15- Me terminar, en cada caso, a matriz x . a) x=[1 & 2 & 5 -1 & 7 & 2 b) x+[1 & 2 5 & 1 c) 2 x=[1 & 1 & 1 2 & 3 & 4 d) 3 x^t=[1 & 1 2 & 7.

Vamos corrigir e detalhar cada caso:<br /><br />a) \( x=\left[\begin{array}{ccc}1 & 2 & 5 \\ -1 & 7 & 2\end{array}\right]^{t} \)<br /><br />Para encontrar \( x \), devemos calcular a transposição da matriz dada:<br /><br />\[ x = \left[\begin{array}{ccc}1 & 2 & 5 \\ -1 & 7 & 2\end{array}\right]^t = \left[\begin{array}{cc}1 & -1 \\ 2 & 7 \\ 5 & 2\end{array}\right] \]<br /><br />Portanto, a matriz \( x \) é:<br /><br />\[ x = \left[\begin{array}{cc}1 & -1 \\ 2 & 7 \\ 5 & 2\end{array}\right] \]<br /><br />b) \( x+\left[\begin{array}{ll}1 & 2 \\ 5 & 1\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 2 & 3\end{array}\right]^{t} \)<br /><br />Primeiro, calculamos a transposição da matriz à direita:<br /><br />\[ \left[\begin{array}{ll}0 & 0 \\ 2 & 3\end{array}\right]^t = \left[\begin{array}{cc}0 & 2 \\ 0 & 3\end{array}\right] \]<br /><br />Agora, vamos encontrar \( x \):<br /><br />\[ x + \left[\begin{array}{ll}1 & 2 \\ 5 & 1\end{array}\right] = \left[\begin{array}{cc}0 & 2 \\ 0 & 3\end{array}\right] \]<br /><br />Subtraindo a matriz \(\left[\begin{array}{ll}1 & 2 \\ 5 & 1\end{array}\right]\) de ambos os lados:<br /><br />\[ x = \left[\begin{array}{cc}0 & 2 \\ 0 & 3\end{array}\right] - \left[\begin{array}{ll}1 & 2 \\ 5 & 1\end{array}\right] = \left[\begin{array}{cc}-1 & 0 \\ -5 & 2\end{array}\right] \]<br /><br />Portanto, a matriz \( x \) é:<br /><br />\[ x = \left[\begin{array}{cc}-1 & 0 \\ -5 & 2\end{array}\right] \]<br /><br />c) \( 2x = \left[\begin{array}{lll}1 & 1 & 1 \\ 2 & 3 & 4\end{array}\right]^{t} \)<br /><br />Primeiro, calculamos a transposição da matriz à direita:<br /><br />\[ \left[\begin{array}{lll}1 & 1 & 1 \\ 2 & 3 & 4\end{array}\right]^t = \left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 2 & 4\end{array}\right] \]<br /><br />Agora, vamos encontrar \( x \):<br /><br />\[ 2x = \left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 2 & 4\end{array}\right] \]<br /><br />Dividindo ambos os lados por 2:<br /><br />\[ x = \frac{1}{2} \left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 2 & 4\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{2} & 1 & \frac{3}{2} \\ \frac{1}{2} & \frac{3}{2} & 2 \\ \frac{1}{2} & 1 & 2\end{array}\right] \]<br /><br />Portanto, a matriz \( x \) é:<br /><br />\[ x = \left[\begin{array}{ccc}\frac{1}{2} & 1 & \frac{3}{2} \\ \frac{1}{2} & \frac{3}{2} & 2 \\ \frac{1}{2} & 1 & 2\end{array}\right] \]<br /><br />d) \( 3x^t = \left[\begin{array}{ll}1 & 1 \\ 2 & 7\end{array}\right] = \left[\begin{array}{ll}1 & 4 \\ 7 & 2\end{