# Matrix equation

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.

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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 $s,t\in\mathbb{R}$.

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Alright, as I feared there is an infinity of solutions then. Thanks – Jukurrpa Apr 25 '12 at 17:56
Yeah, that's usually the case when you have more unknown variables than equations that control the possible solutions. – T. Eskin Apr 25 '12 at 18:00
Btw, you may handle the cases seperately if some of the $a$'s equal zero. If $a_{11}=0$, then $b_{11}$ has no control at all. You may repeat the above reasoning and get all possible solutions by choosing only one "free-variable" $b_{31}=t$. Similarly to any other $a$ being zero. If two $a$'s are zero, then those corresponding $b$'s have no control and can be anything. The remaining $b$ must be zero though. If all $a$'s are zero, then all values for $b$'s satisfy the equation. – T. Eskin Apr 25 '12 at 18:19

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.

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