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**FernandoRevilla** (i) $\displaystyle \begin{pmatrix}{0}&{0}&{0}\\{0}&{0}&{0} \end{pmatrix}\in T$ because in this case $\displaystyle a+c=0+0=0$ and $\displaystyle b+d+f=0+0+0=0$

(ii) If $\displaystyle A=\begin{pmatrix}{a}&{b}&{c}\\{d}&{e}&{f} \end{pmatrix}\in T$ and $\displaystyle A'=\begin{pmatrix}{a'}&{b'}&{c'}\\{d'}&{e'}&{f'} \end{pmatrix}\in T$ then $\displaystyle a+c=0,\;b+d+f=0$ and $\displaystyle a'+c'=0,\;b'+d'+f'=0$ . Now, find $\displaystyle A+A'$ and verify the conditions for belonging to $\displaystyle T$ .

(iii) ...

*Second method*: The elements of $\displaystyle T$ are of the form $\displaystyle \begin{pmatrix}{-c}&{-d-f}&{c}\\{d}&{e}&{f} \end{pmatrix}$ . This means that we can express $\displaystyle T$ in the form $\displaystyle T=L[S]$ with $\displaystyle S\subset T$ which is a subspace by a well known theorem.