 # Two schools P and Q want to award their selected students on the values of Tolerance,

Question:

Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹2,200School Q wants to spend ₹3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.

Solution:

​​Let the award money given for Tolerance, Kindness and Leadership be ₹x, ₹y and ₹respectively.

Since, the total cash award is ₹ 1,200 .

$\therefore x+y+z=1200$     ....(1)

Award money given by school $P$ is ₹ 2,200 .

$\therefore 3 x+2 y+z=2200$              ....(2)

Award money given by school $Q$ is $₹ 3,100$.

$\therefore 4 x+y+3 z=3100$                ....(3)

The above system of equations can be written in matrix form AX = B as

$\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1200 \\ 2200 \\ 3100\end{array}\right]$

Where, $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}1200 \\ 2200 \\ 3100\end{array}\right]$

Now,

$|A|=\left|\begin{array}{lll}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{array}\right|$

$=1(6-1)-1(9-4)+1(3-8)$

$=5-5-5$

$=-5$

Let $C_{i j}$ be the cofactors of elements $a_{i j}$ in $A=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right|=5, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll}3 & 1 \\ 4 & 3\end{array}\right|=-5, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|=-5$

$C_{21}=(-1)^{2+1}\left|\begin{array}{ll}1 & 1 \\ 1 & 3\end{array}\right|=-2, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}1 & 1 \\ 4 & 3\end{array}\right|=-1, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}1 & 1 \\ 4 & 1\end{array}\right|=3$

$C_{31}=(-1)^{3+1}\left|\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right|=-1, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}1 & 1 \\ 3 & 1\end{array}\right|=2, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{ll}1 & 1 \\ 3 & 2\end{array}\right|=-1$

$\operatorname{adj} A=\left[\begin{array}{rrr}5 & -5 & -5 \\ -2 & -1 & 3 \\ -1 & 2 & -1\end{array}\right]^{T}$

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

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

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

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=-\frac{1}{5}\left[\begin{array}{rrr}5 & -2 & -1 \\ -5 & -1 & 2 \\ -5 & 3 & -1\end{array}\right]\left[\begin{array}{l}1200 \\ 2200 \\ 3100\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=-\frac{1}{5}\left[\begin{array}{r}6000-4400-3100 \\ -6000-2200+6200 \\ -6000+6600-3100\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=-\frac{1}{5}\left[\begin{array}{l}-1500 \\ -2000 \\ -2500\end{array}\right]$

$\Rightarrow x=\frac{-1500}{-5}, y=\frac{-2000}{-5}$ and $z=\frac{-2500}{-5}$

$\therefore x=300, y=400$ and $z=500$

Hence, the award money for each value of Tolerance, Kindness and Leadership is ₹300, ₹400 and ₹500.

​One more value which should be considered for award is Honesty.