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The cost of 4 dozen pencils, 3 dozen pens and 2 dozen erasers is ₹ 60. The cost of 2 dozen pencils, 4 dozen pens and 6 dozen erasers is ₹ 90. Whereas the cost of 6 dozen pencils, 2 dozen pens and - Mathematics and Statistics

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प्रश्न

The cost of 4 dozen pencils, 3 dozen pens and 2 dozen erasers is ₹ 60. The cost of 2 dozen pencils, 4 dozen pens and 6 dozen erasers is ₹ 90. Whereas the cost of 6 dozen pencils, 2 dozen pens and 3 dozen erasers is ₹ 70. Find the cost of each item per dozen by using matrices

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उत्तर

Let the cost of 1 dozen pencils, 1 dozen pens and 1 dozen erasers be ₹ x, ₹ y and ₹ z respectively.

According to the given conditions,

4x + 3y + 2z = 60

2x + 4y + 6z = 90   

i.e. x + 2y + 3z = 45

6x + 2y + 3z = 70

Matrix form of the given system of equations is,

`[(4, 3, 2),(1, 2, 3),(6, 2, 3)] [(x),(y),(z)] = [(60),(45),(70)]`

Applying R1 ↔ R2

`[(1, 2, 3),(4, 3, 2),(6, 2, 3)] [(x),(y),(z)] = [(45),(60),(70)]`

Applying R2 → R2 − 4R1, R3 → R3 − 6R1 

`[(1, 2, 3),(0, -5, -10),(0, -10, -15)] [(x),(y),(z)] = [(45),(-120),(-200)]`

Applying R3 → R3 − 2R2,

`[(1, 2, 3),(0, -5, -10),(0, 0, 5)] [(x),(y),(z)] = [(45),(-120),(40)]`

Hence, the original matrix is reduced to an upper triangular matrix.

∴ By equality of matrices, we get

x + 2y + 3z = 45    .......(i)

−5y − 10z = −120

i.e. y + 2z = 24     .......(ii)

5z = 40      .......(iii)

i.e. z = 8

Substituting z = 8 in equation (ii), we get

y + 2(8) = 24

∴ y = 8

Substituting z = 8 and y = 8 in equation (i), we get

x + 2(8) + 3(8) = 45

∴ x + 16 + 24 = 45

∴ x = 5

∴ x = 5, y = 8, z = 8

Thus, the cost of pencils is ₹ 5 per dozen, that of pens is ₹ 8 per dozen and that of erasers is ₹ 8 per dozen.

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