Identify $t\in\mathbb{R}$ such that (basically it is a determinant of a block matrix of size $n$): $$ \left| \begin{array}{cc} \mathbf{x}+t\mathbf{y} & \mathbf{B} \\ \mathbf{c} & \mathbf{D} \end{array} \right| = 0 $$

where $\mathbf{x}\in\mathbb{R}^{m\times1}$, $\mathbf{y}\in\mathbb{R}^{m\times1}$, $\mathbf{c}\in\mathbb{R}^{(n-m)\times1}$ are column vectors, and $\mathbf{B}\in\mathbb{R}^{m\times (n-1)}$, $\mathbf{D}\in\mathbb{R}^{(n-m)\times (n-1)}$ are matrices, and $m$ is an integer with $1 \le m \le n$, and $t\in\mathbb{R}$ is the unknown variable we want to solve.

I am not sure if the solution exists in general cases, but in my cases, solution $t$ exists.

I encountered such a problem when I want to write a computer program that finds a proper value $t$ that makes some points coplanar, such a geometric problem can be finally reduced to the above mathematical problem after some calculation.

The problem is quite challenging, because I would like to find a closed-form solution such that $t=...$ (right hand side is composed of known variables) so that computer can compute the solution $t$, otherwise we have to use symbolic calculation which is unfriendly to a computer. Even an approximate approach (e.g. iterative numerical approach) is also acceptable if the exact solution is too hard to find.

Do you have any thoughts on this? I would be grateful if you could provide some hints on this issue.


1 Answer 1


Let $X:=\begin{pmatrix} \mathbf{x}& \mathbf{B} \\ \mathbf{0}& \mathbf{D} \\ \end{pmatrix}$ and let $Y:=\begin{pmatrix} \mathbf{y}& \mathbf{B} \\ \mathbf{c}& \mathbf{D} \\ \end{pmatrix}$. By multilinearity of the determinant it holds: $$0=\det\begin{pmatrix} \mathbf{x}+t \mathbf y& \mathbf{B} \\ \mathbf{c}& \mathbf{D} \\ \end{pmatrix}= \det X + t \det Y \Leftrightarrow t\det Y=-\det X $$

So if $\det Y=\det X=0$, any $t\in \mathbb R$ works; if $\det Y=0\neq \det X$, no $t\in \mathbb R$ works; if $\det Y\neq 0$, then $t=-\frac{\det X}{ \det Y}$.

  • 2
    $\begingroup$ Thank you very much! The solution looks very elegant! $\endgroup$
    – Karbo Lei
    Feb 24 at 11:23

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