1
$\begingroup$

consider following matrix equation

$\boldsymbol{1}^T \Sigma^{-1} ( \bar{\textbf{x}} - n \boldsymbol{1}. c) = 0 $

here $\boldsymbol{1} , \bar{\textbf{x}}$ are vectors of dimension $n \times 1$ and $\Sigma$ is invertible matrix of dimension $n \times n$ and n,c are constants

is there any way to solve it for scalar c ?

$\boldsymbol{1}^T \Sigma^{-1}\bar{\textbf{x}} = (n.c)\boldsymbol{1}^T \Sigma^{-1}\boldsymbol{1}$

and I am struck at here

$\endgroup$
1
$\begingroup$

$$c = \frac{ \boldsymbol{1}^T \Sigma^{-1}\bar{\textbf{x}} }{ n \boldsymbol{1}^T \Sigma^{-1}\boldsymbol{1} }.$$

Remember that $\boldsymbol{1}^T \Sigma^{-1}\boldsymbol{1}$ and $\boldsymbol{1}^T \Sigma^{-1}\bar{\textbf{x}}$ are scalars.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.