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.
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 $2$.