Question

Me ayudan porfa es de segundo año científico
soy nuevo xd​

Answer 1

Respuesta:

Explicación paso a paso:

1.

1.1.

 $4*\left[\begin{array}{ccc}-2&5&1\\1&2&-2\end{array}\right] =\left[\begin{array}{ccc}-8&20&4\\4&8&-8\end{array}\right]$

1.2

$-3*\left[\begin{array}{ccc}2&4&-1\\-2&1&3\end{array}\right] =\left[\begin{array}{ccc}-6&-12&3\\6&-3&-9\end{array}\right]$

1.3.

$C^{t}=\left[\begin{array}{ccc}2&7\\6&8\\4&2\end{array}\right] \\=>\frac{1}{2}*C^{t} =\frac{1}{2}*\left[\begin{array}{ccc}2&7\\6&8\\4&2\end{array}\right]=\left[\begin{array}{ccc}1&\frac{7}{2} \\3&4\\2&1\end{array}\right]$

2.

$2*D=2*\left[\begin{array}{ccc}175&112&300\\180&240&250\end{array}\right] =\left[\begin{array}{ccc}350&224&600\\360&480&500\end{array}\right]$

3.

$2B+4A= 2*\left[\begin{array}{ccc}-1&0\\2&-5\\8&6\end{array}\right] +4*\left[\begin{array}{ccc}-1&0\\2&3\\9&-4\end{array}\right] \\2B+4A=\left[\begin{array}{ccc}-2&0\\4&-10\\16&12\end{array}\right] +\left[\begin{array}{ccc}-4&0\\8&12\\36&-16\end{array}\right] \\2B+4A=\left[\begin{array}{ccc}-6&0\\12&2\\52&-4\end{array}\right]$