Sets $S$ and $T$ with $1 \times 2$ matrices and $2 \times 1$ matrices as elements, respectively, are as follows. $$S = \{ ( a \; b ) \mid a + b \neq 0 \} , \quad T = \left\{ \left. \binom { p } { q } \right\rvert \, p q \neq 0 \right\}$$ For an element $A$ of set $S$, which of the following statements in are correct? [4 points] ㄱ. For an element $P$ of set $T$, $PA$ does not have an inverse matrix. ㄴ. For an element $B$ of set $S$ and an element $P$ of set $T$, if $PA = PB$, then $A = B$. ㄷ. Among the elements of set $T$, there exists $P$ satisfying $PA \binom { 1 } { 1 } = \binom { 1 } { 1 }$. (1) ㄱ (2) ㄷ (3) ㄱ, ㄴ (4) ㄴ, ㄷ (5) ㄱ, ㄴ, ㄷ
Sets $S$ and $T$ with $1 \times 2$ matrices and $2 \times 1$ matrices as elements, respectively, are as follows.
$$S = \{ ( a \; b ) \mid a + b \neq 0 \} , \quad T = \left\{ \left. \binom { p } { q } \right\rvert \, p q \neq 0 \right\}$$
For an element $A$ of set $S$, which of the following statements in <Remarks> are correct? [4 points]
\textbf{<Remarks>}\\
ㄱ. For an element $P$ of set $T$, $PA$ does not have an inverse matrix.\\
ㄴ. For an element $B$ of set $S$ and an element $P$ of set $T$, if $PA = PB$, then $A = B$.\\
ㄷ. Among the elements of set $T$, there exists $P$ satisfying $PA \binom { 1 } { 1 } = \binom { 1 } { 1 }$.\\
(1) ㄱ\\
(2) ㄷ\\
(3) ㄱ, ㄴ\\
(4) ㄴ, ㄷ\\
(5) ㄱ, ㄴ, ㄷ