Let A = … and B = … be a skew symmetric matrix of order 2 such that b_12 = 1. If AB = I_2 where I_2 is identity matrix of order 2, then

The corner points of the feasible region are $(0, 0)$(0,0), $(2, 4)$(2,4), $(0, 5)$(0,5), and $(4, 0)$(4,0).

The objective function is $z = ax + by$z=ax+by where $a, b > 0$a,b>0.

The maximum value occurs at both $(2, 4)$(2,4) and $(4, 0)$(4,0).

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In Linear Programming, the maximum value of $z = ax + by$z=ax+by always occurs at a corner point of the feasible region.

When the maximum occurs at two corner points simultaneously, both points give the same maximum value of $z$z, and the line $z = ax + by$z=ax+by is parallel to the edge connecting these two points.

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Since the maximum occurs at both $(2, 4)$(2,4) and $(4, 0)$(4,0), the value of $z$z at these points must be equal.

At point $(2, 4)$(2,4):

$$z = a(2) + b(4)$$z=a(2)+b(4)

$$z = 2a + 4b$$z=2a+4b

At point $(4, 0)$(4,0):

$$z = a(4) + b(0)$$z=a(4)+b(0)

$$z = 4a$$z=4a

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Setting the values equal:

$$2a + 4b = 4a$$2a+4b=4a

$$4b = 4a - 2a$$4b=4a−2a

$$4b = 2a$$4b=2a

$$2b = a$$2b=a

Therefore $a = 2b$a=2b.