How do I solve a multivariable equation?
How could I solve for variables $x$, $y$, $z$, and $w$ for the equation
$$ax+by+cz+dw$$
With given values $a$, $b$, $c$, and $d$. For example, how would I find a set of potential values for $x$, $y$, $z$, and $w$ for:
$$-x-3y+7z-3w=0$$
Other than them all being $0$?
$\endgroup$ 12 Answers
$\begingroup$Let $A=\begin{bmatrix}a & b & c & d\end{bmatrix}$ and $\mathbf{x} = \begin{bmatrix}x \\ y \\ z \\ w\end{bmatrix}$.
Thus, we have to solve $A\mathbf{x} = \mathbf{0}$.
We can put $A$ into echelon form by dividing by $a$ (or multiplying by the reciprocal of $a$). $$A \cdot{} \frac{1}{a} = \begin{bmatrix}1 & \frac{b}{a} & \frac{c}{a} & \frac{d}{a}\end{bmatrix}$$ From there we see that there is one pivot column and three free variables. So, the vectors (real numbers can be thought of as 1-dimensional vectors) in $A$ are linearly dependent, meaning there exists a nontrivial (not all entries are zero) vector $\mathbf{x}$ such that the equation holds. Thus, there is an infinite number of solutions.
$A$ augmented with $\mathbf{0}$, yields $\begin{bmatrix}1 & \frac{b}{a} & \frac{c}{a} & \frac{d}{a} & 0\end{bmatrix}$. From that, we see $$x = -\frac{b}{a} - \frac{c}{a} - \frac{d}{a}.$$ Since $y, z,$ and $w$ are free, we can represent the solution as $$\mathbf{x} = \begin{bmatrix}x \\ y \\ z \\ w\end{bmatrix} = t_1\cdot{}\begin{bmatrix}-\frac{b}{a} \\ 1 \\ 0 \\ 0\end{bmatrix}+t_2\cdot{}\begin{bmatrix}-\frac{c}{a} \\ 0 \\ 1 \\ 0\end{bmatrix}+t_3\cdot{}\begin{bmatrix}-\frac{d}{a} \\ 0 \\ 0 \\ 1\end{bmatrix}$$ where $t_1, t_2,$ and $t_3$ are any arbitrary values.
So in your example, the solution vector, $$\mathbf{x} = t_1\cdot{}\begin{bmatrix}-3 \\ 1 \\ 0 \\ 0\end{bmatrix} + t_2\cdot{}\begin{bmatrix}7 \\ 0 \\ 1 \\ 0\end{bmatrix} + t_3\cdot{}\begin{bmatrix}-3 \\ 0 \\ 0 \\ 1\end{bmatrix}$$ where $t_1, t_2,$ and $t_3$ are any arbitrary values.
$\endgroup$ $\begingroup$That equation has infinitely many solutions.
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