If A = \(\begin{bmatrix}-3\\5\\2\end{bmatrix}\) and B = \(\begin{bmatr

1.	If A = \(\begin{bmatrix}-3\\5\\2\end{bmatrix}\) and B = \(\begin{bmatrix}1&6&-4\end{bmatrix}\), then verify that (AB)' = B'A'.
Answer» We have A = \(\begin{bmatrix}-3\\5\\2\end{bmatrix}\) and B = \(\begin{bmatrix}1&6&-4\end{bmatrix}\). ⇒ A' = \(\begin{bmatrix}-3&5&2\end{bmatrix}\) and B' = \(\begin{bmatrix}1\\6\\-4\end{bmatrix}\) Now, AB = \(\begin{bmatrix}-3\\5\\2\end{bmatrix}\)\(\begin{bmatrix}1&6&-4\end{bmatrix}\) = \(\begin{bmatrix}-3&-18&12\\5&30&-20\\2&12&-8\end{bmatrix}\). Therefore, (AB)' = \(\begin{bmatrix}-3&5&2\\-18&30&-20\\12&-20&-8\end{bmatrix}\). Now, B'A' = \(\begin{bmatrix}1\\6\\-4\end{bmatrix}\)\(\begin{bmatrix}-3&5&2\end{bmatrix}\) = \(\begin{bmatrix}-3&5&2\\-18&30&-20\\12&-20&-8\end{bmatrix}\) = (AB)' Hence, (AB)' = B'A'

If A = \(\begin{bmatrix}-3\\5\\2\end{bmatrix}\) and B = \(\begin{bmatrix}1&6&-4\end{bmatrix}\), then verify that (AB)' = B'A'.

Answer»

We have A = \(\begin{bmatrix}-3\\5\\2\end{bmatrix}\) and B = \(\begin{bmatrix}1&6&-4\end{bmatrix}\).

⇒ A' = \(\begin{bmatrix}-3&5&2\end{bmatrix}\) and B' = \(\begin{bmatrix}1\\6\\-4\end{bmatrix}\)

Now, AB = \(\begin{bmatrix}-3\\5\\2\end{bmatrix}\)\(\begin{bmatrix}1&6&-4\end{bmatrix}\) = \(\begin{bmatrix}-3&-18&12\\5&30&-20\\2&12&-8\end{bmatrix}\).

Therefore, (AB)' = \(\begin{bmatrix}-3&5&2\\-18&30&-20\\12&-20&-8\end{bmatrix}\).

Now, B'A' = \(\begin{bmatrix}1\\6\\-4\end{bmatrix}\)\(\begin{bmatrix}-3&5&2\end{bmatrix}\) = \(\begin{bmatrix}-3&5&2\\-18&30&-20\\12&-20&-8\end{bmatrix}\) = (AB)'

Hence, (AB)' = B'A'

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