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

Solution:

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