Question

A coaching institute of English (subject) conducts classes in two batches I and II and fees for rich & poor children are different. In batch I, it has 20 poor and 5 rich children and monthly collection is Rs. 9000 whereas in batch II it has 5 poor and 25 rich children and the total monthly collection is Rs. 26000. Using matrix method, find monthly fees paid by each child of two types.

Answer 1

Hint : Let the money paid by one poor child is x and the money paid by one rich child is y. Convert the given statements into equations. The resulting equations are linear equations in 2 variables. Write these equations in matrices form. And solve these matrices using inverse, determinant whatever is required.

Complete step-by-step answer :
Monthly fee paid by one poor child is x then the money paid by 20 poor children is 20x and 5 poor children is 5x.
Monthly fee paid by one rich child is y then the money paid by 5 rich children is 5y and 25 rich children is 25y.
Monthly fee paid by 20 poor children and 5 rich children is Rs. 9000.
$\Rightarrow 20x + 5y = 9000 \\\ \Rightarrow 4x + y = 1800 \to eq\left( 1 \right) \\\$
Monthly fee paid by 5 poor children and 25 rich children is Rs. 26000.
$\Rightarrow 5x + 25y = 26000 \\\ \Rightarrow x + 5y = 5200 \to eq\left( 2 \right) \\\$
Writing equations 1 and 2 in matrix forms
Let the matrix A consists of the coefficients of x and y in equations 1 and 2, so the matrix A will be \Rightarrow A = \left[ {\begin{array}{*{20}{c}} 4&1 \\\ 1&5 \end{array}} \right] and the matrix B consists of the values of equations 1 and 2, so the matrix B will be \Rightarrow B = \left[ {\begin{array}{*{20}{c}} {1800} \\\ {5200} \end{array}} \right] and the matrix X contains the variables x and y, so the matrix X will be X = \left[ {\begin{array}{*{20}{c}} x \\\ y \end{array}} \right]
Therefore we can write equations 1 and 2 as
$\Rightarrow AX = B \\\ \Rightarrow X = {A^{ - 1}}B \\\$
To find the values of x and y we have to first find the value of inverse of matrix A. If matrix A is
\Rightarrow A = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right],adj\left( A \right) = \left[ {\begin{array}{*{20}{c}} d&{ - b} \\\ { - c}&a; \end{array}} \right] \\\ \Rightarrow {A^{ - 1}} = \dfrac{1}{{\det A}}adj\left( A \right) \\\ A = \left[ {\begin{array}{*{20}{c}} 4&1 \\\ 1&5 \end{array}} \right] \\\ \Rightarrow \det A = \left( {20 - 1} \right) = 19 \\\ \Rightarrow adj\left( A \right) = \left[ {\begin{array}{*{20}{c}} 5&{ - 1} \\\ { - 1}&4 \end{array}} \right] \\\ \Rightarrow {A^{ - 1}} = \dfrac{1}{{19}}\left[ {\begin{array}{*{20}{c}} 5&{ - 1} \\\ { - 1}&4 \end{array}} \right] \\\ \Rightarrow X = {A^{ - 1}}B \\\ \to X = \dfrac{1}{{19}}\left[ {\begin{array}{*{20}{c}} 5&{ - 1} \\\ { - 1}&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {1800} \\\ {5200} \end{array}} \right] \\\ \to \left[ {\begin{array}{*{20}{c}} x \\\ y \end{array}} \right] = \dfrac{1}{{19}}\left[ {\begin{array}{*{20}{c}} {9000 - 5200} \\\ { - 1800 + 20800} \end{array}} \right] = \dfrac{1}{{19}}\left[ {\begin{array}{*{20}{c}} {3800} \\\ {19000} \end{array}} \right] \\\ \to \left[ {\begin{array}{*{20}{c}} x \\\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\left( {\dfrac{{3800}}{{19}}} \right)} \\\ {\left( {\dfrac{{19000}}{{19}}} \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {200} \\\ {1000} \end{array}} \right] \\\ \therefore x = 200,y = 1000 \\\
The monthly fee paid by a poor child is Rs. 200 and by a rich child is Rs. 1000.

Note : In a 2×2 matrix A, the adjoint of A can be found by swapping the diagonal elements a, d and putting negatives in front of b and c. In transpose of a matrix, the rows and columns are reversed whereas the inverse of a matrix is such that when it is multiplied by the original matrix it must result in an identity matrix. So do not confuse between inverse and transpose of a matrix.