Question

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

Answer 1

Let the cost of 1 dozen pencils, 1 dozen pens and 1 dozen erasers be Rs. x, Rs. y and Rs. 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,

Applying R₁ ↔ R₂,

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

Applying  R₃ → R₃ → 2R₂,

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

∴ By equality of matrices, we get

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 Rs. 5 per dozen, that of pens is Rs. 8 per dozen and that of erasers is Rs. 8 per dozen.

Answer 2

Let the cost of one dozen pencils, one dozen pends and one dozen erasers be Rs.x,Rs.y and Rs.z respectively.

 

According to the questions,
 

4x+3y+2z=60
 

2x+4y+6z=90
 

6x+2y+3z=70

The above equations can be expressed in the matrix form as

⎣⎢⎢⎡​426​342​263​⎦⎥⎥⎤​⎣⎢⎢⎡​xyz​⎦⎥⎥⎤​=⎣⎢⎢⎡​609070​⎦⎥⎥⎤​

Using R2​→R2​−21​R1​ and R3​→R3​−23​R1​

⎣⎢⎢⎢⎢⎢⎡​400​325​−25​​250​⎦⎥⎥⎥⎥⎥⎤​⎣⎢⎢⎡​xyz​⎦⎥⎥⎤​=⎣⎢⎢⎡​6060−20​⎦⎥⎥⎤​

Using R3​→R3​+R2​, we get

⎣⎢⎢⎢⎡​400​325​0​255​⎦⎥⎥⎥⎤​⎣⎢⎢⎡​xyz​⎦⎥⎥⎤​=⎣⎢⎢⎡​606040​⎦⎥⎥⎤​

Now the equations can be rewrite in their original form as
 

5z=40....(i)
 

⇒z=8
 

25​y+5z=60....(ii)
 

4x+3y+2z=60....(iii)
 

Putting the value of z in equation (ii),
 

25​y+5(8)=60
 

y=5(60−40)​×2=8
 

⇒y=8
 

Putting the values of y and z in equations (iii), we get
 

4x+3(8)+2(8)=60
 

4x=60−24−16
 

x=420​=5
 

∴x=5
 

∴ The cost of each item per dozen are Rs.5,Rs.8 and Rs.8 respectively.