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.