Question

Determine (A-2B)t

$\left[\begin{array}{ccc}2&3\\5&-1\\0&6\end{array}\right] -2 . \left[\begin{array}{ccc}-5&4\\4&-1\\2&-3\end{array}\right]$

Answer 1

$A \ - 2 * B = \\ \left[\begin{array}{cc}2&3\\5&-1\\0&6\end{array}\right] \ - \ 2 * \ \left[\begin{array}{cc}-5&4\\4&-1\\2&-3\end{array}\right] \ = \\ \\ \\ \left[\begin{array}{cc}2&3\\5&-1\\0&6\end{array}\right] \ - \ \left[\begin{array}{cc}(2 * -5)&(2 *4)\\(2 * 4)&(2 * -1)\\(2 * 2)&(2 * -3)\end{array}\right] \ = \ \\ \\ \\ \left[\begin{array}{cc}2&3\\5&-1\\0&6\end{array}\right] \ - \ \left[\begin{array}{cc}-10&8\\8&-2\\4&-6\end{array}\right] \ =$

$\\ \left[\begin{array}{cc}2 - (-10)&3 - 8\\5 - 8&-1-(-2)\\0 - 4&6-(-6)\end{array}\right] \ = \\ \\ \\ \\ \left[\begin{array}{cc}12&-5\\-3&1\\ -4&12\end{array}\right]$

Como o exercício quer transposta (^T), então trocamos linhas por colunas :

$\\\left[\begin{array}{cc}12&-5\\-3&1\\ -4&12\end{array}\right]^{T} \ = \ \\ \\ \\\left[\begin{array}{ccc}12&-3&-4\\-5&1&12\end{array}\right]$