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.

Explanation

Let the cost of 1 dozen p[encils, 1 dozen of pens and 1 dozen of erasers be Rs. X, Rs. Y and Rs. Z respectively.
`therefore` From the given information,
`4x+3y+2x=60`
`2x+4y+6z=90`
i.e., `6x+2y+3z=70`
The equations can be written in the matrix from as :
`[{:(4,3,2),(1,2,3),(6,2,3):}][{:(x),(y),(z):}]=[{:(60),(45),(70):}]`
On applying `R_(1)harrR_(2),` we get
`[{:(1,2,3),(4,3,2),(6,2,3):}][{:(x),(y),(z):}]=[{:(45),(60),(70):}]`
On applying `R_(2)to R_(2)-4R)_(1)and R_(3)to R_(3)-6R_(1),`
we get
`[{:(1,2,3),(0,-5,-10),(0,-10,-15):}][{:(x),(y),(z):}]=[{:(45),(-120),(-200):}]`
On applying `R_(3)_to R_(3)-2R_(2),` we get
`[{:(1,2,3),(0,-5,-10),(0,0,5):}][{:(x),(y),(z):}]=[{:(45),(120),(40):}]`
`therefore[{:(x+2y+3z),(-5y-10z),(" "5z):}]=[{:(" "45),(-120),(" "40):}]`
`therefore` By equality of materices, we get
`x+2y+3z=45" "...(1)`
`-5y-10z=-120" "...(2)`
`5z=40" "...(3)`
From equation (3)
`z=8`
On substituting `z=8` in equation (2), we get
`-5y=80=-120`
`-5y=-40`
`thereforey=8`
On substituting `z=8` in equation (1), we get
`x+16+24=45`
`x+40=45`
`therefore x=45=44=5`
`therefore x=-5`
`therefore x=5, y=8, z=8`
`therefore` Cost of a dozen pencilis is Rs. 5, of a dozen of pens is Rs. 8 and of a dozen of erasers is Rs. 8.