For the matrices \( A \) and \( B \), verify that \( ( AB )^{\prime}=

1.	For the matrices \( A \) and \( B \), verify that \( ( AB )^{\prime}= B ^{\prime} A ^{\prime} \), where(1) \( A=\left[\begin{array}{c}1 \\ -4 \\ 3\end{array}\right], B=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right] \)(ii) \( A=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right], B=\left[\begin{array}{lll}1 & 5 & 7\end{array}\right] \)
Answer» (i) Given matrices are \(A=\begin{bmatrix}1\\-4\\3\end{bmatrix},B=\begin{bmatrix}-1&2&1\end{bmatrix}\) ∴ \(A'=\begin{bmatrix}1&-4&3\end{bmatrix},B'=\begin{bmatrix}-1\\2\\1\end{bmatrix}\) \(AB=\begin{bmatrix}1\\-4\\3\end{bmatrix}\begin{bmatrix}-1&2&1\end{bmatrix}=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}\) \(\therefore(AB)'=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\) Now, \(B'A'=\begin{bmatrix}-1\\2\\1\end{bmatrix}\begin{bmatrix}1&-4&3\end{bmatrix}=\begin{bmatrix}-1&-4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\) Hence, (AB)' = B'A'. (ii) \(A=\begin{bmatrix}0\\1\\2\end{bmatrix},B=\begin{bmatrix}1&5&7\end{bmatrix}\) ∴ \(A'=\begin{bmatrix}0&1&2\end{bmatrix},B'=\begin{bmatrix}1\\5\\7\end{bmatrix}\) \(AB=\begin{bmatrix}0\\1\\2\end{bmatrix}\begin{bmatrix}1&5&7\end{bmatrix}=\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}\) \((AB)'=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\) Now, \(B'A'=\begin{bmatrix}1\\5\\7\end{bmatrix}\begin{bmatrix}0&1&2\end{bmatrix}=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\) Hence, (AB)' = B'A'.

For the matrices \( A \) and \( B \), verify that \( ( AB )^{\prime}= B ^{\prime} A ^{\prime} \), where(1) \( A=\left[\begin{array}{c}1 \\ -4 \\ 3\end{array}\right], B=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right] \)(ii) \( A=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right], B=\left[\begin{array}{lll}1 & 5 & 7\end{array}\right] \)

Answer»

(i) Given matrices are \(A=\begin{bmatrix}1\\-4\\3\end{bmatrix},B=\begin{bmatrix}-1&2&1\end{bmatrix}\)

∴ \(A'=\begin{bmatrix}1&-4&3\end{bmatrix},B'=\begin{bmatrix}-1\\2\\1\end{bmatrix}\)

\(AB=\begin{bmatrix}1\\-4\\3\end{bmatrix}\begin{bmatrix}-1&2&1\end{bmatrix}=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}\)

\(\therefore(AB)'=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\)

Now, \(B'A'=\begin{bmatrix}-1\\2\\1\end{bmatrix}\begin{bmatrix}1&-4&3\end{bmatrix}=\begin{bmatrix}-1&-4&-3\\2&-8&6\\1&-4&3\end{bmatrix}\)

Hence, (AB)' = B'A'.

(ii) \(A=\begin{bmatrix}0\\1\\2\end{bmatrix},B=\begin{bmatrix}1&5&7\end{bmatrix}\)

∴ \(A'=\begin{bmatrix}0&1&2\end{bmatrix},B'=\begin{bmatrix}1\\5\\7\end{bmatrix}\)

\(AB=\begin{bmatrix}0\\1\\2\end{bmatrix}\begin{bmatrix}1&5&7\end{bmatrix}=\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}\)

\((AB)'=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\)

Now, \(B'A'=\begin{bmatrix}1\\5\\7\end{bmatrix}\begin{bmatrix}0&1&2\end{bmatrix}=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}\)

Hence, (AB)' = B'A'.