For which values of $a, b, c, d,$ and $e$ is the following matrix in reduced row-echelon form?

Question

What are the values of $a, b, c, d,$ and $e$ of the following matrix in reduced row echelon form? I guess $e$ should be $0$.

$$\left[\begin{array}{c}0 & a & 2 & 1 & b\\ 0 & 0 & 0 & c & d\\ 0 & 0 & e & 0 & 0\end{array}\right]\\$$

Answer 1

If it is reduced row echelon form then the first non-zero in each row must equal $1$ and all other entries in that column must be zero. Each "first $1$" in a row must be to the right of any first $1$ in rows above it. Any rows of all zeros must be at the bottom.

So $a=1$ and $c=0$. Either $d=0$ or $d=1$. If $d=0$ then $b$ may take any value, but of $d=1$ then $b=0$.

$e$ must equal $0$ as you suggested.

Thus the two correct solutions are

$\left(\begin{array}{c}0 & 1 & 2 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\end{array}\right)\\\left(\begin{array}{c}0 & 1 & 2 & 1 & b\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\end{array}\right)\\$

Answer 2

$$\left(\begin{array}{c}0 & a & 2 & 1 & b\\ 0 & 0 & 0 & c & d\\ 0 & 0 & e & 0 & 0\end{array}\right)\\$$

$\text{We can see that }x_3=0 \text{ for }e\neq 0 \text{ and }e\in\mathbb{R} \text{ but if }e=0$

$\\\text{ then we end up with a reduced system containing two equations in three variables}\\$

Case I: $e=0$

$$\left(\begin{array}{c}0 & a & 2 & 1 & b\\ 0 & 0 & 0 & c & d\\ 0 & 0 & 0 & 0 & 0\end{array}\right)\\$$

$\text{Solving the system now yields...}\\$

$$x_3=\dfrac{d}{c},c\neq 0\\$$

$$ax_1+2x_2+x_3=b\\$$

$$\implies ax_1+2x_2=b-\dfrac{d}{c}\\$$

$$\text{We have one free variable, let }x_2=s.$$

$$\\ ax_1+2s=\dfrac{bc-d}{c}\\$$

$$\implies ax_1=\dfrac{bc-d}{c}-2s$$

$$\implies ax_1=\dfrac{bc-d-2cs}{c}$$

$$\implies x_1=\dfrac{bc-d-2cs}{ac}$$

Case II: $e\neq 0$

$$\left(\begin{array}{c}0 & a & 2 & 1 & b\\ 0 & 0 & 0 & c & d\\ 0 & 0 & e & 0 & 0\end{array}\right)\\$$

$\text{Solving the system now yields...}\\$

$$x_2=0$$

4$x_3=\dfrac{d}{c},c\neq 0\$$

$$ax_1+2x_2+x_3=b\\$$

$$\implies ax_1=b-\dfrac{d}{c}\\$$

$\text{We have no free variable here, just solve in the usual manner}$

$$\\ ax_1=\dfrac{bc-d}{c}\\$$

$$\implies x_1=\dfrac{bc-d}{ac}$$

Here, I took it as echelon form. For row reduced echelon form, it will take a lot more work and a few more conditions.

$$\left(\begin{array}{c}a & 2 & 1 & b\\ 0 & e & 0 & 0 \\0 & 0 & c & d\end{array}\right)\sim \left(\begin{array}{c}1 & \dfrac{2}{a} & \dfrac{1}{a} & \dfrac{b}{a}\\ 0 & 1 & 0 & 0 \\0 & 0 & 1 & \dfrac{d}{c}\end{array}\right)\sim \left(\begin{array}{c}1 & 0 & 0 & \dfrac{b}{a}-\dfrac{d}{ac}\\ 0 & 1 & 0 & 0 \\0 & 0 & 1 & \dfrac{d}{c}\end{array}\right)$$

$$x_3=\dfrac{d}{c},c\neq 0$$

$$x_2=0$$

$$x_1=\dfrac{b}{a}-\dfrac{d}{ac}=\dfrac{bc-d}{ac},a\neq 0$$

The solutions work as long as $a\neq 0,c\neq 0$

This should work.