Using elementary row operations to solve intersection of two planes

Question

The question I am struggling with is the following: Solve the following using elementary row operations, and interpret each system of equations geometrically: \begin{align*} x - 3y + 2z &= 8\\ 3x - 9y + 2z &= 4 \end{align*} The answer given in the book is $x = -2+3t, y = t, z = 5$, and the planes meet in a line.

I put this in matrix form, so $$\left(\begin{array}{c c c|c} 1 & -3 & 2 & 8\\ 3 & -9 & 2 & 4 \end{array}\right) $$ Then I subtracted R1 from R2 $$\left(\begin{array}{c c c|c} 1 & -3 & 2 & 8\\ 2 & -6 & 0 & -4 \end{array}\right) $$ Then R1 - $\frac{1}{2}$R2 $$\left(\begin{array}{c c c|c} 0 & 0 & 2 & 10\\ -2 & -6 & 0 & -4 \end{array}\right) $$ So I got that $2z =10$, so $z = 5$, but I am stumped on how to continue. How can I get $x$ and $y$ and interpret the results in terms of plane intersection?

Answer 1

$2z = 10$, so $z = 5$.

$2x - 6y = -4 \Rightarrow2x = -4 + 6y \Rightarrow x = -2 + 3y$. Let $y = t$, we get $x = -2 + 3t$, $y = t$, $z = 5$, as desired.

Answer 2

Are you confused because you expect one result (one point in space)? If the result is a plane or line, there are infinitely many solutions.

Thus you need to introduce some new parameter - t. Depending on what you set t to it will point to different points in 3D space with coordinates:

x = 2 - 3t
y = t
z = 5

In order to get there, if you keep using vector notation, you first need to come up with a Particular Solution and the Null Space:

$$R=\left(\begin{array}{} 2\\ 0\\ 5 \end{array}\right), N=\left(\begin{array}{} -3\\ 1\\ 0 \end{array}\right) $$

So the General Solution looks like this:

$$ \left(\begin{array}{} 2\\ 0\\ 5 \end{array}\right)+t \left(\begin{array}{} -3\\ 1\\ 0 \end{array}\right) $$

Or you can keep working with polynomials and get to the result as described by @Matheus Andrade.