Question

Высшая математика, можете написать кратко:

Answer 1

Ответ:

Пошаговое объяснение:

НУ, если коротко, то

Answer 2

1.

$A+B=\left(\begin{array}{ccc}5&5&1\\-2&0&1\\4&5&3\end{array}\right) + \left(\begin{array}{ccc}5&4&1\\9&9&3\\1&0&0\end{array}\right)=\left(\begin{array}{ccc}10&9&2\\7&9&4\\5&5&3\end{array}\right)\\-2C=-2\left(\begin{array}{cc}3&2\\4&9\\0&2\end{array}\right) =\left(\begin{array}{cc}-6&-4\\-8&-18\\0&-4\end{array}\right)$$A-2B=\left(\begin{array}{ccc}5&5&1\\-2&0&1\\4&5&3\end{array}\right) -2 \left(\begin{array}{ccc}5&4&1\\9&9&3\\1&0&0\end{array}\right)=\\=\left(\begin{array}{ccc}5&5&1\\-2&0&1\\4&5&3\end{array}\right) + \left(\begin{array}{ccc}-10&-8&-2\\-18&-18&-6\\-2&0&0\end{array}\right)=\left(\begin{array}{ccc}-5&-3&-1\\-20&-18&-5\\2&5&3\end{array}\right)$

$A\cdot C=\left(\begin{array}{ccc}5&5&1\\-2&0&1\\4&5&3\end{array}\right) \cdot \left(\begin{array}{cc}3&2\\4&9\\0&2\end{array}\right) =\left(\begin{array}{cc}15+20&10+45+2\\-6&-4+2\\12+20&8+45+6\end{array}\right)=\left(\begin{array}{cc}35&57\\-6&-2\\32&59\end{array}\right)$

2.

$2A-B=2\left[\begin{array}{ccc}1&-5&3\\4&-1&2\\0&2&1\end{array}\right] - \left[\begin{array}{ccc}1&3&2\\-1&-2&0\\3&0&1\end{array}\right]=\\=\left[\begin{array}{ccc}2&-10&6\\8&-2&4\\0&4&2\end{array}\right] + \left[\begin{array}{ccc}-1&-3&-2\\1&2&0\\-3&0&-1\end{array}\right]=\left[\begin{array}{ccc}1&-13&4\\9&0&4\\-3&4&1\end{array}\right]$

$A^2=\left[\begin{array}{ccc}1&-5&3\\4&-1&2\\0&2&1\end{array}\right]\cdot\left[\begin{array}{ccc}1&-5&3\\4&-1&2\\0&2&1\end{array}\right]=\left[\begin{array}{ccc}-19&6&-4\\0&-15&12\\8&0&5\end{array}\right]$

$(2A-B)\cdot A^2=\left[\begin{array}{ccc}1&-13&4\\9&0&4\\-3&4&1\end{array}\right]\cdot \left[\begin{array}{ccc}-19&6&-4\\0&-15&12\\8&0&5\end{array}\right] =\\=\left[\begin{array}{ccc}13&201&-140\\-139&54&-16\\65&-78&65\end{array}\right]$