RREF Matrix with unknown variable d

Question

I have this question, although I've used Maple 16 to get the answer i obviously need the working out, and thus i don't know what to do with the constant.

1. The following is a system of linear equations in unknowns $x,\ y,\ z$ and $d$ is a scalar.

$x-8y+10z\ =\ 15$

$2y-3z\ =\ -3$

$x-4y+4z\ =\ d$

(a) Write down the augmented matrix associated with this linear system of equations.

(b) Using only elementary row operations find the row reduced echelon form for the augmented matrix of the above system of equations.

(c) For which value of d is the above system of equations consistent? For that value of d find the solution of the system of equations, expressed in vector parametric form

ANSWERS

(a) Okay so for a

$\left[ \begin{array}{cccc} 1 & -8 & 10 & 15\\ 0 & 2 & -3 & -3\\ 1 & -4 & 4 & d\\ \end{array} \right]$

(b) Using Maple i get the answer to be

$\left[ \begin{array}{cccc} 1 & 0 & -2 & 0\\ 0 & 1 & -\frac{3}{2} & 0\\ 0 & 0 & 0 & 1\\ \end{array} \right]$

Although i could write my working out, my problem is.. where has the d disappeared to?

(c) Considering i can't do (b) I'm not sure what to do here either.

Answer 1

$$\left[ \begin{array}{cccc} 1 & -8 & 10 & 15\\ 0 & 2 & -3 & -3\\ 1 & -4 & 4 & d\\ \end{array} \right] \overset{}{\rightarrow}\left[ \begin{array}{cccc} 1 & -8 & 10 & 15\\ 0 & 2 & -3 & -3\\ 0 & 4 & -6 & d-15\\ \end{array} \right]\overset{}{\rightarrow}\left[ \begin{array}{cccc} 1 & -8 & 10 & 15\\ 0 & 2 & -3 & -3\\ 0 & 0 & 0 & d-9\\ \end{array} \right] $$

The row reductions had a single element left over in the last row. This gets scaled regardless of the value of $d$ (unless $d=9$ in which case it should be zero)

This means that the row reduced form is as you have shown. The operations to reach that form do however still depend on the value of $d$.

Step one was, row $3$ gets one of row $1$ subtracted. Step two was row $3$ gets minus two of row $2$. Then (not shown) row $3$ is divided by $d-9$ to reach the final form you have shown. (Row $2$ gets divided by two also, and the one in row $3$ is the "pivot" to zero that final column to reach the exact form you show.)