{ll}6-x & -z \\-7 & 36-r \\ -w & -8\end{array}\right]\]

Step 5 ：The result is a matrix, which is consistent with the original problem. Therefore, the result is correct. \(\boxed{6A^{T}-C^{T}=\left[\begin{array}{ll}6-x & -z \\-7 & 36-r \\ -w & -8\end{array}\right]}\)