Matrix equation

Question

I would like to know if there is a way to resolve an equation such as this one:
$$ (\begin{matrix} a_{11} & a_{12} & a_{13} \end{matrix}) \times (\begin{matrix} b_{11} \\ b_{21} \\ b_{31} \end{matrix}) = (\begin{matrix}0\end{matrix}) $$ Where we only know $a_{11}$, $a_{12}$ and $a_{13}$ ? I guess there are can be an infinity of solutions, starting with all elements of B being equal to zero. But I would like to know what would be the way to find one, if there is any.

Answer 1

By applying the matrix product, this equality comes down to:

$a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31}=0$

Assuming that we know all the $a$'s, we may choose $b_{21}=s$ and $b_{31}=t$ where $s,t\in\mathbb{R}$ are "free-variables".

So we get:

• $b_{11}=\frac{-a_{12}s-a_{13}t}{a_{11}}$

• $b_{21}=s$

• $b_{31}=t$

Hence all the solution to that problem are vectors of the form $(b_{11},b_{21},b_{31})=s(-\frac{a_{12}}{a_{11}},1,0)+t(-\frac{a_{13}}{a_{11}},0,1)$ where $s,t\in\mathbb{R}$.

Answer 2

This is essentially a dot product of vectors $(a_{11}, a_{12}, a_{13})$ and $(b_{11}, b_{21}, b_{31})$ in three-dimensional space. A dot product is zero when the two vectors are orthogonal (i.e. at right angles), so if the $a$ vector is nonzero, the $b$ vector may lie anywhere in the plane it is perpendicular to; of course if the $a$ vector is zero, the $b$ vector can be anything at all.