$R$ 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.
 A: Assuming that you are using the same $\bf{b}$ in $[A | \bf{b}]$ and $[R | \bf{b}]$ (i.e. you have not row reduced the right hand side column), then $Q$ need NOT be true. For example, we can have the last entry of $\bf{b}$ as (say) $1$, then the rank of $[R | \bf{b}]$ is more than that of $[A | \bf{b}]$.
Here is a simple example that can explain it better:
$$[A | \bf{b}] \rightarrow [R | \bf{c}] \implies \left[\begin{array}{lll|l}1&1&1&1\\1&1&1&1\end{array}\right] \rightarrow \left[\begin{array}{lll|l}1&1&1&1\\0&0&0&0\end{array}\right].$$
Clearly $[A|\bf{b}]$ has rank $1$. But if we look at $[R | \color{red}{\bf{b}}]$, then we have
$$\left[\begin{array}{lll|l}1&1&1&\color{red}{1}\\0&0&0&\color{red}{1}\end{array}\right],$$
which has rank $2$.
A: 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.
