<s>[INST] $R$ be the row reduced echelon form of a $4 \times 4$ real matrix $A$ Let $R$ be the row reduced echelon form of a $4 \times 4$ real matrix $A$ and
let the
third column of $R$ be $\left[\begin{array}{l}0 \\ 1 \\ 0 \\ 0\end{array}\right]$. Then which is true?
P): If $\left[\begin{array}{l}\alpha \\ \beta \\ \gamma \\ 0\end{array}\right]$ is a solution of $A \mathrm{x}=0$, then $\boldsymbol{\gamma}=0$.
Q): For all $\mathrm{b} \in \mathbb{R}^{4}, \operatorname{rank}[A \mid \mathrm{b}]=\operatorname{rank}[R \mid \mathbf{b}]$.
For $P$ \begin{aligned}
&{\left[\begin{array}{llll}
a & b & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & \phi \\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
\alpha \\
\beta \\
\gamma \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0\\
0
\end{array}\right]} \\
&a \alpha+b \beta+0+0=0 \\
&0+0+\gamma+0=0 \\
&\Rightarrow \gamma=0
\end{aligned}
For $Q$
Rank of a matrix and rank of Row echelon matrix is same . So Q is also correct.
So both the statements are true. Is my approach correct?
Thanks in advance.
 [/INST] For $P$ \begin{aligned}
&{\left[\begin{array}{llll}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
0 \\
-2 \\
2 \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0\\
0
\end{array}\right]} \\
 \\
 \\
&\Rightarrow \gamma=2
\end{aligned}
The statement $P$ is also false.
</s>