The management committee of a residential colony decided to award some of its members (say x) for honesty,

Question:

The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.

Solution:

As per the information given in the question, the following equations hold true:

$x+y+z=12,2 x+3(y+z)=33$ and $x+z-2 y=0$

The above three equations can be represented in the form of a matrix as

$\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]$

Or $A X=B$, where, $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]$

$|A|=3 \neq 0$ Thus, $A$ is non-singular. Therefore, its inverse exists.

Adj $A$ is given by $\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right] \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]$

$X=A^{-1} B=\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]=\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right]$

$\therefore x=3, y=4$, and $z=5 .$

Therefore, the number of awardees for Honesty, Cooperation and Supervision are 3, 4, and 5 respectively.

One more value which the management of the colony must include for awards may be Sincerity.

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