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# For the matrices $A$ and $B$, verify that $(AB)' = B'A'$ , where $$\text{ (ii) } A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \text{ , } B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$$

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Toolbox:
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
(ii)$A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$
$B=\begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$
LHS:
$AB= \begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix}\begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 0\times 1 & 0\times 5 & 0\times 7\\1\times 1 & 1\times 5 & 1\times 7\\2\times 1 & 2\times 5 & 2\times 7 \end{bmatrix}=\begin{bmatrix}0& 0 &0\\1 & 5 &7\\2& 10 & 14\end{bmatrix}$
Interchange the rows column to get the transpose.
$\;\;\; (AB)'=\begin{bmatrix}0& 1 &2\\0 & 5 &10\\0& 7 & 14\end{bmatrix}$
RHS:
$B'A'=\begin{bmatrix} 1 & 5 & 7 \end{bmatrix}'\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}'$
$B'A'=\begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}'\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 1\times 0 & 1\times 1 & 1\times 2\\5\times 0 & 5\times 1 & 5\times 2\\7\times 0 & 7\times 1 & 7\times 2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix}0& 1 &2\\0 & 5 &10\\0& 7 & 14\end{bmatrix}$
LHS=RHS

answered Mar 6, 2013
edited Mar 13, 2013

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