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If A = $\quad \begin{bmatrix} 2 & 5 & 6\\ 0 & 1 & 2 \end{bmatrix}, B = \quad \begin{bmatrix} 6 & 1 \\ 0 & 4 \\ 5 & 7 \end{bmatrix}$, Verify that $(AB)^{T} = B^{T}A^{T}$
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written 5.2 years ago by |
AB = \begin{bmatrix} 2 & 5 & 6 \ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 6 & 1 \ 0 & 4 \ 5 & 7 \end{bmatrix}
AB = \begin{bmatrix} 12+0+30 & 2+20+42 \ 0+0+10 & 0+4+14 \end{bmatrix}
AB = \begin{bmatrix} 42 & 64 \ 10 & 18 \end{bmatrix}
$(AB^{T}) = \begin{bmatrix} 42 & 10 \\ 64 & 18 \end{bmatrix}$
$B^{T}A^{T} = \quad \begin{bmatrix} 6 & 0 & 5 \\ 1 & 4 & 7 \end{bmatrix} \quad \begin{bmatrix} 2 & 0 \\ 5 & 1 \\ 6 & 2 \end{bmatrix}$
$B^{T}A^{T} = \begin{bmatrix} 12+0+30 & 0+0+10 \\ 2+20+42 & 0+4+14 \end{bmatrix}$
$B^{T}A^{T} = \begin{bmatrix} 42 & 10 \\ 64 & 18 \end{bmatrix}$
$Hence, (AB^{T}) = B^{T}A^{T}$
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