Question

Решение матрицы Математика

Answer 1

Ответ:

$B*(A+2B)-3AB=\left[\begin{array}{ccc}111&-44&-11\\22&-1&7\\8&-25&0\end{array}\right] .$

Пошаговое объяснение:

$A=\left[\begin{array}{ccc}7&-3&0\\1&-1&0\\2&0&3\end{array}\right] ,B=\left[\begin{array}{ccc}-4&2&1\\1&0&1\\3&2&1\end{array}\right] ;$

$B(A+2B)-3AB=\left[\begin{array}{ccc}-4&2&1\\1&0&1\\3&2&1\end{array}\right] *\Bigg(\left[\begin{array}{ccc}7&-3&0\\1&-1&0\\2&0&3\end{array}\right] +$

$+2*\left[\begin{array}{ccc}-4&2&1\\1&0&1\\3&2&1\end{array}\right] \Bigg)-3*\left[\begin{array}{ccc}7&-3&0\\1&-1&0\\2&0&3\end{array}\right] *\left[\begin{array}{ccc}-4&2&1\\1&0&1\\3&2&1\end{array}\right]=$

$=\left[\begin{array}{ccc}-4&2&1\\1&0&1\\3&2&1\end{array}\right]\Bigg(\left[\begin{array}{ccc}7&-3&0\\1&-1&0\\2&0&3\end{array}\right] +\left[\begin{array}{ccc}2*(-4)&2*2&2*1\\2*1&2*0&2*1\\2*3&2*2&2*1\end{array}\right]\Bigg)-3*$

$*\left[\begin{array}{ccc}-28-3&14&7-3+0\\-4-1&2&1-1+0\\-8+0+9&4+6&2+0+3\end{array}\right] =$

$=\left[\begin{array}{ccc}-4&2&1\\1&0&1\\3&2&1\end{array}\right]\Bigg(\left[\begin{array}{ccc}7&-3&0\\1&-1&0\\2&0&3\end{array}\right] +\left[\begin{array}{ccc}-8&4&2\\2&0&2\\6&4&2\end{array}\right] \Bigg)-3*$

$*\left[\begin{array}{ccc}-31&14&4\\-5&2&0\\1&10&5\end{array}\right] =\left[\begin{array}{ccc}-4&2&1\\1&0&1\\3&2&1\end{array}\right]*\left[\begin{array}{ccc}7-8&-3+4&0+2\\1+2&-1+0&0+2\\2+6&0+4&3+2\end{array}\right] -$

$-\left[\begin{array}{ccc}3*(-31)&3*14&3*4\\3*(-5)&3*2&3*0\\3*1&3*10&3*5\end{array}\right] =\left[\begin{array}{ccc}-4&2&1\\1&0&1\\3&2&1\end{array}\right] *$

$*\left[\begin{array}{ccc}-1&1&2\\3&-1&2\\8&4&5\end{array}\right] -\left[\begin{array}{ccc}-93&42&12\\-15&6&0\\3&30&15\end{array}\right] =$

$=\left[\begin{array}{ccc}4+6+8&-4-2+4&-8+4+5\\-1+8&1+4&2+5\\-3+6+8&3-2+4&6+4+5\end{array}\right] -$

$-\left[\begin{array}{ccc}-93&42&12\\-15&6&0\\3&30&15\end{array}\right] =\left[\begin{array}{ccc}18&-2&1\\7&5&7\\11&5&15\end{array}\right] -\left[\begin{array}{ccc}-93&42&12\\-15&6&0\\3&30&15\end{array}\right] =$

$\left[\begin{array}{ccc}18-(-93)&-2-42&1-12\\7-(-15)&5-6&7-0\\11-3&5-30&15-15\end{array}\right] =$

$=\left[\begin{array}{ccc}111&-44&-11\\22&-1&7\\8&-25&0\end{array}\right]$