A school wants to award its students for the values of Honesty,

Let the award money given for Honesty, Regularity and Hard work be x, y and z respectively.

Since total cash award is Rs 6,000.

∴ x + y + z = Rs 6,000 ...(1)

Three times the award money for Hard work and Honesty is Rs 11,000.

∴ x + 3 z = Rs 11,000

⇒ x + 0.y + 3 z = Rs 11,000 ...(2)

Award money for Honesty and Hard work is double the one given for regularity.

∴ x + z = 2y

⇒ x − 2y + z = 0 ...(3)

The above system can be written in matrix form as,

$\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}6000 \\ 11000 \\ 0\end{array}\right]$

Or $A X=B$, where

$A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}6000 \\ 11000 \\ 0\end{array}\right]$

$|A|=6 \neq 0$

Thus, A is non-singular. Hence, it is invertible.

$\operatorname{Adj} A=\left[\begin{array}{ccc}6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1\end{array}\right]$

$\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{6}\left[\begin{array}{ccc}6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1\end{array}\right]$

$X=A^{-1} B=\frac{1}{6}\left[\begin{array}{ccc}6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1\end{array}\right]\left[\begin{array}{l}6000 \\ 11000 \\ 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}500 \\ 2000 \\ 3500\end{array}\right]$

Hence, $x=500, y=2000$, and $z=3500$.

Thus, award money given for Honesty, Regularity and Hard work are Rs 500, Rs 2000 and Rs 3500 respectively.

School can include sincerity for awards.