Question

Two schools $P$ and $Q$ want to award their selected students on the values of Discipline, Politeness, and Punctuality. The school $P$ wants to award Rs. $x$ each, Rs. $y$ each and Rs. $z$ each for the three respectively values to its $3,2$ and $1$ students with total award money of Rs. $1,000$. School $Q$ wants to spend Rs. $1,500$ to award it is $4,1$ and $3$ students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is Rs. $600$, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.

Answer 1

Let $x,y$ and $z$ be the prize amount per student for Discipline, Politeness and Punctuality respectively.

$3x+2y+z=1000$

$4x+y+3z=1500$

$x+y+z=600$

These three equations can be written as

$\left[\begin{array}{ccc}3& 2& 1\\ 4& 1& 3\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1000\\ 1500\\ 600\end{array}\right]$

$AX=B$

$\left|A\right|=3(1–3)–2(4–3)+1(4–1)$

$=-6-2+3=-5$

Hence, the unique solution given by $X=A^{-1}B$

$C_{11}={(-1)}^{1+1}\left|\begin{array}{cc}1& 3\\ 1& 1\end{array}\right|=1-3=-2$

$C_{12}={(-1)}^{1+2}\left|\begin{array}{cc}4& 3\\ 1& 1\end{array}\right|=-(4-3)=-1$

$C_{13}={(-1)}^{1+3}\left|\begin{array}{cc}4& 1\\ 1& 1\end{array}\right|=4-1=3$

$C_{21}={(-1)}^{2+1}\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|=-(2-1)=-1$

$C_{22}={(-1)}^{2+2}\left|\begin{array}{cc}3& 1\\ 1& 1\end{array}\right|=3-1=2$

$C_{23}={(-1)}^{2+3}\left|\begin{array}{cc}3& 2\\ 1& 1\end{array}\right|=-(3-2)=-1$

$C_{31}={(-1)}^{3+1}\left|\begin{array}{cc}2& 1\\ 1& 3\end{array}\right|=6-1=5$

$C_{32}={(-1)}^{3+2}\left|\begin{array}{cc}3& 1\\ 4& 3\end{array}\right|=-(9-4)=-5$

$C_{33}={(-1)}^{3+3}\left|\begin{array}{cc}3& 2\\ 4& 1\end{array}\right|=3-8=-5$

$adjA={\left[\left[\begin{array}{ccc}-2& -1& 3\\ -1& 2& -1\\ 5& -5& -5\end{array}\right]\right]}^T$

$=\left[\begin{array}{ccc}-2& -1& 5\\ -1& 2& -5\\ 3& -1& -5\end{array}\right]$

$X=A^{-1}B=\frac{1}{\left|A\right|}\left(adj\left(A\right)\right)B$

$X=-\frac{1}{5}\left[\begin{array}{ccc}-2& -1& 5\\ -1& 2& -5\\ 3& -1& -5\end{array}\right]\left[\begin{array}{c}1000\\ 1500\\ 600\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{-1}{5}\left[\begin{array}{c}-2000-1500+3000\\ -1000+3000-3000\\ 3000-1500-3000\end{array}\right]$

$=\frac{-1}{5}\left[\begin{array}{c}-500\\ -1000\\ -1500\end{array}\right]$

$=\left[\begin{array}{c}100\\ 200\\ 300\end{array}\right]$

$\therefore x=100,y=200,z=300$

VALUE:Excellence in extra curricular activities should be another value considered for an award.