Question

On her birthday Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got Rs.10 more. However, if there were 16 children more, everyone would have got Rs.10 less. Using the matrix method, find the number of children and the amount distributed by Seema. What values are reflected by Seema?

Answer 1

Hint : Consider no. of children as x and the amount for each child is y. Total amount becomes xy. According to given conditions subtract and add to ‘x’ and ‘y’ and find their corresponding total amounts. When the children are 8 less, the amount for each child will get is 10 more, total amount will be $\left( {x - 8} \right)\left( {y + 10} \right)$ and when the children are 16 more, the amount each child will get is 10 less, total amount becomes $\left( {x + 16} \right)\left( {y - 10} \right)$. Write these equations in matrices form. And solve these matrices using inverse, determinant whatever is required.

Complete step-by-step answer :
We have to calculate the no. of children and total money each child got using the matrix method.
Let the no. of children is x and money each child got is y.
Total amount all the children will get is $xy$
When the no. of children are 8 less $x - 8$, then the money each child got would be Rs. 10 more $y + 10$
Then the total money would become $\left( {x - 8} \right)\left( {y + 10} \right) \to (1)$
When the no. of children are 16 more $x + 16$, then the money each child got would be Rs.10 less $y - 10$
Then the total money would become $\left( {x + 16} \right)\left( {y - 10} \right) \to (2)$
The money is not changed so the equations 1 and 2 will be equal to $xy$
$\left( {x - 8} \right)\left( {y + 10} \right) = xy \\\ \to xy + 10x - 8y - 80 = xy \\\ \to 10x - 8y = 80 \\\ \to 5x - 4y = 40 \to eq(3) \\\ \left( {x + 16} \right)\left( {y - 10} \right) = xy \\\ \to xy - 10x + 16y - 160 = xy \\\ \to - 10x + 16y = 160 \\\ \to - 5x + 8y = 80 \\\ \to 5x - 8y = - 80 \to eq(4) \\\$
Writing equations 3 and 4 in matrix forms
Let the matrix A consists of the coefficients of x and y in equations 3 and 4, so the matrix A will be A = \left[ {\begin{array}{*{20}{c}} 5&{ - 4} \\\ 5&{ - 8} \end{array}} \right] and the matrix B consists of the values of equations and 4, so the matrix B will be B = \left[ {\begin{array}{*{20}{c}} {40} \\\ { - 80} \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 3 and 4 as
$AX = B \\\ 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
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] \\\ {A^{ - 1}} = \dfrac{1}{{\det A}}adj\left( A \right) \\\ A = \left[ {\begin{array}{*{20}{c}} 5&{ - 4} \\\ 5&{ - 8} \end{array}} \right] \\\ \det A = \left( { - 40 + 20} \right) = - 20 \\\ adj\left( A \right) = \left[ {\begin{array}{*{20}{c}} { - 8}&{ - \left( { - 4} \right)} \\\ { - 5}&5 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 8}&4 \\\ { - 5}&5 \end{array}} \right] \\\ {A^{ - 1}} = \dfrac{1}{{ - 20}}\left[ {\begin{array}{*{20}{c}} { - 8}&4 \\\ { - 5}&5 \end{array}} \right] = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}} 8&{ - 4} \\\ 5&{ - 5} \end{array}} \right] \\\ X = {A^{ - 1}}B \\\ \to X = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}} 8&{ - 4} \\\ 5&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {40} \\\ { - 80} \end{array}} \right] \\\ \to \left[ {\begin{array}{*{20}{c}} x \\\ y \end{array}} \right] = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}} {320 + 320} \\\ {200 + 400} \end{array}} \right] = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}} {640} \\\ {600} \end{array}} \right] \\\ \to \left[ {\begin{array}{*{20}{c}} x \\\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\left( {\dfrac{{640}}{{20}}} \right)} \\\ {\left( {\dfrac{{600}}{{20}}} \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {32} \\\ {30} \end{array}} \right] \\\ \therefore x = 32,y = 30 \\\
The no. of children is 32, money each student got initially is Rs.30
Total amount= $32 \times 30 = Rs.960$

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.