# Row equivalent matrices and proving

(a) Let A = ($a_1$ $a_2$ ... $a_n$) and B = ($b_1$ $b_2$ ... $b_n$) be row equivalent matrices where $a_i$ and $b_i$ are the ith columns of A and B respectively.

For $c_1$ $c_2$ ... $c_n$ $\in$ $R$, show that $c_1a_1$+$c_2a_2$+...+$c_na_n$ = 0 if and only if $c_1b_1$+$c_2b_2$+...+$c_nb_n$ = 0.

(b) Let A = ($a_1$ $a_2$ $a_3$ $a_4$ $a_5$) be a 4 x 5 matrix such that the columns $a_1$, $a_2$, $a_4$ are linearly independent while $a_3$ = $a_1 - 2a_2$ and $a_5$ = $a_1-a_2+a_4$.

Write down the reduced row-echelon form of A.

For part(a), should I use the row-echelon form of A and B which is the same?

## 1 Answer

Suppose that $$A = \begin{bmatrix} \alpha_1 \\ \vdots \\ \alpha_k \end{bmatrix}$$ where $\alpha_i$ is the $i$th row of $A$. If $\sum\limits_{j=1}^n c_j \, a_j = 0$, then $\alpha_i \cdot c = 0$ for every $i = 1, ..., k$, where $c = [c_1, c_2, ..., c_n]^{\operatorname{T}}$. In other words, $c$ is in the orthogonal complement of the row space of $A$...