<s>[INST] Find numerical matrices to remove the same coefficients on the main diameter $M=\left(
\begin{array}{cccc}
\frac{1}{4}m_{11} & m_{12} & ... & m_{1n} \\
m_{12} & \frac{1}{4}m_{22} & ... & m_{2n} \\
\colon & \colon &  &  \colon\\
m_{1n} & ... &  & \frac{1}{4}m_{nn}
\end{array}
\right)$
$M$ is a Symmetric matrix That is, M has the $\frac{1}{4}$ coefficient along its main diagonal. How can I find the $M_1$ matrix?
$M_1=\left(
\begin{array}{cccc}
m_{11} & m_{12} & ... & m_{1n} \\
m_{12} & m_{22} & ... & m_{2n} \\
\colon & \colon &  &  \colon\\
m_{1n} & ... &  & m_{nn}
\end{array}
\right)$
Can it be done by simple transformations?? by multiply from left and right by the specific matrix.
 [/INST] Consider
$$M=\left(
\begin{array}{cccc}
\frac{1}{4} & \frac{1}{4} \\
\frac{1}{4} & \frac{1}{4} 
\end{array}
\right)$$
For this matrix, you have
$$M_1=\left(
\begin{array}{cccc}
1 & \frac{1}{4} \\
\frac{1}{4} & 1
\end{array}
\right)$$
If there existed $P$, $Q$ such that $PMQ=M_1$, then taking the determinant, you would get, because $\mathrm{det}(M)=0$, that
$$\mathrm{det}(M_1)=0$$
which is not true.
</s>