Solution for a matrix commutation I am new to linear algebra and having trouble with the matrix commutation logic below.
Given, $(\mu - \gamma)'\Sigma^{-1}1 = \lambda$ where $\lambda \in R$ and $\mu, 1, \gamma \in R^N$ and $\Sigma^{-1}$ is a NxN matrix.
I am trying to express $\gamma \text{ in terms of } \mu, \Sigma^{-1}, \lambda$.
Is this achievable and if yes, any advise on how I should proceed?
Thanks in advance.
 A: I think you can't. Here is what I tried and why I thought that it is impossible.
I assume $\gamma, \mu$ to be a 1$\times$n matrix, 1 is a n$\times$1 matrix whose entries are identically 1 and  $\Sigma^{−1}$ is an invertible matrix.
Now we can manipulate given formula as this :
$\mu\Sigma^{−1}1 - \gamma\Sigma^{−1}1=λ$
$\Rightarrow \gamma\Sigma^{−1}1 = \mu\Sigma^{−1}1 - λ$
And if I could find a 1$\times$n matrix $\alpha$ such that $\Sigma^{−1}1 \times \alpha = I$(here I put $\times$ to emphasize that it is a matrix multiplication, and $I$ represents n$\times$n identity matrix), then by right multiplying it both sides, we get
$\gamma = (\mu\Sigma^{−1}1 - λ)\alpha$.
But this is impossible.
Suppose there is such $\alpha$, that is, $\Sigma^{−1}1 \alpha = I$. Then since $\Sigma^{−1}$ is invertible matirx, $1 \alpha = \Sigma I = \Sigma$.
Hence $1 \alpha$ has to be an invertible matirx. But $rank(1 \alpha) = 1$ or $0$ for any 1$\times$n matrix $\alpha$. This leads to a contradiction unless you are dealing with 1$\times$1 matrices, which is just an ordinary real number arithmetic.
