Let the award money given for honesty, regularity and hard work be Rs. x, Rs. y and Rs. z respectively.
Since total cash award is Rs. 6,000.
∴ x + y + z = 6,000 ...(1)
Three times the award money for hard work and honesty amounts to Rs. 11,000.
∴ x + 3z = 11,000
⇒ x + 0×y + 3 z = 11,000 ...(2)
Award money for honesty and hard work is double that given for regularity.
∴ x + z = 2y
⇒ x – 2y + z = 0 ...(3)
The above system of equations can be written in matrix form AX = B as:
[

1	1	1
1	0	3
1	−2	1

][

x

y

z

] = [

6000

11000

0

]
Here,
A = [

1	1	1
1	0	3
1	−2	1

] , X = [

x

y

z

] and B = [

6000

11000

0

]
|A| = 1 (0 + 6) - 1 (1 - 3) + 1 (- 2 - 0) = 6 ≠ 0
Thus, A is non-singular. Hence, it is invertible.
Adj A = [

6	−3	3
2	0	−2
−2	3	−1

]
∴ A−1 =

1

|A|

adj A =

1

6

[

6	−3	3
2	0	−2
−2	3	−1

]
X = A−1B = =

1

6

[

36000−33000+0

12000+0−0

−12000+33000−0

] =

1

6

[

3000

12000

21000

]
⇒ [

x

y

z

] = [

500

2000

3500

]
Hence, x = 500, y = 2000, and z = 3500.
Thus, award money given for honesty, regularity and hard work is Rs. 500, Rs. 2000 and Rs. 3500 respectively.
The school can include awards for obedience.