solved prove that transpose(AB)=transpose(B)transpose(A) for the following matrices

Question : Prove that transpose(AB)=transpose(B)transpose(A) for the following matrices Solution : Given two matrices \[A=\left(\begin{matrix}1&0&1\\0&1&1\end{matrix}\right)\ ,\ B=\left(\begin{matrix}1&4&3\\6&2&0\\-2&0&1\end{matrix}\right)\] Their transpose are \[A^T=\left(\begin{matrix}1&0\\0&1\\1&1\end{matrix}\right),\ B^T=\left(\begin{matrix}1&6&-2\\4&2&0\\3&0&1\end{matrix}\right)\] Product of A and B \[AB=\left(\begin{matrix}-1&4&4\\4&2&1\end{matrix}\right)\] Hence the transpose of AB \[\left(AB\right)^T=\left(\begin{matrix}-1&4\\4&2\\4&1\end{matrix}\right)\] and the product of transposes of…

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