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Maths-

General

Easy

Question

If A = $open square brackets table row 1 2 row cell negative 1 end cell 2 end table close square brackets$ and B = $open square brackets table row 3 4 row 2 cell negative 2 end cell end table close square brackets$ , then (AB)^T is

$[\begin{matrix} 11 & - 2 \\ 5 & - 6 \end{matrix}]$
$[\begin{matrix} 11 & 5 \\ - 2 & - 6 \end{matrix}]$
$[\begin{matrix} 7 & 1 \\ 0 & - 8 \end{matrix}]$
$[\begin{matrix} 7 & 0 \\ 1 & - 8 \end{matrix}]$

The correct answer is: $open square brackets table row 7 1 row 0 cell negative 8 end cell end table close square brackets$

AB = $open square brackets table row cell 1 space space 2 end cell row cell negative 1 space space 2 space space space end cell end table close square brackets space open square brackets table row cell 3 space space space 4 end cell row cell 2 space space minus 2 end cell end table close square brackets equals space open square brackets table row cell 3 plus 4 space space space space space space space 4 minus 4 end cell row cell negative 3 plus 4 space space minus 4 minus 4 end cell end table close square brackets A B space equals space open square brackets table row cell 7 space space space space 0 end cell row cell 1 space space space minus 8 end cell end table close square brackets A B to the power of T equals open square brackets table row cell 7 space space space space 1 end cell row cell 0 space space minus 8 end cell end table close square brackets$

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