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Questions  >>  CBSE XII  >>  Math  >>  Model Papers
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Q)

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 in 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.

This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

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A)
$\textbf{Step 1}$:
Given that $x$ = # of members awarded for honesty, $y$ = # of members awarded for for helping others and $z$ = members awarded for supervising the workers to keep the colony neat and clean .
By the given condition; the equations are:
$x+y+z = 12\; (1)$
$2x+3y+3z=33\;(2)$
$x+z=2y \;(3)$
We can express this in the form: $\begin{bmatrix} 1 & 1 & 1\\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end {bmatrix}=\begin{bmatrix} 12 \\ 33 \\ 0 \end{bmatrix}$
This is of the form $Ax=B$
Evaluating $A = 1 (3 \times 1 - (-2) \times 3) - 1 (2 \times 1 - 1 \times 3) + 1 (2 \times -2 - 1 \times 3) = =9+1-7=3$
Therefore $A^{-1}$ exists.
$\textbf{Step 2}$:
$A_{11}=3+6=9$
$A_{12}=-2(2-3)=+1$
$A_{13}=-4-3=-7$
$A_{21}=-(1+2)=-3$
$A_{22}=(1-1)=0$
$A_{23}=-(-2-1)=3$
$A_{31}=(3-3)=0$
$A_{32}=-(3-2)=-1$
$A_{33}=(3-2)=1$
Therefore Adjoint of A=$\begin{bmatrix} 9 & -3 & 0\\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}$
Therefore $A^{-1}=\large \frac{1}{3}$$\begin{bmatrix} 9 & -3 & 0\\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}$
$\textbf{Step 3}$:
$AX=B\qquad => X=A^{-1}B$
$\begin{bmatrix} x \\ y \\ z \end {bmatrix} = \frac{1}{3} \begin{bmatrix} 9 & -3 & 0\\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}\begin{bmatrix} 12 \\ 33 \\ 0 \\ \end {bmatrix}=\begin{bmatrix} \frac{9}{3} \\ \frac{12}{3} \\ \frac{15}{3} \\ \end {bmatrix}$
Hence $\begin{bmatrix} x \\ y \\ z \end {bmatrix}=\begin{bmatrix} 3 \\ 4 \\ 5 \end {bmatrix}$
Therefore $x=3,y=4,z=5$
Apart from honesty, cooperaton and supervision, another value which can be added is punctuality among the workers
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