matrices Solve single equation with parameters

Question

So I have this equation $$-2x+3y+9z=3$$ and I am supposed to solve it using parameters since there is no unique solution to this. that would be vector form?

I know how to do it when there is at least 2 equations. https://www.khanacademy.org/math/algebra-home/alg-matrices/alg-row-echelon-and-gaussian-elimination/v/matrices-reduced-row-echelon-form-2

but what about the one?

tank u !

Answer 1

whenever you fix the value for $2$ of the unknown, you can recover the third.

$$z=\frac{3+2x-3y}{9}$$

$$\begin{pmatrix} x \\ y \\ z\end{pmatrix}=\begin{pmatrix} s \\ t \\ \frac{3+2s-3t}{9}\end{pmatrix}$$

where $s$ and $t$ are parameters.

Answer 2

In fact, since you have one equation and three unknowns, you can complete the system by adding other two "dummy" equations.
In these, for the coefficients of the unknowns, you can fix any value with the only bound that the resulting determinant be non-null.
You complete the column of the "known" values with two parameters.
Since the determinant is not null, you are assured that for any values of the two parameters you have a unique vector satisfying the system and thus the first equation.
Conversely for the parameters spanning all their range you get all the solution vectors.

Example

You can put, as it is usual done, $$ \left( {\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ { - 2} & 3 & 9 \\ \end{array} } \right)\left( {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} s \\ t \\ 3 \\ \end{array} } \right) $$ and which will give the solution indicated by Siong.
But you can also put $$ \left( {\begin{array}{*{20}c} 1 & 0 & 0 \\ 1 & 1 & 0 \\ { - 2} & 3 & 9 \\ \end{array} } \right)\left( {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} s \\ t \\ 3 \\ \end{array} } \right)\quad \Rightarrow \quad \left\{ \begin{gathered} x = s \hfill \\ y = - s + t \hfill \\ z = 1/9\left( {5s - 3t + 3} \right) \hfill \\ \end{gathered} \right. $$ or $$ \left( {\begin{array}{*{20}c} 1 & 1 & 1 \\ { - 1} & 0 & 1 \\ { - 2} & 3 & 9 \\ \end{array} } \right)\left( {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} s \\ t \\ 3 \\ \end{array} } \right)\quad \Rightarrow \quad \left\{ \begin{gathered} x = - 3s - 6t + 3 \hfill \\ y = 7s + 11t - 6 \hfill \\ z = - 3s - 5t + 3 \hfill \\ \end{gathered} \right. $$