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Prove that points on a line satisfy a system of linear equations

by RoRi

Let

    \[ P = (1,1,1), \qquad A = (2,1,-1) \]

and consider the line L = \{ P + tA \}. Show that each point (x,y,z) on L satisfies the system of linear equations:

    \begin{align*} x - y + z &= 1, \\ x + y + 3z &= 5, \\ 3x + y + 7z &= 11. \end{align*}

Proof. First, we solve the system of equations. (I’m using matrices to do this since it’s faster and easier. Apostol may not have actually covered this yet, but you can solve it by eliminating variables in basically the same way.) We have,

    \begin{align*} \begin{pmatrix*}[r] 1 & -1 & 1 & 1 \\ 1 & 1 & 3 & 5 \\ 3 & 1 & 7 & 11 \end{pmatrix*} &\xrightarrow{\substack{R_2 \mapsto R_2 - R_1 \\ R_3 \mapsto R_3 - 3R_1}} \begin{pmatrix*}[r] 1 & -1 & 1 & 1 \\ 0 & 2 & 2 & 4 \\ 0 & 4 & 4 & 8 \end{pmatrix*} \\[9pt] &\xrightarrow{\substack{R_2 \mapsto \frac{1}{2}R_2 \\ R_3 \mapsto R_3 - 2R_2}} \begin{pmatrix*}[r] 1 & -1 & 1 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix*} \\[9pt] &\xrightarrow{R_1 \mapsto R_1 + R_2} \begin{pmatrix*}[r] 1 & 0 & 2 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix*}. \end{align*}

There, the variable z is free and we have x = 3-z, y = 2-z. Letting z = 1-t we have

    \[ x = 2t+1, \qquad y = t+1, \qquad z = 1-t. \]

But these are exactly the points

    \[ \{ (2t+1, t+1, t-1) \} = \{ (1,1,1) + t(2,1,-1) \} = \{ P + tA \}. \qquad \blacksquare \]