Resposta:
Explicação passo-a-passo:
[tex]A^{T} = \left[\begin{array}{ccc}2&1&-3\\3&1&0\ \end{array}\right] \\\\A^{T} . B =\left[\begin{array}{ccc}2&1&-3\\3&1&0\ \end{array}\right] . \left[\begin{array}{ccc}1&1\\2&-2\\3&4\end{array}\right] =\\\\= \left[\begin{array}{ccc}2+2-9&2-2-12\\3+2+0&3-2+0\ \end{array}\right] = \left[\begin{array}{ccc}-5&-12\\5&1\ \end{array}\right]\\\\A^{T} . B + C = \left[\begin{array}{ccc}-5&-11\\7&4\ \end{array}\right]\\\[/tex]
[tex]M^{-1} = \left[\begin{array}{ccc}-5&-11\\7&4\ \end{array}\right]\ . \left[\begin{array}{ccc}X&Y\\Z&W\ \end{array}\right] = \left[\begin{array}{ccc}1&0\\0&1\ \end{array}\right][/tex]
- 5X - 11 Z = 1
- 5Y - 0W = 0 => - 5y = 0 => Y = 0
7 X + 4 Z = 0
7 Y + 4 W = 1
7 . 0 + 4 w = 1
4 W = 1 => [tex]W = \frac{1}{4}[/tex]
- 5X - 11 Z = 1 . (7) = - 35 X - 77 Z = 7
7 X + 4 Z = 0 .(5) = 35 X + 20 Z = 0
0 X - 57 Z = 7
[tex]Z = \frac{7}{57}[/tex]
7 X + 4 Z = 0
7 X + 4 .(7/57) = 0
7 X = 28/57
[tex]X = \frac{4}{57}[/tex]
[tex]M^{-1} = \left[\begin{array}{ccc}\frac{4}{57} &0\\\\\\\frac{7}{57} &\frac{1}{4} \ \end{array}\right][/tex]
