Evaluate the given expression. Take \[ \begin{array}{l} A=\left[\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 6 & -1 \end{array}\right] \text { and } C=\left[\begin{array}{lll} x & 1 & w \\ z & r & 2 \end{array}\right] \\ 6 A^{T}-C^{T} \end{array} \]
Solution
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. \(\boxed{6A^{T}-C^{T}=\left[\begin{array}{ll}6-x & -z \\-7 & 36-r \\ -w & -8\end{array}\right]}\)
