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Matrices

Question
CBSEENMA12035633

A typist charges Rs. 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and 10 Hindi pages are Rs. 180. Using matrices, find the charges of typing one English and one Hindi page separately. However, typist charged only Rs. 2 per page from a poor student Shyam for 5 Hindi pages. How much less was charged from this poor boy? Which values are reflected in this problem?

Solution

Let charges for typing one English page be Rs. x.
Let charges for typing one Hindi page be Rs.y.
Thus from the given statements, we have,
10x  + 3y = 145
3x + 10y = 180
Thus the above system can be written as,
open square brackets table row 10 3 row 3 10 end table close square brackets open square brackets table row straight x row straight y end table close square brackets space space equals space open square brackets table row 145 row 180 end table close square brackets
rightwards double arrow space AX space equals space straight B comma space where comma space straight A space equals space open square brackets table row 10 3 row 3 10 end table close square brackets comma space straight X space space equals space open square brackets table row straight x row straight y end table close square brackets space and space open square brackets table row 145 row 180 end table close square brackets
Multiply space straight A to the power of negative 1 end exponent space on space both space the space sides comma space we space have comma

straight A to the power of negative 1 end exponent space straight x space AX space equals space straight A to the power of negative 1 end exponent straight B

rightwards double arrow space IX space space equals space straight A to the power of negative 1 end exponent straight B

rightwards double arrow space straight X space equals space straight A to the power of negative 1 end exponent straight B

Thus comma space we space need space to space find space the space inverse space of space the space matrix space straight A.

We space know space that comma space if space space straight P space equals space open square brackets table row straight a straight b row straight c straight d end table close square brackets space then space straight p to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator ad minus bc end fraction space open square brackets table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close square brackets

Thus comma space straight A to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator 10 space straight x space 10 minus 3 space straight x space 3 end fraction open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets

equals fraction numerator 1 over denominator 100 minus 9 end fraction open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets

equals 1 over 91 open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets

therefore space space straight X space equals space 1 over 91 open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets space space open square brackets table row 145 row 180 end table close square brackets

equals 1 over 91 open square brackets table row cell 10 space straight x space 145 space minus end cell cell 3 straight x space 180 end cell row cell negative 3 space straight x space 145 space plus end cell cell 10 straight x 180 end cell end table close square brackets

space equals 1 over 91 open square brackets table row 910 row 1365 end table close square brackets

equals open square brackets table row 10 row 15 end table close square brackets

rightwards double arrow space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row 10 row 15 end table close square brackets

therefore comma

straight x space equals 10 space and space straight y space equals space 15

Amount space taken space from space shyam space equals 2 space straight x space 5 space equals Rs. space 10

Actual space rate space equals space 15 space straight x space 5 space equals Rs. space 75

Difference space amount space equals space 75 minus 10 space equals space 65

Rs. space 65 space was space less space charged space from space the space poor space boy space Shyam.
Humanity is reflected in this problem.