Помогите. 3 любых примера.

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$1)XA=B=\ \textgreater \ X=BA^{-1}\\(A|E)=\ \textgreater \ (E|A^{-1})$
$\left(\begin{array}{cc|cc|c}1&2&1&0&\downarrow\\3&4&0&1&-3\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|cc|c}1&2&1&0&\\0&-2&-3&1&:-2\end{array}\right)=\ \textgreater \ \\\left(\begin{array}{cc|cc|c}1&2&1&0&-2\\0&1&1,5&-0,5&\uparrow\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|cc|c}1&0&-2&1&\\0&1&1,5&-0,5&\end{array}\right)\\X=\left(\begin{array}{cc}0&0\\0&0\end{array}\right)\left(\begin{array}{cc}-2&1\\1,5&-0,5\end{array}\right)=\left(\begin{array}{cc}0&0\\0&0\end{array}\right)$

$2)XA=B=\ \textgreater \ X=BA^{-1}\\(A|E)=\ \textgreater \ (E|A^{-1})$
$\left(\begin{array}{cc|cc|c}4&3&1&0&:4\\-5&-4&0&1&\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|cc|c}1&\frac{3}{4}&\frac{1}{4}&0&\downarrow\\-5&-4&0&1&5\end{array}\right)=\ \textgreater \ \\\left(\begin{array}{cc|cc|c}1&\frac{3}{4}&\frac{1}{4}&0&\\0&-\frac{1}{4}&\frac{5}{4}&1&*-4\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|cc|c}1&\frac{3}{4}&\frac{1}{4}&0&-\frac{3}{4}\\0&1&-5&-4&\uparrow\end{array}\right)=\ \textgreater \ \\\left(\begin{array}{cc|cc|c}1&0&4&3&\\0&1&-5&-4&\end{array}\right)\\$
$X=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\left(\begin{array}{cc}4&3\\-5&-4\end{array}\right)= \left(\begin{array}{cc}4&3\\-5&-4\end{array}\right)$

3)Так как определитель равен нулю,перепишем матрицу учитывая, что есть Х:
$X=\left(\begin{array}{c}x_1\\x_2\end{array}\right)\\\left(\begin{array}{c}x_1+x_2\\x_1+x_2\end{array}\right)= \left(\begin{array}{c}2\\3\end{array}\right)\\\begin{cases}x_1+x_2=2\\x_1+x_2=3\end{cases}\\ \left(\begin{array}{cc|c|c}1&1&2&\downarrow\\1&1&3&-1\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|c|c}1&1&2&\\0&0&1&\end{array}\right)$
Т.к. получена строка 0x₁+0x₂=1, следует что решений у уравнения нет.

4)Так как определитель равен нулю,перепишем матрицу учитывая, что есть Х:
$X=\left(\begin{array}{c}x_1\\x_2\end{array}\right)\\\left(\begin{array}{c}x_1+x_2\\x_1+x_2\end{array}\right)= \left(\begin{array}{c}2\\2\end{array}\right)\\\begin{cases}x_1+x_2=2\\x_1+x_2=2\end{cases}\\ \left(\begin{array}{cc|c|c}1&1&2&\downarrow\\1&1&2&-1\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|c|c}1&1&2&\\0&0&0&\end{array}\right)$
Получена нулевая строка, а значит уравнение имеет бесконечное число решений

$5)AXB=C=\ \textgreater \ X=A^{-1}CB^{-1}\\(A|E)=\ \textgreater \ (E|A^{-1})\\\left(\begin{array}{cc|cc|c}1&-1&1&0&\downarrow\\2&3&0&1&-2\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|cc|c}1&-1&1&0&\\0&5&-2&1&:5\end{array}\right)=\ \textgreater \ \\\left(\begin{array}{cc|cc|c}1&-1&1&0&1\\0&1&-\frac{2}{5}&\frac{1}{5}&\uparrow\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|cc|c}1&0&\frac{3}{5}&\frac{1}{5}&\\0&1&-\frac{2}{5}&\frac{1}{5}&\end{array}\right)$
<img src="https://tex.z-dn.net/?f=%28B%7CE%29%3D%5C+%5Ctextgreater+%5C+%28E%7CB%5E%7B-1%7D%29%5C%5C%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7Ccc%7Cc%7D-5%266%261%260%26%3A-5%5C%5C-4%265%260%261%26%5Cend%7Barray%7D%5Cright%29%3D%5C+%5Ctextgreater+%5C+%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7Ccc%7Cc%7D1%26-%5Cfrac%7B6%7D%7B5%7D%26-%5Cfrac%7B1%7D%7B5%7D%260%26%5Cdownarrow%5C%5C-4%265%260%261%264%5Cend%7Barray%7D%5Cright%29%3D%5C+%5Ctextgreater+%5C+%5C%5C%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7Ccc%7Cc%7D1%26-%5Cfrac%7B6%7D%7B5%7D%26-%5Cfrac%7B1%7D%7B5%7D%260%26%5C%5C0%26%5Cfrac%7B1%7D%7B5%7D%26-%5Cfrac%7B4%7D%7B5%7D%261%26%2A5%5Cend%7Barray%7D%5Cright%29%3D%5C+%5Ctextgreater+%5C+%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7Ccc%7Cc%7D1%26-%5Cfrac%7B6%7D%7B5%7D%26-%5Cfrac%7B1%7D%7B5%7D%260%26%5Cfrac%7B6%7D%7B5%7D%5C%5C0%261%26-4%265%26%5Cuparrow%5Cend%7Barray%7D%5Cright%29%3D%5C+%5Ctextgreater+%5C+" id="TexFormula9" title="(B|E)=\ \textgreater \ (E|B^{-1})\\\left(\begin{array}{cc|cc|c}-5&6&1&0&:-5\\-4&5&0&1&\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|cc|c}1&-\frac{6}{5}&-\frac{1}{5}&0&\downarrow\\-4&5&0&1&4\end{array}\right)=\ \textgreater \ \\\left(\begin{array}{cc|cc|c}1&-\frac{6}{5}&-\frac{1}{5}&0&\\0&\frac{1}{5}&-\frac{4}{5}&1&*5\end{array}\right)=\ \textgreater \ \left(\begin{array}{cc|cc|c}1&-\frac{6}{5}&-\frac{1}{5}&0&\frac{6}{5}\\0&1&-4&5&\uparrow\end{array}\right)=\ \textgreater \ " alt="(B|E)=\ \textgreater \ (E|B^{-1})\\\left(\begin{array}{cc|cc|c}-5&6&1&0&:-5\\-4&5&0&1&\end{array}\right)=\ \textgreater \ \left(\begin{array}{

Alexаndr_zn (72.9k баллов) 16 Май, 18