Q. 325.0( 1 Vote )

# A shopkeeper has

Let the varieties of pen A, B and C be x, y and z respectively.

As per the Data we get,

x + y + z = 21

4x + 3y + 2z = 60

6x + 2y + 3z = 70

These three equations can be written as

A X = B

|A| = 1(9 – 4) – 1(12 – 12) + 1(8 – 18)

= 1(5) – 1(0) + 1(– 10)

= 5 – 0 – 10

= – 5

Hence, the unique solution given by x = A – 1B

C11 = (– 1)1 + 1 (9 – 4) = 5

C12 = (– 1)1 + 2 (12 – 12) = 0

C13 = (– 1)1 + 3 (8 – 18) = – 10

C21 = (– 1)2 + 1 (3 – 2) = – 1

C22 = (– 1)2 + 2 (3 – 6) = – 3

C23 = (– 1)2 + 3 (2 – 6 ) = 4

C31 = (– 1)3 + 1 (2 – 3) = – 1

C32 = (– 1)3 + 2 (2 – 4) = 2

C33 = (– 1)3 + 3 (3 – 4) = – 1

X = A – 1 B =

X =

X =

=

=

Hence, A = Rs 5, B = Rs 8 and C = Rs 8

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