Answer
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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.
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.
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