question_answer

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs. 6000. Three times the award money for Hard work added to that given for Honesty amounts to Rs. 11000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, by using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards,        

Answer:

Consider x, y and z are the award of honesty, regularity and hard work and form the system of equations. Then, write them in matrix form as AX = B. Now, the solution is given by \[X={{A}^{-\,1}}B,\] put the values of \[{{A}^{-\,1}},\,\,X\]and B and calculate the required values. Let award for honesty = Rs. x Award for regularity = Rs. y and award for hard work = Rs. z According to first condition, \[x+y+z=6000\] According to second condition, \[3z+x=11000\] According to third condition, \[x+z=2y\] Now, the above equations can be rewritten in standard form of linear equations as \[x+y+z=6000\]                       ?(i) \[x+0y+3z=11000\]                 ?(ii) and       \[x-2y+z=0\]                          ?(iii) We can represent these equations using matrices as i.e.\[AX=B\] where, and its solution is given by \[X={{A}^{-\,1}}B\]   ?(iv) Now, \[=1\,(0+6)-\,(1-3)+1\,(-\,2-0)\] \[=6+2-2=6\ne 0\] \[\therefore \]\[{{A}^{-\,1}}\]exists, because A is a non-singular matrix. Now, cofactors of \[|A|\]are \[{{A}_{11}}={{(-\,1)}^{1\,\,+\,\,1}}\left| \begin{matrix}    0 & 3  \\    -\,2 & 1  \\ \end{matrix} \right|=+\,\,(0+6)=+\,\,6\] \[{{A}_{12}}={{(-\,1)}^{1\,\,+\,\,2}}\left| \begin{matrix}    1 & 3  \\    1 & 1  \\ \end{matrix} \right|=-\,(1-3)=2\] \[{{A}_{13}}={{(-\,1)}^{1\,\,+\,\,3}}\left| \begin{matrix}    1 & 0  \\    1 & -\,2  \\ \end{matrix} \right|=+\,(-\,2)=-\,2\] \[{{A}_{21}}={{(-\,1)}^{2\,\,+\,\,1}}\left| \begin{matrix}    1 & 1  \\    -\,2 & 1  \\ \end{matrix} \right|=-\,(1+2)=-\,3\] \[{{A}_{22}}={{(-\,1)}^{2\,\,+\,\,2}}\left| \begin{matrix}    1 & 1  \\    1 & 1  \\ \end{matrix} \right|=+\,(1-1)=0\] \[{{A}_{23}}={{(-\,1)}^{2\,\,+\,\,3}}\left| \begin{matrix}    1 & 1  \\    1 & -\,2  \\ \end{matrix} \right|=-\,(-\,2-1)=3\] \[{{A}_{31}}={{(-\,1)}^{3\,\,+\,\,1}}\left| \begin{matrix}    1 & 1  \\    0 & 3  \\ \end{matrix} \right|=+\,(3-0)=3\] \[{{A}_{32}}={{(-\,1)}^{3\,\,+\,\,2}}\left| \begin{matrix}    1 & 1  \\    1 & 0  \\ \end{matrix} \right|=-\,(3-1)=-\,2\] \[{{A}_{33}}={{(-\,1)}^{3\,\,+\,\,3}}\left| \begin{matrix}    1 & 1  \\    1 & 0  \\ \end{matrix} \right|=+\,(0-1)=-\,1\] Now \[\therefore \]      Then, Now, \[\Rightarrow \]               On comparing, we get                         \[x=500\],                         \[y=2000\]and                         \[z=3500\]             Hence, award for honesty = Rs. 500             Award for regularity = Rs. 2000 and award for hard work = Rs. 3500 Value The school must include punctuality for award.