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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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

$(AB)^{T} = B^{T}A^{T}$

written 5.6 years ago by

teamques10 ★ 69k

modified 3.0 years ago by

pedsangini276 • 4.8k

engineering mathematics

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written 5.6 years ago by

teamques10 ★ 69k

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 =

$\begin{bmatrix} 42 & 64 \ 10 & 18 \end{bmatrix}$ …

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