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Q5.    For the matrices A and B, verify that (AB)' = B'A', where

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

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   A = \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix}B = \begin{bmatrix} 1 & 5&7 \end{bmatrix}

To prove : (AB)' = B'A'

L.H.S : (AB)'

AB = \begin{bmatrix} 0\\1 \\2 \end{bmatrix}\begin{bmatrix} 1& 5 &7 \end{bmatrix}

AB = \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}

R.H.S : B'A'

B' = \begin{bmatrix} 1\\5 \\7 \end{bmatrix}

A' = \begin{bmatrix} 0& 1 &2 \end{bmatrix}

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

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

Heence, L.H.S =R.H.S  i.e.(AB)' = B'A'.

Posted by

seema garhwal

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