The management committee of a residential colony decided to award some of its members (say x) for honesty,

As per the information given in the question, the following equations hold true:

$x+y+z=12,2 x+3(y+z)=33$ and $x+z-2 y=0$

The above three equations can be represented in the form of a matrix as

$\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]$

Or $A X=B$, where, $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]$

$|A|=3 \neq 0$ Thus, $A$ is non-singular. Therefore, its inverse exists.

Adj $A$ is given by $\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right] \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]$

$X=A^{-1} B=\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]=\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right]$

$\therefore x=3, y=4$, and $z=5 .$

Therefore, the number of awardees for Honesty, Cooperation and Supervision are 3, 4, and 5 respectively.

One more value which the management of the colony must include for awards may be Sincerity.