Adding a scalar multiple of one column to another leaves the column and row ranks unaltered and deduce from this that row rank = column rank

1. Show that adding a scalar multiple of one column to another leaves both the column rank and the row rank of a matrix unaltered.
2. Deduce from 1 that the row rank equals the column rank.

HINT: 1. Use the fact that $$\varphi((x_1,x_2,...,x_n))=(x_1+\beta x_2,x_2,...x_n)$$ (where $$\beta$$ is a fixed scalar) is an isomorphism from $$F^n$$ to itself.
2. Reduce the columns not belonging to a column basis to null vectors.

My attempt:
$$1\le k,p\le n$$ are fixed numbers. Let $$A_{m\times n}$$ and $$B_{m\times n}$$ be matrices such that
$$B_{*j}= \begin{cases} A_{*j} &\quad\text{if, } j\ne k\\ A_{*k}+A_{*p} &\quad\text{if, } j=k\\ \end{cases}$$ $$\quad\forall j=1,2,...n$$

Let $$S_1=\mathcal{R}(A)$$ and $$S_2=\mathcal{R}(B)$$ where $$\mathcal{R}$$ represents the rowspace of a matrix. $$V_1,V_2$$ are vector spaces over $$F^n$$.

We define $$\varphi: F^n\longrightarrow F^n$$ such that $$\varphi((a_{i1},a_{i2},...,a_{ik},...,a_{in}))=(a_{i1},a_{i2},...,a_{ik}+a_{ip},...,a_{in})$$ $$\Rightarrow\varphi(A_{i*})=B_{i*}$$

Using the hint, we know that $$\varphi$$ is an isomorphism from $$F^n$$ to itself
$$\Rightarrow S_1$$ is a subspace of $$F^n$$ so $$\varphi(S_1)=S_2$$ is a subspace of $$F^n$$ in the same dimension
$$\Rightarrow \dim(\mathcal{R}(A))=\dim(S_1)=\dim(S_2)=\dim(\mathcal{R}(B))$$

Similarly, $$\dim(\mathcal{C}(A))=\dim(\mathcal{C}(B))$$ where, $$\mathcal{C}$$ represents the column space of a matrix.

a) Is my attempt correct? Is there a better approach?
b) I know a general proof to the 2 but can't understand how to deduce it from 1.

b) Since row operation is dual to column operation, reduce $$A$$ to its RREF, which has equal row rank and column rank, or further column-reudce its RREF to most column-reduced form $$C$$ such that it is eminant that its column rank equals its row rank (by counting remaining non-zero entries in the matrix), hence column rank of A= column rank of C= row rank of C= row rank of A.
Note that though only one type of operations (adding a scalar multiple to others) is available, it still convert $$A$$ to its RREF except all leading entries may not be 1.