Question

Two school $A$ and $B$ want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school $A$ wants to award $Rs.x$ each, $Rs.y$ each and $Rs.z$ each for the three respective values to $3,2$ and $1$ students respectively with a total award money of $Rs.1,600$. School $B$ wants to spend $Rs.2,300$ to award its $4,1$ and $3$ students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is $Rs.900$, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.

Answer 1

Let Matrix $D$ represents number of students receiving prize for the three categories:
$X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ where $x,y$ and $z$ are rupees mentioned as it is the question, for sincerity, truthfulness and helpfulness respectively.
$E=\left[\begin{array}{c}1600\\ 2300\\ 900\end{array}\right]$ is a matrix representing total award money for school A, B and for one prize for each value.
We can represent the given question in matrix multiplication as:
$DX=E$
$\Rightarrow \left|\begin{array}{ccc}3& 2& 1\\ 4& 1& 3\\ 1& 1& 1\end{array}\right|\left|\begin{array}{c}x\\ y\\ z\end{array}\right|=\left|\begin{array}{c}1600\\ 2300\\ 900\end{array}\right|$
Solution of the matrix equtaion exist if $|D|\ne 0$
i.e., $\left|\begin{array}{ccc}3& 2& 1\\ 4& 1& 3\\ 1& 1& 1\end{array}\right|=3(1-3)-2(4-3)+1(4-1)$
$=-6-2+3$
=$-5$
Therefore, the solution of the matrix equation is
$X=D^{-1}E$
To find $D^{-1}$:
$D^{-1}=\frac{1}{|D|}adj\left(D\right)$
Cofactor matrix of D $=\left|\begin{array}{ccc}-2& -1& 3\\ -1& 2& -1\\ 5& -5& -5\end{array}\right|$
Transpose of cofactor matrix = Adj D $\Rightarrow \left|\begin{array}{ccc}-2& -1& 5\\ -1& 2& -5\\ 3& -1& -5\end{array}\right|$
Therefore, $D^{-1}=\frac{1}{-5}\left[\begin{array}{ccc}-2& -1& 5\\ -1& 2& -5\\ 3& -1& -5\end{array}\right]$
Now, $X=D^{-1}E$
$=\frac{1}{-5}\left[\begin{array}{ccc}-2& -1& 5\\ -1& 2& -5\\ 3& -1& -5\end{array}\right]\left|\begin{array}{c}1600\\ 2300\\ 900\end{array}\right|$
$\left|\begin{array}{c}x\\ y\\ z\end{array}\right|=\left|\begin{array}{c}200\\ 300\\ 400\end{array}\right|$
Therefore, $x=200,y=300,z=400$.