Seema decided to donate some money …

Ans. Let Number of Children = x

Amount received by one child = y

total amount distributed = xy  ........(1)

According to first Condition : 

If No of children = (x - 8)

Amoutn received by one = (y+10)

total amount = (x-8)(y+10) 

=> xy +10x -8y - 80  ....... (2)

put (1) = (2)  {both are total amount}

xy = xy +10x -8y + 80

=> 10x - 8y = 80  

=> 5x-4y = 40 ........(3)

According to 2nd condition :

If No of children = (x +16)

Amount received by one = (y-10)

total amount = (x+16)(y-10) 

=> xy -10x + 16y - 160  ......... (4)

put (1) = (4)  {both are total amount}

=> xy = xy -10x + 16y - 160

=> -10y + 16y = 160

=> -5y + 8y = 80  ........ (5)

We can express (3) and (5) in Matrix as follow :

$A = \left(\begin{array}{cc}5& -4\\ -5& 8\end{array}\right), B = \left(\begin{array}{c}40\\ 80\end{array}\right), X = \left(\begin{array}{c}x\\ y\end{array}\right)$

As we know, 
AX = B

=> X = A^-1B          ........... (6)

To find A^-1, We need coffector of A

A₁₁ = 8,       A₁₂= 5,  

A₂₁= 4,     A₂₂= 5

Adjoint of A = $\left(\begin{array}{cc}8& 4\\ 5& 5\end{array}\right)$

A^-1 = $\frac{1}{\left(A\right)}Adjoint of A$

=> $\left(A\right) = \left(40-20\right) = 20$

=> $A^{-1}= \frac{1}{20}\left(\begin{array}{cc}8& 4\\ 5& 5\end{array}\right)$

=> $A^{-1}B = \frac{1}{20}\left(\begin{array}{cc}8& 4\\ 5& 5\end{array}\right)\left(\begin{array}{c}40\\ 80\end{array}\right)$

=> $A^{-1}B = \frac{1}{20}\left(\begin{array}{c}320 +320\\ 200+400\end{array}\right)$

=> $A^{-1}B = \left(\begin{array}{c}32\\ 30\end{array}\right) ......... \left(7\right)$

From (6) and (7), we get 

$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}32\\ 30\end{array}\right)$

So No of Children = 32 

Amount got by one child = 30

Total Amount = 960