Reduced row-echelon form of a matrix with variables I have been staring at this for an hour. How would you reduce such a matrix?
\begin{bmatrix}
  p & 0 & a \\
  b & 0 & 0 \\
  q & c & r 
\end{bmatrix}
$abc\neq0$
 A: Because $a,b,c \ne 0$ then
$$
\left( {\begin{array}{*{20}{c}}
   p & 0 & a  \\
   b & 0 & 0  \\
   q & c & r  \\
\end{array}} \right) \to \left( {\begin{array}{*{20}{c}}
   b & 0 & 0  \\
   q & c & r  \\
   p & 0 & a  \\
\end{array}} \right)\mathop  \to \limits_{ - \frac{p}{b}{\rho _1} + {\rho _3}}^{ - \frac{q}{b}{\rho _1} + {\rho _2}} \left( {\begin{array}{*{20}{c}}
   b & 0 & 0  \\
   0 & c & r  \\
   0 & 0 & a  \\
\end{array}} \right)\mathop  \to \limits^{ - \frac{r}{a}{\rho _3} + {\rho _2}} \left( {\begin{array}{*{20}{c}}
   b & 0 & 0  \\
   0 & c & 0  \\
   0 & 0 & a  \\
\end{array}} \right)\mathop  \to \limits^{\frac{1}{b}{\rho _1},\frac{1}{c}{\rho _2},\frac{1}{a}{\rho _3}} \left( {\begin{array}{*{20}{c}}
   1 & 0 & 0  \\
   0 & 1 & 0  \\
   0 & 0 & 1  \\
\end{array}} \right).
$$
A: If $abc\neq 0$, then $a,b,c\neq 0$. Divide the three rows by $a,b,c$ respectively and continue with other elementary row operations.
A: First you want a non-zero entry in the $(1, 1)$ position. There is only one element of the first column which you know is non-zero, so do a row swap to make sure that it is in the $(1, 1)$ position. Once in place, use this non-zero entry to eliminate the other entries in the first column using row operations of the form $R_i \mapsto R_i + kR_1$. Then you can divide the first row by a constant to ensure that the $(1, 1)$ entry is a $1$.
In the second column you want a non-zero entry in the $(2, 2)$ position if possible. As there is only one non-zero entry, do the appropriate row swap to put it in the $(2, 2)$ position. Again, you can divide the second row by a constant to ensure that the $(2, 2)$ entry is a $1$.
In the third column you want a non-zero entry in the $(3, 3)$ position if possible. If done correctly, you should have the appropraite entry already in place. A row operation of the form $R_i \mapsto R_i + kR_3$ should eliminate the other entry in the third column. Finally, dividing the third column by a constant should leave you with a familiar matrix.
A: Perhaps this is "cheating", but $$\det
\begin{bmatrix}
  p & 0 & a \\
  b & 0 & 0 \\
  q & c & r 
\end{bmatrix}
=abc \neq 0.$$
Since the determinant is non-zero, its reduced row echelon form is the identity matrix.
